Free Math Worksheets

- What must be true about the two function values at an intersection? - Move all x-terms to one side before solving. - Multiplying by a common denominator may make the fraction equation easier.

1. Set the function values equal: \(\frac{2}{5}x + 3 = -\frac{1}{2}x - 6\). 2. Combine the x-terms: \(\frac{2}{5}x + \frac{1}{2}x = -9\), so \(\frac{9}{10}x = -9\). 3. Solving gives \(x = -10\). 4. Substitute into \(f\): \(y = \frac{2}{5} \cdot (-10) + 3 = -1\). 5. Therefore, \(S = (-10, -1)\).

\(S = (-10, -1)\)

- What must be true about the two function values at an intersection? - What y-coordinate does every point on the x-axis have? - Use your result from part a to justify part b.

1. Set the function values equal: \(3x - 9 = -x + 3\). 2. Then \(4x = 12\), so \(x = 3\). 3. Substitute into either function: \(k(3) = 3 \cdot 3 - 9 = 0\). Thus the intersection point is \((3, 0)\). 4. A point lies on the x-axis exactly when its y-coordinate is \(0\). Since the intersection has y-coordinate \(0\), it lies on the x-axis.

a) \((3, 0)\) b) The intersection lies on the x-axis because its y-coordinate is \(0\).

- What must be true about the two function values at an intersection? - After finding the common x-value, how can you find the y-coordinate? - Recall the signs of coordinates in each quadrant.

1. Set the function values equal: \(1.5x - 2 = -x + 3\). 2. Then \(2.5x = 5\), so \(x = 2\). 3. Substitute into \(g\): \(g(2) = -2 + 3 = 1\). Thus \(S = (2, 1)\). 4. Both coordinates are positive, so \(S\) lies in Quadrant I.

\(S = (2, 1)\), in Quadrant I.

- How can you graph a line when you know its slope and \(y\)-intercept? - What does the intersection of two lines represent for a system of equations? - How can you check whether a point lies on a line?

1. Graph equation (I), which has slope \(m_1 = 1.5\) and \(y\)-intercept \(b_1 = -2\). 2. Graph equation (II), which has slope \(m_2 = -0.5\) and \(y\)-intercept \(b_2 = 2\). 3. Read the intersection from the graph: \((2, 1)\). 4. Check equation (I): \(1 = 1.5 \cdot 2 - 2 = 3 - 2 = 1\), so the point satisfies equation (I). 5. Check equation (II): \(1 = -0.5 \cdot 2 + 2 = -1 + 2 = 1\), so the point satisfies equation (II).

The solution is \((2, 1)\). Substitution confirms that the point satisfies both equations.

- First rewrite both equations so that \(y\) is alone on one side. - Use the \(y\)-intercept and slope to graph each line. - Where do the two lines intersect? - Check your result by substituting the values of \(x\) and \(y\) into the original equations.

1. Rewrite each equation in slope-intercept form: (I) \(y = -x + 5\) (II) \(y = 2x - 1\) 2. Graph equation (I), which has \(y\)-intercept \(5\) and slope \(-1\). 3. Graph equation (II), which has \(y\)-intercept \(-1\) and slope \(2\). 4. The lines intersect at \((2, 3)\), so the solution set is \(\{(2, 3)\}\).

The solution set is \(\{(2, 3)\}\).

- Can you rewrite each equation in the form \(y = mx + b\)? - What does the point where two lines intersect represent for the system? - How can you use the coordinates of the intersection to check the accuracy of your graph? - Which axis intercepts or other easy points can help you graph each line?

1. Rewrite equation (II) as \(y = -x + 4\). 2. Graph the first line using its \(y\)-intercept \(1\) and slope \(0.5\), for example with the points \((0, 1)\) and \((2, 2)\). 3. Graph the second line using its \(y\)-intercept \(4\) and slope \(-1\), for example with the points \((0, 4)\) and \((4, 0)\). 4. The lines intersect at \((2, 2)\). 5. Check the point in both original equations: \(2 = 0.5 \cdot 2 + 1\) and \(2 + 2 = 4\). Both equations are true.

The solution is \((2, 2)\).

- Find where each line crosses the vertical axis to identify its \(y\)-intercept. - Use two clearly readable points on each line to determine its slope. - The intersection of the two lines is the ordered pair that satisfies both equations.

1. Line \(g\) has \(y\)-intercept \(1\). Using the points \((0, 1)\) and \((1, 2)\), its slope is \(1\), so its equation is \(y = x + 1\). 2. Line \(h\) has \(y\)-intercept \(4\). Using the points \((0, 4)\) and \((2, 3)\), its slope is \(-0.5\), so its equation is \(y = -0.5x + 4\). 3. The lines intersect at \((2, 3)\). 4. Therefore, the solution set is \(\{(2, 3)\}\).

(I) \(y = x + 1\) (II) \(y = -0.5x + 4\) Solution set: \(\{(2, 3)\}\)

- The y-intercepts give the initial heights. - The magnitude of each negative slope is the burn rate. - The intersection represents equal heights at the same time.

1. Candle A begins at \(6\) inches and decreases at \(0.75\,\text{in./h}\), so \(A(x)=-0.75x+6\). Candle B begins at \(9\) inches and decreases at \(1.5\,\text{in./h}\), so \(B(x)=-1.5x+9\). 2. Set the functions equal: \(-0.75x+6=-1.5x+9\). Then \(0.75x=3\), so \(x=4\). Substitution gives \(y=3\). The candles are both \(3\) inches tall after \(4\) hours.

a) Candle A: initial height \(6\,\text{in.}\), burn rate \(0.75\,\text{in./h}\) Candle B: initial height \(9\,\text{in.}\), burn rate \(1.5\,\text{in./h}\) b) After \(4\) hours, both are \(3\) inches tall.

- The y-intercepts are the starting balances. - The slopes are the weekly saving rates. - The intersection represents equal balances.

1. Sarah starts with \(\$40\), while Luke starts with \(\$10\), so Sarah has more initially. Luke's balance increases by \(\$5\) per week, while Sarah's increases by \(\$2\) per week, so Luke saves more each week. 2. Their equations are \(L(x)=5x+10\) and \(S(x)=2x+40\). Set them equal: \(5x+10=2x+40\), so \(3x=30\) and \(x=10\). The common balance is \(5\cdot10+10=60\), or \(\$60\).

a) Sarah starts with more; Luke saves more each week, \(\$5\) per week versus \(\$2\) per week. b) After \(10\) weeks; \(\$60\)

- What does it mean graphically for a point to satisfy one equation? - What condition must a point meet to solve the entire system of three equations? - Is there one point that lies on all three lines?

1. A solution of the entire system must satisfy all three equations at the same time. 2. On the graph, this is the point where all three lines intersect. 3. The three lines have one common intersection at \((2, 2)\). 4. Therefore, \((2, 2)\) is the only solution of the system.

\((2, 2)\)

- At an intersection, the two lines have the same y-value for the same x-value. - Set the two expressions for \(y\) equal to create an equation with one variable. - Solve for \(x\), then substitute that value into either equation to find \(y\).

1. Set the expressions equal: \(x - 2 = -2x + 4\). 2. Add \(2x\) to both sides: \(3x - 2 = 4\). 3. Add \(2\) to both sides: \(3x = 6\). 4. Divide by \(3\): \(x = 2\). 5. Substitute \(x = 2\) into \(g_1\): \(y = 2 - 2 = 0\). 6. Therefore, the intersection point is \(P = (2, 0)\). The graph confirms this intersection.

\(P = (2, 0)\)

- First read the \(y\)-intercept and slope of each line. - Write an equation for each line from the graph. - A solution must satisfy both equations.

1. Line \(g\) passes through \((0, 3)\) and \((1, 2)\). Its slope is \(-1\), so \(y = -x + 3\), or equivalently \(x + y = 3\). 2. Line \(h\) passes through \((0, -1)\) and \((1, 2)\). Its slope is \(3\), so \(y = 3x - 1\), or equivalently \(-3x + y = -1\). 3. The graph shows the intersection \((1, 2)\). 4. Check: \(1 + 2 = 3\) and \(-3 \cdot 1 + 2 = -1\). The point satisfies both equations.

(I) \(x + y = 3\) (II) \(-3x + y = -1\) The intersection is \((1, 2)\), and it satisfies both equations.

- Translate each purchase into an equation. - Use a system-solving method that eliminates one variable. - Scale one equation so one pair of variable coefficients matches.

1. Let \(x\) be the price of one watermelon and \(y\) the price of one mango, in dollars. 2. Write the system \(4x + 6y = 52\) and \(5x + 3y = 47\). 3. Multiply the second equation by \(2\): \(10x + 6y = 94\). 4. Subtract the first equation: \(6x = 42\), so \(x = 7\).

d) \(\$7\)

- How far does Alex travel in one hour? Use the constant rate to build the table. - Remember that Blake starts \(2\) miles ahead. - The catch-up time occurs when the two position values are equal. - Can you write each runner's position as a function of \(t\)?

1. The positions are \(s_A(t) = 8t\) and \(s_B(t) = 2 + 6t\). 2. The table is \(\begin{array}{c|ccccc} t\text{ (h)} & 0 & 1 & 2 & 3 & 4 \\ \hline s_A(t)\text{ (mi)} & 0 & 8 & 16 & 24 & 32 \\ s_B(t)\text{ (mi)} & 2 & 8 & 14 & 20 & 26 \end{array}\). 3. The table values agree at \(t = 1\), when both positions are \(8\). Thus, Alex catches Blake after \(1\,\text{h}\) at mile marker \(8\). 4. Verify algebraically: \(8t = 2 + 6t\). Subtracting \(6t\) gives \(2t = 2\), so \(t = 1\). 5. At \(t = 1\), both positions equal \(8\,\text{mi}\), so both runners are \(8\,\text{mi}\) from mile marker \(0\).

a) \(\begin{array}{c|ccccc} t\text{ (h)} & 0 & 1 & 2 & 3 & 4 \\ \hline s_A(t)\text{ (mi)} & 0 & 8 & 16 & 24 & 32 \\ s_B(t)\text{ (mi)} & 2 & 8 & 14 & 20 & 26 \end{array}\) b) \(s_A(t) = 8t\) and \(s_B(t) = 2 + 6t\) c) Alex catches Blake after \(1\,\text{h}\) at mile marker \(8\). d) After \(1\,\text{h}\), both runners are \(8\,\text{mi}\) from mile marker \(0\).

- Write a cost function for each plan. - What do the monthly fee and per-minute rate represent in a linear equation? - Set the two cost expressions equal to find the break-even time. - What does it mean when the two plans have equal costs?

1. For \(30\) minutes, QuickRide costs \(30 \cdot \$0.30 = \$9.00\). CityCar costs \(\$10.00 + 30 \cdot \$0.10 = \$13.00\). 2. For \(60\) minutes, QuickRide costs \(60 \cdot \$0.30 = \$18.00\). CityCar costs \(\$10.00 + 60 \cdot \$0.10 = \$16.00\). 3. The cost functions are \(Q(x) = 0.30x\) and \(C(x) = 10.00 + 0.10x\). Find the break-even point by solving \(0.30x = 10.00 + 0.10x\). 4. Then \(0.20x = 10.00\), so \(x = 50\). At \(50\) minutes, both plans cost \(\$15.00\). For more than \(50\) minutes, CityCar is less expensive. 5. The point \((50, 15)\) represents the monthly driving time and cost at which the two plans have the same total price.

a) At \(30\) minutes: QuickRide \(\$9.00\), CityCar \(\$13.00\). At \(60\) minutes: QuickRide \(\$18.00\), CityCar \(\$16.00\). b) CityCar is less expensive for more than \(50\) minutes of driving per month. c) Both plans cost \(\$15.00\) at \(50\) minutes of monthly driving.

- Which part of each equation changes with the number of miles, and which part stays fixed? - How can you find when two plans cost exactly the same amount? - Compare the per-mile rates after finding the break-even distance.

1. The constants \(45\) and \(25\) are the fixed rental fees before any mileage charge is added. 2. For \(80\) miles, Plan A costs \(0.30 \cdot 80 + 45 = 69\), so \(\$69.00\). Plan B costs \(0.50 \cdot 80 + 25 = 65\), so \(\$65.00\). Plan B is less expensive. 3. Find the break-even distance by setting the costs equal: \(0.30x + 45 = 0.50x + 25\). 4. Solving gives \(20 = 0.20x\), so \(x = 100\). 5. Plan A has the smaller per-mile rate, so it is less expensive for trips longer than \(100\) miles.

a) \(45\) and \(25\) are the fixed rental fees, in dollars. b) Plan A: \(\$69.00\); Plan B: \(\$65.00\). Plan B is less expensive. c) Plan A is less expensive for trips longer than \(100\) miles.

- What must be true about the y-values where two lines intersect? - What does a point with x-coordinate \(0\) tell you about the y-intercept? - How can you use two points to determine a slope?

1. Set the equations of \(g_1\) and \(g_2\) equal: \(2x - 1 = -x + 5\). Then \(3x = 6\), so \(x = 2\). 2. Substitute \(x = 2\) into either equation: \(y = 2 \cdot 2 - 1 = 3\). Thus \(S = (2, 3)\). 3. For \(g_3\), point \(P(0, 1)\) gives the y-intercept \(b = 1\). 4. Use \(S(2, 3)\): \(3 = 2m + 1\). Then \(2 = 2m\), so \(m = 1\). 5. Therefore, \(g_3\) has equation \(y = x + 1\).

a) \(S = (2, 3)\) b) \(y = x + 1\)

- What equation can you write when two functions have the same output at the same x-value? - How does the sign of the slope tell you whether a line increases or decreases? - What is the y-coordinate at an x-intercept?

1. At the intersection, \(h(x) = k(x)\). Solve \(0.8x + 2 = -1.2x + 6\). This gives \(2x = 4\), so \(x = 2\). Then \(h(2) = 0.8 \cdot 2 + 2 = 3.6\). The intersection is \((2, 3.6)\). 2. Line \(h\) is increasing because its slope \(0.8\) is positive. Line \(k\) is decreasing because its slope \(-1.2\) is negative. 3. At the x-intercept of \(k\), \(y = 0\). Solve \(-1.2x + 6 = 0\), which gives \(x = 5\). The x-intercept is \((5, 0)\).

a) \((2, 3.6)\) b) \(h\) is increasing; \(k\) is decreasing. c) \((5, 0)\)

- At an intersection point, both functions have the same input and output. - Compare the two function values for each listed x-value. - Solve the equation for \(x\), then substitute to find \(y\). - Check the point in both equations.

1. The values are \(\begin{array}{c|cccc}x&0&1&2&3\\\hline g(x)&4&3&2&1\\h(x)&-2&0&2&4\end{array}\). Both functions have value \(2\) when \(x = 2\), so the intersection is \((2, 2)\). 2. Verify algebraically: \(-x + 4 = 2x - 2\). 3. Add \(x\) and then add \(2\): \(6 = 3x\), so \(x = 2\). 4. Substitute into either function: \(g(2) = -2 + 4 = 2\). Thus, the intersection point is \((2, 2)\).

The intersection point is \((2, 2)\).

- How do you find the slope between two points? - Which given point reveals the y-intercept directly? - What condition do the function values satisfy at an intersection?

1. The slope of \(g\) is \(m = \frac{5 - 1}{2 - 0} = 2\). 2. Point \(P(0, 1)\) gives the y-intercept \(b = 1\), so \(g\) has equation \(y = 2x + 1\). 3. Set the two equations equal: \(2x + 1 = -x + 7\). 4. Then \(3x = 6\), so \(x = 2\). 5. Substitute into \(h\): \(y = -2 + 7 = 5\). Thus \(S = (2, 5)\).

a) \(y = 2x + 1\) b) \(S = (2, 5)\)

- If you know the x-coordinate of the intersection, how can you find its y-coordinate? - What must be true about that point for the second line? - Substitute the point into the equation containing the unknown slope.

1. Find the y-value of the intersection using \(f\): \(f(-2) = \frac{1}{2} \cdot (-2) + 1 = 0\). 2. Thus the intersection point is \((-2, 0)\). 3. Substitute this point into \(g(x) = mx + 4\): \(0 = -2m + 4\). 4. Then \(2m = 4\), so \(m = 2\).

\(m = 2\)

- Start by finding where the first two lines have equal function values. - How can you check whether a given point lies on another line? - Substitute the x-coordinate of the intersection into \(h\) and compare the y-value.

1. Set \(f(x) = g(x)\): \(3x - 4 = -x + 8\). 2. Then \(4x = 12\), so \(x = 3\). 3. Find the y-value: \(f(3) = 3 \cdot 3 - 4 = 5\). Thus \(S = (3, 5)\). 4. Check \(h\) at \(x = 3\): \(h(3) = 2 \cdot 3 - 1 = 5\). 5. Since \(h(3) = 5\), the graph of \(h\) also passes through \(S\).

The intersection of \(f\) and \(g\) is \(S = (3, 5)\). The graph of \(h\) also passes through \(S\).

- Use the fully known function to find the y-value at \(x = 4\). - What must be true about both functions at their intersection? - Use the y-value you found to solve for the missing parameter.

1. Use \(g\) to find the y-value of the intersection: \(g(4) = -1.2 \cdot 4 + 10 = 5.2\). 2. Since the point is on both lines, \(f(4) = 5.2\). 3. Substitute into \(f\): \(0.8 \cdot 4 + b = 5.2\). 4. Then \(3.2 + b = 5.2\), so \(b = 2\). 5. The intersection point is \((4, 5.2)\).

\(b = 2\) and the intersection point is \((4, 5.2)\).

- What must be true about the coordinates of a point where two lines intersect? - What algebraic step can help you find the common x-coordinate? - After finding the x-coordinate, how can you find the corresponding y-coordinate? - In a linear equation, what information is already fixed by a known slope and one point?

1. Set the equations equal: \(1.5x + 2 = -0.5x + 6\). Then \(2x + 2 = 6\), so \(x = 2\). 2. Substitute \(x = 2\): \(y = 1.5 \cdot 2 + 2 = 5\). Thus \(S = (2, 5)\). 3. Write \(k\) as \(y = -2x + b\). Use \(S\): \(5 = -2 \cdot 2 + b\). 4. Then \(b = 9\), so \(k\) has equation \(y = -2x + 9\).

1. \(S = (2, 5)\) 2. \(y = -2x + 9\)

- First write an equation for each diagonal from its two endpoints. - How do you find the slope of a line through two points? - Once you have both equations, set them equal to find their common point.

1. For diagonal \(AC\), the slope is \(m_{AC} = \frac{3 - 0}{4 - (-2)} = \frac{1}{2}\). Using point \(A(-2, 0)\), its equation is \(y = 0.5x + 1\). 2. For diagonal \(BD\), the slope is \(m_{BD} = \frac{0 - 4}{4 - 0} = -1\). Since \(B(0, 4)\) is the y-intercept, its equation is \(y = -x + 4\). 3. Set the equations equal: \(0.5x + 1 = -x + 4\). 4. Then \(1.5x = 3\), so \(x = 2\). 5. Substitute into either diagonal: \(y = -2 + 4 = 2\). Therefore, \(S = (2, 2)\).

\(S = (2, 2)\)

- First write an equation for the path determined by the two given points. - At the crossing point, what must be true about the two y-values? - Check the point you find in both equations.

1. For Car 1, the slope is \(m = \frac{1 - 4}{5 - (-1)} = -\frac{1}{2}\). 2. Use point \(Q(5, 1)\) to find the y-intercept: \(1 = -0.5 \cdot 5 + b\), so \(b = 3.5\). Thus the first path is \(y = -0.5x + 3.5\). 3. Set the two path equations equal: \(-0.5x + 3.5 = 2x - 4\). 4. Then \(7.5 = 2.5x\), so \(x = 3\). 5. Substitute into the second equation: \(y = 2 \cdot 3 - 4 = 2\). Therefore, \(S = (3, 2)\).

\(S = (3, 2)\)

- At an intersection, what must be true about the coordinates on both lines? - How does the sign of the coefficient of \(x\) tell whether a line rises or falls? - What do parallel lines have in common? - How can a given point be used to determine a missing y-intercept?

1. At the intersection, \(-2x + 6 = 0.5x - 4\). Solving gives \(10 = 2.5x\), so \(x = 4\). Then \(y = -2\). The intersection is \((4, -2)\). 2. Line \(g\) has slope \(-2\), so it is decreasing. Line \(h\) has slope \(0.5\), so it is increasing. 3. A line parallel to \(g\) must have slope \(-2\). Write \(k(x) = -2x + b\). Using \((1, 1)\), \(1 = -2 + b\), so \(b = 3\). Thus, \(k(x) = -2x + 3\).

a) \((4, -2)\) b) \(g\) is decreasing; \(h\) is increasing. c) \(k(x) = -2x + 3\)

- Use the y-intercept and slope to identify points on each line. - At an intersection, what must be true about the two function values? - Which coordinate of the intersection gives the solution for \(x\)?

1. The graph of \(f\) passes through \((0, -3)\) and \((2, 1)\). The graph of \(g\) passes through \((0, 6)\) and \((6, 0)\). 2. Solve \(2x - 3 = -x + 6\). Adding \(x\) and then adding \(3\) gives \(3x = 9\), so \(x = 3\). Substitution gives \(y = 3\), so the intersection is \((3, 3)\). 3. At the intersection, both functions have the same input and the same output. Here, both equal \(3\) when \(x = 3\).

a) One possible set is \(f: (0, -3), (2, 1)\) and \(g: (0, 6), (6, 0)\). b) \(x = 3\), and the intersection point is \((3, 3)\). c) The intersection shows the input where the two function values are equal.

- When a variable represents a number of tickets, interpret its coefficient as a price per ticket. - A discount reduces the individual ticket price. - Represent the total number of people with a second equation. - Solve one equation for a variable and substitute.

1. The coefficient \(12\) is the price in dollars of one youth ticket, and \(18\) is the price of one adult ticket. 2. With a \(\$2\) youth discount, the youth ticket price is \(12 - 2 = 10\), so the new equation is \(10x + 18y = 360\). 3. For part c, use the system \(12x + 18y = 360\) and \(x + y = 25\). 4. Solve the second equation for \(x\): \(x = 25 - y\). 5. Substitute: \(12(25 - y) + 18y = 360\). 6. Simplify: \(300 - 12y + 18y = 360\), so \(6y = 60\) and \(y = 10\). 7. Then \(x = 25 - 10 = 15\).

a) \(12\) is the youth ticket price in dollars, and \(18\) is the adult ticket price in dollars. b) The new equation is \(10x + 18y = 360\). c) There are \(15\) youth tickets and \(10\) adult tickets.

- Write one equation for the number of cases and one for the number of bottles. - Account for the bottles in each type of case. - Solve the two equations simultaneously. - Check that the case counts add to \(11\).

1. Let \(x\) be the number of lemonade cases and \(y\) the number of water cases. 2. The system is \(x + y = 11\) and \(12x + 10y = 122\). 3. Solve the first equation for \(y\): \(y = 11 - x\). 4. Substitute: \(12x + 10(11 - x) = 122\). 5. Simplify: \(2x + 110 = 122\), so \(x = 6\). 6. Then \(y = 11 - 6 = 5\).

There were \(6\) cases of lemonade and \(5\) cases of water.

- How can you rearrange an equation so that \(y\) is alone on one side? - What do the slopes tell you about how the two lines compare? - When do two lines have no intersection or infinitely many intersections?

1. Rewrite equation (II): \(2y = -2x + 2\), so \(y = -x + 1\). 2. Graph \(y = x - 3\) and \(y = -x + 1\). 3. The lines intersect at \((2, -1)\). 4. The first line has slope \(m_1 = 1\), and the second line has slope \(m_2 = -1\). Because \(m_1 \ne m_2\), the lines are not parallel and intersect at exactly one point.

a) \(y = -x + 1\) b) \((2, -1)\) c) The slopes \(1\) and \(-1\) are different, so the lines intersect at exactly one point.

- Use the perimeter formula for a rectangle. - Express the two adjusted side lengths. - What must be true about the sides of a square? - Add or subtract the resulting equations.

1. Let \(l\) be the longer side and \(w\) the shorter side, in centimeters. 2. The perimeter equation is \(2(l + w) = 30\), so \(l + w = 15\). 3. The adjusted sides of the square are equal, so \(l - 3 = w + 2\), or \(l - w = 5\). 4. Add the equations: \((l + w) + (l - w) = 15 + 5\). 5. Then \(2l = 20\), so \(l = 10\). 6. Substitute into \(l + w = 15\): \(10 + w = 15\), so \(w = 5\).

The original rectangle has side lengths \(10\,\text{cm}\) and \(5\,\text{cm}\).

- Write and simplify the perimeter equation. - Write expressions for the original and new areas. - Be careful with signs when subtracting the new area. - After simplifying, solve the resulting linear system.

1. Let \(l\) be the original length and \(w\) the original width, in feet. 2. The perimeter equation is \(2(l + w) = 40\), so \(l + w = 20\). 3. The original area is \(lw\), and the new area is \((l + 2)(w - 2)\). 4. Because the area decreases by \(12\,\text{ft}^2\), \(lw - (l + 2)(w - 2) = 12\). 5. Expand and simplify: \(lw - (lw - 2l + 2w - 4) = 12\), so \(2l - 2w + 4 = 12\), or \(l - w = 4\). 6. Add \(l + w = 20\) and \(l - w = 4\): \(2l = 24\), so \(l = 12\). 7. Then \(w = 20 - 12 = 8\).

The garden bed was originally \(12\,\text{ft}\) long and \(8\,\text{ft}\) wide.

- Choose variables for the base fee and the price per gigabyte. - Compare how the total cost changes when the data use changes. - Dividing the difference in cost by the difference in data use gives the per-gigabyte price. - Subtracting the equations eliminates the base fee.

1. Let \(G\) be the monthly base fee and \(p\) the price per gigabyte, in dollars. 2. Write the system \(G + 5p = 17.50\) and \(G + 12p = 31.50\). 3. Subtract the first equation from the second: \(7p = 14\), so \(p = 2\). 4. Substitute into the first equation: \(G + 5 \cdot 2 = 17.50\). 5. Solve: \(G + 10 = 17.50\), so \(G = 7.50\).

The monthly base fee is \(\$7.50\), and the price is \(\$2.00\) per gigabyte.

- Identify the cost that does not depend on the number of pages. - Describe how the total cost changes as the page count increases. - Write a system with one variable for the fixed fee and one for the per-page cost. - Subtract the equations to eliminate the fixed fee.

1. Let \(B\) be the monthly service fee and \(s\) the price per page, in dollars. 2. Write the system \(B + 200s = 18\) and \(B + 500s = 33\). 3. Subtract the first equation from the second: \(300s = 15\). 4. Solve: \(s = \frac{15}{300} = 0.05\), so each page costs \(\$0.05\). 5. Substitute into the first equation: \(B + 200 \cdot 0.05 = 18\). 6. Simplify: \(B + 10 = 18\), so \(B = 8\).

The monthly service fee is \(\$8.00\), and the price is \(\$0.05\) per page.

- Identify the part of the fare that stays the same for every trip. - Determine the cost of the additional miles between the two trips. - Write one cost equation for each ride. - Divide the difference in total cost by the difference in distance.

1. Let \(G\) be the base fee and \(m\) the price per mile, in dollars. 2. Write the system \(G + 8m = 23\) and \(G + 15m = 40.50\). 3. Subtract the first equation from the second: \(7m = 17.50\). 4. Solve: \(m = 2.50\), so the mileage charge is \(\$2.50\) per mile. 5. Substitute into the first equation: \(G + 8 \cdot 2.50 = 23\). 6. Simplify: \(G + 20 = 23\), so \(G = 3\).

The base fee is \(\$3.00\), and the price is \(\$2.50\) per mile.

- Use the perimeter formula for a rectangle. - Translate “\(3\) less than twice” into an algebraic equation. - Substitute the expression for the longer side into the perimeter equation.

1. Let \(l\) be the longer side and \(w\) the shorter side, in centimeters. 2. The perimeter equation is \(2(l + w) = 54\), so \(l + w = 27\). 3. The side-length relationship is \(l = 2w - 3\). 4. Substitute into the perimeter equation: \((2w - 3) + w = 27\). 5. Simplify: \(3w - 3 = 27\), so \(3w = 30\) and \(w = 10\). 6. Then \(l = 2 \cdot 10 - 3 = 17\).

The side lengths are \(17\,\text{cm}\) and \(10\,\text{cm}\).

- Use one equation for the total number of tickets and another for the total revenue. - Include the price of each ticket type in the revenue equation. - Choose an efficient method for solving the system.

1. Let \(x\) be the number of adult tickets and \(y\) the number of child tickets. 2. Write the system \(x + y = 120\) and \(8x + 5y = 735\). 3. Solve the first equation for \(y\): \(y = 120 - x\). 4. Substitute: \(8x + 5(120 - x) = 735\). 5. Simplify: \(8x + 600 - 5x = 735\), so \(3x = 135\) and \(x = 45\). 6. Then \(y = 120 - 45 = 75\).

The school sold \(45\) adult tickets and \(75\) child tickets.

- Choose variables for the two individual prices. - Translate each purchase into an equation. - Multiply the equations so one variable can be eliminated.

1. Let \(a\) be the price of one apple and \(b\) the price of one banana, in dollars. 2. Write the system \(3a + 2b = 3.25\) and \(2a + 3b = 3.00\). 3. Multiply the first equation by \(3\): \(9a + 6b = 9.75\). 4. Multiply the second equation by \(2\): \(4a + 6b = 6.00\). 5. Subtract: \(5a = 3.75\), so \(a = 0.75\). 6. Substitute into the first equation: \(3 \cdot 0.75 + 2b = 3.25\). 7. Simplify: \(2.25 + 2b = 3.25\), so \(b = 0.50\).

One apple costs \(\$0.75\), and one banana costs \(\$0.50\).

- Express who has saved more and by how much. - Represent doubling and tripling with your variables. - One equation already gives one variable in terms of the other, so consider substitution.

1. Let \(x\) be the amount Jordan has saved and \(y\) the amount Casey has saved. 2. Write the system: \(x = y + 20\) and \(2x + 3y = 240\). 3. Substitute \(y + 20\) for \(x\): \(2(y + 20) + 3y = 240\). 4. Solve: \(2y + 40 + 3y = 240\), so \(5y = 200\) and \(y = 40\). 5. Then \(x = 40 + 20 = 60\).

Jordan has saved \(\$60\), and Casey has saved \(\$40\).

- Calculate the downstream and upstream speeds from distance and time. - Downstream speed is the boat speed plus the current speed. - Upstream speed is the boat speed minus the current speed. - Add the two equations to eliminate the current speed.

1. Let \(v_b\) be the boat’s speed in still water and \(v_c\) the speed of the current, in miles per hour. 2. Downstream, the boat’s speed is \(v_b + v_c = \frac{36}{2} = 18\). 3. Upstream, the boat’s speed is \(v_b - v_c = \frac{36}{3} = 12\). 4. Add the equations: \((v_b + v_c) + (v_b - v_c) = 18 + 12\). 5. Then \(2v_b = 30\), so \(v_b = 15\). 6. Substitute into \(v_b + v_c = 18\): \(15 + v_c = 18\), so \(v_c = 3\).

The boat’s speed in still water is \(15\,\text{mph}\), and the current’s speed is \(3\,\text{mph}\).

- Convert both times to hours. - Use distance divided by time to find each ground speed. - A tailwind adds to the plane’s still-air speed, while a headwind subtracts from it. - Add the two equations to eliminate the wind speed.

1. Convert the travel times to hours: \(1\) hour \(30\) minutes is \(1.5\) hours, and \(1\) hour \(48\) minutes is \(1.8\) hours. 2. With the tailwind, the ground speed is \(\frac{450}{1.5} = 300\,\text{mph}\). 3. With the headwind, the ground speed is \(\frac{450}{1.8} = 250\,\text{mph}\). 4. Let \(v_p\) be the plane’s speed in still air and \(v_w\) the wind speed. Then \(v_p + v_w = 300\) and \(v_p - v_w = 250\). 5. Add the equations: \(2v_p = 550\), so \(v_p = 275\). 6. Substitute into \(v_p + v_w = 300\): \(275 + v_w = 300\), so \(v_w = 25\).

The plane’s speed in still air is \(275\,\text{mph}\), and the wind speed is \(25\,\text{mph}\).

- Compare how much more was paid in the second month and how many additional visits occurred. - Match the difference in cost to the difference in visits. - After finding the per-visit fee, subtract the visit charges from either total.

1. Let \(G\) be the monthly membership fee and \(s\) the cost of one sauna visit, in dollars. 2. Write the system \(G + 6s = 56\) and \(G + 10s = 72\). 3. Subtract the first equation from the second: \(4s = 16\), so \(s = 4\). 4. Substitute into the first equation: \(G + 6 \cdot 4 = 56\). 5. Simplify: \(G + 24 = 56\), so \(G = 32\).

The monthly membership fee is \(\$32\), and each sauna visit costs \(\$4\).

- Multiply one equation so both equations have the same notebook coefficient. - After finding the individual prices, calculate the regular price of the bundle quantities. - Compare the regular total with the sale price.

1. Let \(h\) be the price of one notebook and \(s\) the price of one pen, in dollars. 2. Write the system \(4h + 3s = 11\) and \(2h + 5s = 9\). 3. Multiply the second equation by \(2\): \(4h + 10s = 18\). 4. Subtract the first equation: \(7s = 7\), so \(s = 1\). 5. Substitute into \(2h + 5s = 9\): \(2h + 5 = 9\), so \(h = 2\). 6. At regular prices, \(6\) notebooks and \(8\) pens cost \(6 \cdot 2 + 8 \cdot 1 = 20\) dollars. 7. Since \(18.50 < 20\), the sale bundle saves \(20 - 18.50 = 1.50\) dollars.

a) One notebook costs \(\$2\), and one pen costs \(\$1\). b) Yes. The regular price is \(\$20.00\), so the bundle saves \(\$1.50\).

- Translate each statement about the total amount into an equation. - Choose variables for the two amounts being saved. - Show how each amount changes in the hypothetical situation. - Solve the resulting system of two equations.

1. Let \(L\) be the amount Leon has saved and \(M\) the amount Mia has saved. 2. The actual total gives \(L + M = 30\). 3. The hypothetical amounts give \(2L + \frac{1}{2}M = 36\). 4. Solve the first equation for \(M\): \(M = 30 - L\). 5. Substitute: \(2L + \frac{1}{2}(30 - L) = 36\). 6. Simplify: \(2L + 15 - \frac{1}{2}L = 36\), so \(\frac{3}{2}L = 21\) and \(L = 14\). 7. Then \(M = 30 - 14 = 16\).

Leon has saved \(\$14\), and Mia has saved \(\$16\).

- Write a perimeter equation for the original rectangle. - Express the adjusted length and width. - Write and simplify a second perimeter equation. - Subtract the equations to eliminate one variable.

1. Let \(x\) be the original length and \(y\) the original width, in centimeters. 2. The original perimeter gives \(2x + 2y = 50\), or \(x + y = 25\). 3. The new perimeter gives \(2(x + 5) + 2(2y) = 80\). 4. Simplify: \(2x + 10 + 4y = 80\), so \(x + 2y = 35\). 5. Subtract \(x + y = 25\) from \(x + 2y = 35\): \(y = 10\). 6. Then \(x + 10 = 25\), so \(x = 15\).

The original length is \(15\,\text{cm}\), and the original width is \(10\,\text{cm}\).

- Express a \(20\%\) increase and a \(20\%\) decrease using decimal multipliers. - Write one equation for the current total and one for the projected total. - Solve the system using substitution. - Check that both student counts are whole numbers.

1. Let \(a\) be the current number of students in Homeroom A and \(b\) the current number in Homeroom B. 2. The current total gives \(a + b = 60\). 3. Apply the projected percent changes: \(1.2a + 0.8b = 62\). 4. Solve the first equation for \(b\): \(b = 60 - a\). 5. Substitute: \(1.2a + 0.8(60 - a) = 62\). 6. Simplify: \(1.2a + 48 - 0.8a = 62\), so \(0.4a = 14\) and \(a = 35\). 7. Then \(b = 60 - 35 = 25\).

Homeroom A currently has \(35\) students, and Homeroom B currently has \(25\) students.

- Write one equation for each condition. - One equation expresses one variable directly in terms of the other, which makes substitution useful. - Substitute the first value you find into either original equation to determine the other value.

1. The first condition gives \(x = 1.5y\), and the second gives \(x + y = 35\). 2. Substitute \(1.5y\) for \(x\) in the second equation: \(1.5y + y = 35\). 3. Combine like terms: \(2.5y = 35\), so \(y = 14\). 4. Substitute into the first equation: \(x = 1.5 \cdot 14 = 21\). 5. Check: \(21 + 14 = 35\) and \(21 = 1.5 \cdot 14\).

The equations are \(x = 1.5y\) and \(x + y = 35\). The ordered pair is \((21, 14)\).

- Choose variables for both current ages. - Express each person’s age \(5\) years from now. - Write one equation for each relationship in the problem. - Substitute the expression for one age into the other equation.

1. Let \(s\) be Sarah’s current age and \(f\) her father’s current age. 2. The first relationship gives \(f = 3s\). 3. In \(5\) years, their ages will be \(s + 5\) and \(f + 5\), so \(f + 5 = 2(s + 5) + 6\). 4. Substitute \(f = 3s\): \(3s + 5 = 2(s + 5) + 6\). 5. Simplify: \(3s + 5 = 2s + 16\), so \(s = 11\). 6. Then \(f = 3 \cdot 11 = 33\).

Sarah is \(11\) years old, and her father is \(33\) years old.

- Write the relationship between the original length and width. - Write formulas for the original and new perimeters. - Subtract the original perimeter from the new perimeter. - Use the side-length relationship after finding one dimension.

1. The relationship between the original sides is \(l = w + 4\). 2. The original perimeter is \(2l + 2w\). 3. After the length is doubled, the new perimeter is \(2(2l) + 2w = 4l + 2w\). 4. The increase in perimeter gives \((4l + 2w) - (2l + 2w) = 20\). 5. Simplify: \(2l = 20\), so \(l = 10\). 6. Substitute into \(l = w + 4\): \(10 = w + 4\), so \(w = 6\).

The original length is \(10\,\text{cm}\), and the original width is \(6\,\text{cm}\).

- Choose variables for the two ticket counts. - Use one equation for the total number of tickets and another for total revenue. - Include the ticket price with each unknown count. - Choose an efficient method for solving the system.

1. Let \(x\) be the number of adult tickets and \(y\) the number of child tickets. 2. Write the system \(x + y = 20\) and \(12x + 7y = 190\). 3. Solve the first equation for \(x\): \(x = 20 - y\). 4. Substitute: \(12(20 - y) + 7y = 190\). 5. Simplify: \(240 - 12y + 7y = 190\), so \(-5y = -50\) and \(y = 10\). 6. Then \(x = 20 - 10 = 10\).

There were \(10\) adults and \(10\) children.

- Write one equation for each class’s purchase. - Choose variables for the two case prices. - Multiply the equations so one variable can be eliminated. - Make sure you find both prices.

1. Let \(x\) be the price of one case of sparkling water and \(y\) the price of one case of apple juice, in dollars. 2. Write the system \(5x + 3y = 105\) and \(3x + 4y = 96\). 3. Multiply the first equation by \(4\): \(20x + 12y = 420\). 4. Multiply the second equation by \(-3\): \(-9x - 12y = -288\). 5. Add the equations: \(11x = 132\), so \(x = 12\). 6. Substitute into \(3x + 4y = 96\): \(36 + 4y = 96\), so \(y = 15\).

A case of sparkling water costs \(\$12\), and a case of apple juice costs \(\$15\).

- Track how each person’s amount changes when money is transferred. - Translate “twice as much” and “the same amount” into equations. - Write one equation for each statement. - Simplify both equations before choosing a solution method.

1. Let \(a\) be Anna’s savings and \(b\) Ben’s savings. 2. If Ben gives Anna \(\$15\), the relationship is \(a + 15 = 2(b - 15)\). 3. If Anna gives Ben \(\$5\), the relationship is \(a - 5 = b + 5\). 4. Rewrite the equations as \(a - 2b = -45\) and \(a - b = 10\). 5. From the second equation, \(a = b + 10\). Substitute into the first: \((b + 10) - 2b = -45\). 6. Solve: \(-b = -55\), so \(b = 55\). 7. Then \(a = 55 + 10 = 65\).

Anna has saved \(\$65\), and Ben has saved \(\$55\).

- Recall the sum of the interior angles of a triangle. - Use the fact that the base angles of an isosceles triangle are congruent. - Write an equation relating the vertex angle to a base angle. - Check that the three angles total \(180^\circ\).

1. Let \(b\) be the measure of each base angle and \(v\) the measure of the vertex angle, in degrees. 2. The angle sum gives \(2b + v = 180\). 3. The relationship between the angles gives \(v = b + 15\). 4. Substitute into the angle-sum equation: \(2b + (b + 15) = 180\). 5. Simplify: \(3b + 15 = 180\), so \(3b = 165\) and \(b = 55\). 6. Then \(v = 55 + 15 = 70\).

The two base angles are \(55^\circ\) each, and the vertex angle is \(70^\circ\).

- How many intersections do you need to find the three vertices of a triangle formed by three lines? - Which pairs of lines should you intersect? - On the line \(x = 2\), the x-coordinate of every point is already known. How can you use that?

1. Find the intersection of \(g_1\) and \(g_2\): \(2x - 4 = -0.5x + 6\). Then \(2.5x = 10\), so \(x = 4\). Substituting into \(g_1\) gives \(y = 2 \cdot 4 - 4 = 4\), so one vertex is \((4, 4)\). 2. For the intersection of \(g_1\) and \(g_3\), use \(x = 2\). Then \(y = 2 \cdot 2 - 4 = 0\), so a second vertex is \((2, 0)\). 3. For the intersection of \(g_2\) and \(g_3\), use \(x = 2\). Then \(y = -0.5 \cdot 2 + 6 = 5\), so the third vertex is \((2, 5)\).

The vertices are \((4, 4)\), \((2, 0)\), and \((2, 5)\).

- Solve a system for each pair of lines. - Choose substitution or elimination based on the form of each pair. - Check each ordered pair in both equations that produced it.

1. Find the intersection of \(L_1\) and \(L_2\). Add \(x - y = -2\) and \(2x + y = 8\): \(3x = 6\), so \(x = 2\). Substitute into \(L_1\): \(2 - y = -2\), so \(y = 4\). This vertex is \((2, 4)\). 2. Find the intersection of \(L_1\) and \(L_3\). From \(L_1\), \(y = x + 2\). Substitute into \(L_3\): \(x + 2(x + 2) = 4\). Then \(3x + 4 = 4\), so \(x = 0\) and \(y = 2\). This vertex is \((0, 2)\). 3. Find the intersection of \(L_2\) and \(L_3\). From \(L_2\), \(y = 8 - 2x\). Substitute into \(L_3\): \(x + 2(8 - 2x) = 4\). Then \(-3x + 16 = 4\), so \(x = 4\) and \(y = 0\). This vertex is \((4, 0)\).

The triangle’s vertices are \((2, 4)\), \((0, 2)\), and \((4, 0)\).

- Write an equation for the rectangle’s perimeter. - Relate the square’s side length to the rectangle’s longer side. - Use the difference between the two perimeters. - Substitute the longer side into the rectangle equation.

1. Let \(l\) be the length and \(w\) the width of the rectangular field, in yards. 2. The rectangle’s perimeter gives \(2(l + w) = 300\), so \(l + w = 150\). 3. The square’s side length is \(l\), and its perimeter is \(300 + 160 = 460\). Therefore, \(4l = 460\). 4. Solve: \(l = 115\). 5. Substitute into \(l + w = 150\): \(115 + w = 150\), so \(w = 35\).

The rectangular field is \(115\,\text{yd}\) long and \(35\,\text{yd}\) wide.

- Write an equation for the perimeter. - Set the adjusted area equal to the original area. - Expand the product \((x + 3)(y - 2)\). - Use substitution after simplifying both equations.

1. Let \(x\) be the side that is increased and \(y\) the side that is decreased, in centimeters. 2. The perimeter gives \(2(x + y) = 40\), so \(x + y = 20\). 3. The unchanged area gives \((x + 3)(y - 2) = xy\). 4. Expand and simplify: \(xy - 2x + 3y - 6 = xy\), so \(-2x + 3y = 6\). 5. From \(x + y = 20\), write \(x = 20 - y\). 6. Substitute: \(-2(20 - y) + 3y = 6\). 7. Simplify: \(-40 + 5y = 6\), so \(5y = 46\) and \(y = 9.2\). 8. Then \(x = 20 - 9.2 = 10.8\).

The original side lengths are \(10.8\,\text{cm}\) and \(9.2\,\text{cm}\).

- Choose a variable for each type of animal. - Write one equation for the total number of heads and another for the total number of legs. - For part b, consider whether a sum of even numbers can be odd.

1. Let \(x\) be the number of chickens and \(y\) the number of sheep. 2. Write the system: \(x + y = 25\) and \(2x + 4y = 72\). 3. Solve the first equation for \(x\): \(x = 25 - y\). 4. Substitute: \(2(25 - y) + 4y = 72\). 5. Simplify: \(50 - 2y + 4y = 72\), so \(2y = 22\) and \(y = 11\). 6. Then \(x = 25 - 11 = 14\). 7. For part b, every animal has an even number of legs. A sum of even numbers must be even, but \(75\) is odd.

a) There are \(14\) chickens and \(11\) sheep. b) The claim is impossible because every animal contributes an even number of legs, so the total cannot be odd.

- Choose variables for the amounts of the two juices. - Write one equation for the total volume. - Write another equation for the total cost using each price per quart.

1. Let \(x\) be the amount of Juice A and \(y\) the amount of Juice B, in quarts. 2. Write the system \(x + y = 20\) and \(1.20x + 2.00y = 32.80\). 3. Solve the first equation for \(y\): \(y = 20 - x\). 4. Substitute: \(1.2x + 2(20 - x) = 32.8\). 5. Simplify: \(1.2x + 40 - 2x = 32.8\), so \(-0.8x = -7.2\) and \(x = 9\). 6. Then \(y = 20 - 9 = 11\).

The class should use \(9\,\text{qt}\) of Juice A and \(11\,\text{qt}\) of Juice B.

- Translate each sentence into an equation with two variables. - Choose a solution method that can eliminate one variable efficiently. - Multiply the equations so one pair of variable terms becomes opposites. - Check both numbers in both original equations.

1. Let the numbers be \(x\) and \(y\). Write the system \(2x + 3y = 50\) and \(5x - 2y = 11\). 2. Multiply the first equation by \(2\): \(4x + 6y = 100\). 3. Multiply the second equation by \(3\): \(15x - 6y = 33\). 4. Add the equations: \(19x = 133\), so \(x = 7\). 5. Substitute into \(2x + 3y = 50\): \(14 + 3y = 50\), so \(y = 12\). 6. Check: \(2 \cdot 7 + 3 \cdot 12 = 50\) and \(5 \cdot 7 - 2 \cdot 12 = 11\).

The first number is \(7\), and the second number is \(12\).

- Write one equation for the total weight of the blend. - Find the total dollar value of the finished blend. - Write a second equation using the value contributed by each coffee. - Solve the resulting system.

1. Let \(m\) be the number of pounds of mild coffee and \(b\) the number of pounds of bold coffee. 2. The total weight gives \(m + b = 20\). 3. The total value is \(20 \cdot 16.50 = 330\) dollars, so \(12m + 18b = 330\). 4. Solve the first equation for \(m\): \(m = 20 - b\). 5. Substitute: \(12(20 - b) + 18b = 330\). 6. Simplify: \(240 - 12b + 18b = 330\), so \(6b = 90\) and \(b = 15\). 7. Then \(m = 20 - 15 = 5\).

The blend requires \(5\,\text{lb}\) of mild coffee and \(15\,\text{lb}\) of bold coffee.

- Their distances traveled must add to the original separation. - Use distance equals rate times time. - Express Ben’s speed in terms of Anna’s speed. - Substitute that relationship into the distance equation.

1. Let \(v_A\) be Anna’s speed and \(v_B\) Ben’s speed, in miles per hour. 2. Ben’s speed is \(v_B = v_A + 6\). 3. Their distances after \(1.5\) hours add to \(45\) miles, so \(1.5v_A + 1.5v_B = 45\). 4. Substitute \(v_B = v_A + 6\): \(1.5v_A + 1.5(v_A + 6) = 45\). 5. Simplify: \(3v_A + 9 = 45\), so \(3v_A = 36\) and \(v_A = 12\). 6. Then \(v_B = 12 + 6 = 18\).

Anna’s average speed is \(12\,\text{mph}\), and Ben’s average speed is \(18\,\text{mph}\).

- Recall the sum of the interior angles of a triangle. - Express both \(\beta\) and \(\gamma\) in terms of \(\alpha\). - Substitute into the angle-sum equation. - Check that the three results total \(180^\circ\).

1. The angle sum is \(\alpha + \beta + \gamma = 180^\circ\). 2. The given relationships are \(\beta = 2\alpha\) and \(\gamma = \beta - 30^\circ = 2\alpha - 30^\circ\). 3. Substitute into the angle-sum equation: \(\alpha + 2\alpha + (2\alpha - 30^\circ) = 180^\circ\). 4. Simplify: \(5\alpha - 30^\circ = 180^\circ\), so \(5\alpha = 210^\circ\) and \(\alpha = 42^\circ\). 5. Then \(\beta = 2 \cdot 42^\circ = 84^\circ\). 6. Finally, \(\gamma = 84^\circ - 30^\circ = 54^\circ\).

\(\alpha = 42^\circ\), \(\beta = 84^\circ\), and \(\gamma = 54^\circ\).

- Use the fact that the base angles of an isosceles triangle are congruent. - Express the vertex angle in terms of one base angle. - Use the sum of the interior angles of a triangle.

1. Because the triangle is isosceles, \(\alpha = \beta\). 2. The angle sum is \(2\alpha + \gamma = 180^\circ\). 3. The given relationship is \(\gamma = 1.5(\alpha + \beta) = 1.5(2\alpha) = 3\alpha\). 4. Substitute into the angle-sum equation: \(2\alpha + 3\alpha = 180^\circ\). 5. Then \(5\alpha = 180^\circ\), so \(\alpha = 36^\circ\). 6. Therefore, \(\beta = 36^\circ\) and \(\gamma = 3 \cdot 36^\circ = 108^\circ\).

\(\alpha = 36^\circ\), \(\beta = 36^\circ\), and \(\gamma = 108^\circ\).

- Recall the sum of the two acute angles in a right triangle. - Translate the comparison between the two angles into an equation. - Use substitution to solve the system.

1. The two acute angles of a right triangle are complementary, so \(\alpha + \beta = 90^\circ\). 2. The second relationship is \(\alpha = 3\beta + 12^\circ\). 3. Substitute into the first equation: \((3\beta + 12^\circ) + \beta = 90^\circ\). 4. Simplify: \(4\beta + 12^\circ = 90^\circ\), so \(4\beta = 78^\circ\) and \(\beta = 19.5^\circ\). 5. Then \(\alpha = 90^\circ - 19.5^\circ = 70.5^\circ\).

\(\alpha = 70.5^\circ\) and \(\beta = 19.5^\circ\).

- Choose variables for the two unknown weights. - Translate each weighing into an equation. - Consider multiplying the equations so one variable has matching coefficients. - Verify the weights in both original combinations.

1. Let \(g\) be the weight of one gold bead and \(s\) the weight of one silver coin, in grams. 2. Write the system \(2g + 3s = 48\) and \(3g + 2s = 57\). 3. Multiply the first equation by \(3\): \(6g + 9s = 144\). 4. Multiply the second equation by \(2\): \(6g + 4s = 114\). 5. Subtract the second new equation from the first: \(5s = 30\), so \(s = 6\). 6. Substitute into \(2g + 3s = 48\): \(2g + 18 = 48\), so \(g = 15\).

One gold bead weighs \(15\,\text{g}\), and one silver coin weighs \(6\,\text{g}\).

- Represent a two-digit number using its tens digit and ones digit. - Translate the digit sum into an equation. - Write an expression for the number with its digits reversed. - Use the stated difference to form a second equation.

1. Let \(x\) be the tens digit and \(y\) the ones digit. The original number is \(10x + y\). 2. The digit sum gives \(x + y = 12\). 3. The reversed number is \(10y + x\), so \(10y + x = 10x + y + 18\). 4. Simplify the second equation: \(-x + y = 2\). 5. Add \(x + y = 12\) and \(-x + y = 2\): \(2y = 14\), so \(y = 7\). 6. Then \(x = 12 - 7 = 5\), so the original number is \(10 \cdot 5 + 7 = 57\).

The original number is \(57\).

- Choose variables for the two ticket prices. - Write one cost equation for each group. - Multiply the equations so one variable has matching coefficients.

1. Let \(x\) be the price of one adult ticket and \(y\) the price of one child ticket, in dollars. 2. Write the system \(3x + 4y = 88\) and \(2x + 5y = 82\). 3. Multiply the first equation by \(2\): \(6x + 8y = 176\). 4. Multiply the second equation by \(3\): \(6x + 15y = 246\). 5. Subtract the first new equation from the second: \(7y = 70\), so \(y = 10\). 6. Substitute into \(3x + 4y = 88\): \(3x + 40 = 88\), so \(x = 16\).

An adult ticket costs \(\$16\), and a child ticket costs \(\$10\).

- Use the first scale to find the weight of one wooden block. - Carry that block weight into the equation for the second scale. - Write one equation for each balanced scale. - On the second scale, find what must be added to the two blocks to total \(20\,\text{oz}\).

1. Let \(x\) be the weight of one wooden block in ounces. The first scale gives \(3x + 8 = 32\). 2. Subtract \(8\): \(3x = 24\). Divide by \(3\): \(x = 8\). 3. Use the second scale: \(2x + y = 20\). 4. Substitute \(x = 8\): \(2 \cdot 8 + y = 20\). 5. Simplify and solve: \(16 + y = 20\), so \(y = 4\).

The unknown weight is \(4\,\text{oz}\).

- Identify the two speeds that combine to produce the ground speed. - Add the speeds when Julia walks with the walkway. - Subtract the walkway speed when she walks against it. - Add the equations to eliminate one variable.

1. Let \(v_J\) be Julia’s walking speed and \(v_W\) the walkway’s speed, in meters per second. 2. Walking with the walkway gives \(v_J + v_W = 2.5\). 3. Walking against the walkway gives \(v_J - v_W = 0.7\). 4. Add the equations: \(2v_J = 3.2\), so \(v_J = 1.6\). 5. Substitute into the first equation: \(1.6 + v_W = 2.5\), so \(v_W = 0.9\).

Julia’s walking speed is \(1.6\,\text{m/s}\), and the walkway’s speed is \(0.9\,\text{m/s}\).

- Use the total number of vehicles to express one vehicle count in terms of the other. - Account for the number of wheels contributed by each type of vehicle. - Write one equation for the total number of wheels. - Check that the two vehicle counts add to \(22\).

1. Let \(x\) be the number of cars. Then the number of motorcycles is \(22 - x\). 2. Write the wheel-count equation \(4x + 2(22 - x) = 72\). 3. Distribute and combine like terms: \(4x + 44 - 2x = 72\), so \(2x + 44 = 72\). 4. Subtract \(44\): \(2x = 28\). Divide by \(2\): \(x = 14\). 5. The number of motorcycles is \(22 - 14 = 8\).

There are \(14\) cars and \(8\) motorcycles.

- Choose variables for the two amounts being mixed. - Write one equation for the total volume. - Write another equation for the amount of pure fruit juice. - Solve the system using substitution or elimination.

1. Let \(x\) be the number of gallons of Drink A and \(y\) the number of gallons of Drink B. 2. The total volume gives \(x + y = 60\). 3. The amount of fruit juice gives \(0.40x + 0.70y = 0.50 \cdot 60\), or \(0.4x + 0.7y = 30\). 4. Solve the first equation for \(y\): \(y = 60 - x\). 5. Substitute: \(0.4x + 0.7(60 - x) = 30\). 6. Simplify: \(0.4x + 42 - 0.7x = 30\), so \(-0.3x = -12\) and \(x = 40\). 7. Then \(y = 60 - 40 = 20\).

The company should mix \(40\,\text{gal}\) of Drink A and \(20\,\text{gal}\) of Drink B.

- Write one equation for the total mass. - Write a second equation for the amount of pure gold in the mixture. - Express each percentage as a decimal multiplier. - Eliminate or substitute one variable to solve the system.

1. Let \(m_1\) be the mass of the \(60\%\) alloy and \(m_2\) the mass of the \(90\%\) alloy, in grams. 2. The total mass gives \(m_1 + m_2 = 300\). 3. The amount of gold gives \(0.60m_1 + 0.90m_2 = 0.80 \cdot 300\), or \(0.6m_1 + 0.9m_2 = 240\). 4. Solve the first equation for \(m_1\): \(m_1 = 300 - m_2\). 5. Substitute: \(0.6(300 - m_2) + 0.9m_2 = 240\). 6. Simplify: \(180 - 0.6m_2 + 0.9m_2 = 240\), so \(0.3m_2 = 60\) and \(m_2 = 200\). 7. Then \(m_1 = 300 - 200 = 100\).

The jeweler should use \(100\,\text{g}\) of the \(60\%\) alloy and \(200\,\text{g}\) of the \(90\%\) alloy.

- Use one equation for the total number of tickets. - Write a second equation for total revenue. - Express one ticket count in terms of the other. - For equal ticket counts, divide the fixed total equally between the two types.

1. Let \(x\) be the number of adult tickets and \(y\) the number of student tickets. 2. Write the system \(x + y = 240\) and \(8x + 5y = 1440\). 3. Solve the first equation for \(y\): \(y = 240 - x\). 4. Substitute: \(8x + 5(240 - x) = 1440\). 5. Simplify: \(8x + 1200 - 5x = 1440\), so \(3x = 240\) and \(x = 80\). 6. Then \(y = 240 - 80 = 160\). 7. For equal numbers, each type would have \(240 \div 2 = 120\) tickets. 8. The revenue would be \(120 \cdot 8 + 120 \cdot 5 = 1560\) dollars.

a) The festival sold \(80\) adult tickets and \(160\) student tickets. b) The revenue would have been \(\$1560\).

- Identify the two parts that make up the total weight. - After three-fourths of the apples are sold, what fraction remains? - Write one equation for the original weight and one for the later weight. - Consider subtracting the equations to eliminate the crate’s weight.

1. Let \(x\) be the weight of the empty crate and \(y\) the original weight of the apples, in pounds. 2. Write the system \(x + y = 31\) and \(x + \frac{1}{4}y = 10\). 3. Subtract the second equation from the first: \(\frac{3}{4}y = 21\), so \(y = 28\). 4. Substitute into \(x + y = 31\): \(x + 28 = 31\), so \(x = 3\).

The empty crate weighs \(3\,\text{lb}\).

- Choose variables for the two unknown weights. - Write an equation for each combination of items. - Decide which system-solving method is most efficient. - Check the weights in both original combinations.

1. Let \(r\) be the weight of one backpack and \(s\) the weight of one sleeping bag, in pounds. 2. Write the system \(r + 2s = 11\) and \(2r + s = 10\). 3. Solve the second equation for \(s\): \(s = 10 - 2r\). 4. Substitute into the first equation: \(r + 2(10 - 2r) = 11\). 5. Simplify: \(r + 20 - 4r = 11\), so \(-3r = -9\) and \(r = 3\). 6. Then \(s = 10 - 2 \cdot 3 = 4\).

One backpack weighs \(3\,\text{lb}\), and one sleeping bag weighs \(4\,\text{lb}\).

- Write an expression for the amount remaining in each tank after a given time. - Translate “equal amounts” into an equation. - Write a system with the initial total and the equal-amount condition. - First determine how much water drains from each tank in \(50\) minutes.

1. Let \(x\) be the initial amount in Tank 1 and \(y\) the initial amount in Tank 2, in gallons. 2. The initial total gives \(x + y = 900\). 3. After \(50\) minutes, Tank 1 has lost \(3 \cdot 50 = 150\) gallons and Tank 2 has lost \(5 \cdot 50 = 250\) gallons. 4. Equal remaining amounts give \(x - 150 = y - 250\), or \(x = y - 100\). 5. Substitute into the total equation: \((y - 100) + y = 900\), so \(2y = 1000\) and \(y = 500\). 6. Then \(x = 500 - 100 = 400\).

Tank 1 originally contained \(400\,\text{gal}\), and Tank 2 originally contained \(500\,\text{gal}\).

- One person’s amount decreases each week, while the other person’s amount increases. - Determine how much Paul spends over \(12\) weeks. - Determine how much Marie adds over \(12\) weeks. - Write an equation showing that their ending amounts are equal.

1. Let \(p\) be Paul’s starting amount and \(m\) Marie’s starting amount. 2. Their starting total gives \(p + m = 150\). 3. After \(12\) weeks, Paul has \(p - 4 \cdot 12 = p - 48\), and Marie has \(m + 2 \cdot 12 = m + 24\). 4. Equal ending amounts give \(p - 48 = m + 24\), or \(p - m = 72\). 5. Add the equations \(p + m = 150\) and \(p - m = 72\): \(2p = 222\), so \(p = 111\). 6. Then \(m = 150 - 111 = 39\).

Paul started with \(\$111\), and Marie started with \(\$39\).

- Choose variables for the two individual prices. - Translate each piece of information into an equation. - Express the more expensive item in terms of the less expensive item. - Substitute so the equation contains only one unknown.

1. Let \(b\) be the price of one sandwich and \(j\) the price of one juice box, in dollars. 2. Write the system \(2b + 3j = 15.60\) and \(b = j + 1.80\). 3. Substitute into the first equation: \(2(j + 1.80) + 3j = 15.60\). 4. Simplify: \(2j + 3.60 + 3j = 15.60\), so \(5j = 12\) and \(j = 2.40\). 5. Then \(b = 2.40 + 1.80 = 4.20\).

One sandwich costs \(\$4.20\), and one juice box costs \(\$2.40\).

- Choose variables for both current ages. - Express both ages \(12\) years from now. - Translate the future age relationship into an equation. - Substitute one age expression into the other equation.

1. Let \(f\) be the father’s current age and \(s\) the son’s current age. 2. Their current ages satisfy \(f = 4s\). 3. In \(12\) years, their ages satisfy \(f + 12 = 2(s + 12)\). 4. Substitute \(f = 4s\): \(4s + 12 = 2(s + 12)\). 5. Simplify: \(4s + 12 = 2s + 24\), so \(2s = 12\) and \(s = 6\). 6. Then \(f = 4 \cdot 6 = 24\).

The father is \(24\) years old, and the son is \(6\) years old.

- For the first equation, choose pairs whose coordinates add to \(10\). - Substitute the expression for \(y\) from the second equation into the first. - A common solution lies on both lines.

1. For equation (I), two examples are \((0, 10)\) and \((5, 5)\). 2. For equation (II), two examples are \((0, 2)\) and \((2, 4)\). 3. Substitute \(y = x + 2\) into equation (I): \(x + x + 2 = 10\). 4. Solve: \(2x = 8\), so \(x = 4\). 5. Then \(y = 4 + 2 = 6\). The common solution is \((4, 6)\).

a) For (I): \((0, 10)\), \((5, 5)\). For (II): \((0, 2)\), \((2, 4)\). b) \((4, 6)\)

- First rearrange each equation so that \(y\) is alone on one side. - When solving the second equation for \(y\), pay close attention to the signs when dividing by a negative number. - Choose convenient points for each graph by substituting simple values for \(x\). - Read the coordinates of the intersection carefully.

1. Solve the first equation for \(y\): \(y = -2x + 4\). 2. Solve the second equation for \(y\): \(x - 2y = -3\), so \(-2y = -x - 3\) and \(y = \frac{1}{2}x + \frac{3}{2}\). 3. Graph \(y = -2x + 4\), using points such as \((0, 4)\) and \((2, 0)\). 4. Graph \(y = \frac{1}{2}x + \frac{3}{2}\), using points such as \((1, 2)\) and \((3, 3)\). 5. The lines intersect at \((1, 2)\), so this ordered pair is the solution.

The solution set is \(\{(1, 2)\}\).

- Choose variables for the two prices per quart. - Write one equation for each blend. - Multiply one equation so a variable can be eliminated. - Check both prices in both original equations.

1. Let \(x\) be the price per quart of orange juice and \(y\) the price per quart of apple juice, in dollars. 2. Write the system \(3x + 2y = 10.10\) and \(5x + 4y = 18.10\). 3. Multiply the first equation by \(2\): \(6x + 4y = 20.20\). 4. Subtract the second equation: \(x = 2.10\). 5. Substitute into the first equation: \(3 \cdot 2.10 + 2y = 10.10\). 6. Simplify: \(6.30 + 2y = 10.10\), so \(2y = 3.80\) and \(y = 1.90\).

Orange juice costs \(\$2.10\) per quart, and apple juice costs \(\$1.90\) per quart.

- Choose variables for the two prices per pound. - Write one equation for each blend. - Simplify an equation before solving the system when possible. - Find both prices before calculating their difference.

1. Let \(a\) be the price per pound of Arabica and \(r\) the price per pound of Robusta, in dollars. 2. Write the system \(4a + 6r = 152\) and \(5a + 5r = 155\). 3. Divide the second equation by \(5\): \(a + r = 31\), so \(r = 31 - a\). 4. Substitute into the first equation: \(4a + 6(31 - a) = 152\). 5. Simplify: \(4a + 186 - 6a = 152\), so \(-2a = -34\) and \(a = 17\). 6. Then \(r = 31 - 17 = 14\). 7. Arabica costs \(17 - 14 = 3\) dollars more per pound.

Arabica costs more. It costs \(\$3\) more per pound than Robusta.

- Choose variables for the two unknown weights. - Translate the two relationships between the weights into equations. - One equation already expresses one variable in terms of the other, so consider substitution. - Verify that the elephant and four zebras have the stated total weight.

1. Let \(z\) be the weight of one zebra and \(e\) the weight of the elephant, in kilograms. 2. Write the system \(e + 4z = 5500\) and \(e = 12z + 100\). 3. Substitute the second equation into the first: \((12z + 100) + 4z = 5500\). 4. Simplify: \(16z + 100 = 5500\), so \(16z = 5400\) and \(z = 337.5\). 5. Then \(e = 12 \cdot 337.5 + 100 = 4150\).

One zebra weighs \(337.5\,\text{kg}\), and the elephant weighs \(4150\,\text{kg}\).

- Write the perimeter formula for a rectangle. - Translate the relationship between length and width into an equation. - Substitute one equation into the other. - Compare the calculated width with \(15\,\text{ft}\).

1. Let \(w\) be the width and \(l\) the length, in feet. 2. The perimeter gives \(2l + 2w = 84\), or \(l + w = 42\). 3. The length relationship is \(l = 3w - 6\). 4. Substitute into the perimeter equation: \((3w - 6) + w = 42\). 5. Simplify: \(4w - 6 = 42\), so \(4w = 48\) and \(w = 12\). 6. Then \(l = 3 \cdot 12 - 6 = 30\). 7. The width is \(12\,\text{ft}\), which is less than \(15\,\text{ft}\), so the apprentice is not correct.

The flower bed is \(30\,\text{ft}\) long and \(12\,\text{ft}\) wide. The apprentice’s claim is false.

- Choose variables for the two container capacities. - Translate each relationship into an equation. - One variable is already isolated in one equation, which makes substitution useful. - Check your capacities in both original statements.

1. Let \(x\) be the capacity of one large container and \(y\) the capacity of one small container, in gallons. 2. Write the system \(2x + 6y = 54\) and \(x = 2y + 2\). 3. Substitute the second equation into the first: \(2(2y + 2) + 6y = 54\). 4. Simplify: \(4y + 4 + 6y = 54\), so \(10y = 50\) and \(y = 5\). 5. Then \(x = 2 \cdot 5 + 2 = 12\).

A large container holds \(12\,\text{gal}\), and a small container holds \(5\,\text{gal}\).

- Choose variables for the two ticket prices. - Write one revenue equation for each night. - Simplify the equations before solving. - Find the regular price for the group and compare it with the package price.

1. Let \(s\) be the student ticket price and \(a\) the adult ticket price, in dollars. 2. Write the system \(60s + 20a = 440\) and \(45s + 30a = 480\). 3. Simplify the equations to \(3s + a = 22\) and \(3s + 2a = 32\). 4. Subtract the first simplified equation from the second: \(a = 10\). 5. Substitute into \(3s + a = 22\): \(3s + 10 = 22\), so \(s = 4\). 6. At regular prices, the group would pay \(5 \cdot 4 + 2 \cdot 10 = 40\) dollars. 7. Since \(35 < 40\), the package saves \(5\) dollars.

a) A student ticket costs \(\$4\), and an adult ticket costs \(\$10\). b) Yes. The package costs \(\$5\) less than the \(\$40\) regular price.

- Choose variables for the two ticket prices. - Translate the total cost into an equation. - Translate “costs \(\$12\) more than” carefully. - Multiply the equations so one variable can be eliminated. - Check that both prices satisfy both conditions.

1. Let \(x\) be the adult ticket price and \(y\) the student ticket price, in dollars. 2. Write the system \(3x + 5y = 75\) and \(2x = 3y + 12\). 3. Rewrite the second equation as \(2x - 3y = 12\). 4. Multiply the first equation by \(3\): \(9x + 15y = 225\). 5. Multiply the second equation by \(5\): \(10x - 15y = 60\). 6. Add the equations: \(19x = 285\), so \(x = 15\). 7. Substitute into \(2x - 3y = 12\): \(30 - 3y = 12\), so \(y = 6\).

An adult ticket costs \(\$15\), and a student ticket costs \(\$6\).

- Use distance equals rate times time for each observation. - Write one equation for the combined distances. - Carefully translate which train travels \(140\) miles farther in the second comparison. - Substitute a repeated distance expression into the first equation.

1. Let \(v_A\) and \(v_B\) be the speeds of Trains A and B, in miles per hour. 2. The first condition gives \(4v_A + 3v_B = 620\). 3. The second condition gives \(2v_A = 3v_B - 140\), or \(3v_B = 2v_A + 140\). 4. Substitute into the first equation: \(4v_A + (2v_A + 140) = 620\). 5. Simplify: \(6v_A + 140 = 620\), so \(6v_A = 480\) and \(v_A = 80\). 6. Then \(3v_B = 2 \cdot 80 + 140 = 300\), so \(v_B = 100\).

Train A travels at \(80\,\text{mph}\), and Train B travels at \(100\,\text{mph}\).

- Identify what remains unchanged in both packing arrangements. - Choose variables for the number of crates and the number of apples. - When two crates are empty, how many crates are actually used? - Write an expression for the total number of apples in each arrangement.

1. Let \(x\) be the number of crates and \(y\) the number of apples. 2. The first packing arrangement gives \(y = 12x + 15\). 3. In the second arrangement, only \(x - 2\) crates are used, so \(y = 15(x - 2)\). 4. Set the expressions equal: \(12x + 15 = 15(x - 2)\). 5. Simplify: \(12x + 15 = 15x - 30\), so \(3x = 45\) and \(x = 15\). 6. Then \(y = 12 \cdot 15 + 15 = 195\).

There are \(195\) apples and \(15\) crates.

- Convert the early and late times from minutes to hours. - Write a distance expression for each speed scenario. - The distance is the same in both scenarios. - Solve for the scheduled time, then calculate the distance.

1. Let \(t\) be the scheduled travel time in hours and \(d\) the distance in miles. 2. Convert the time differences: \(10\) minutes is \(\frac{1}{6}\) hour, and \(5\) minutes is \(\frac{1}{12}\) hour. 3. The two scenarios give \(d = 60\left(t + \frac{1}{6}\right)\) and \(d = 80\left(t - \frac{1}{12}\right)\). 4. Set the distance expressions equal: \(60t + 10 = 80t - \frac{20}{3}\). 5. Solve: \(20t = 10 + \frac{20}{3} = \frac{50}{3}\), so \(t = \frac{5}{6}\) hour, or \(50\) minutes. 6. Find the distance: \(d = 60\left(\frac{5}{6} + \frac{1}{6}\right) = 60\).

The destination is \(60\,\text{mi}\) away, and the scheduled travel time is \(50\) minutes.

- Write an expression for the gift price in each situation. - Combine the students’ contributions with the amount short or left over. - Set the two expressions for the same gift price equal. - Divide the total cost equally for part b.

1. Let \(x\) be the number of students and \(y\) the price of the gift, in dollars. 2. The first situation gives \(y = 4.50x + 12\). 3. The second situation gives \(y = 5.50x - 12\). 4. Set the expressions equal: \(4.50x + 12 = 5.50x - 12\). 5. Solve: \(24 = x\), so there are \(24\) students. 6. The gift price is \(y = 4.50 \cdot 24 + 12 = 120\) dollars. 7. The exact contribution is \(120 \div 24 = 5\) dollars per student.

a) There are \(24\) students, and the gift costs \(\$120\). b) Each student should contribute \(\$5.00\).

- Account for the additional people in the second situation. - Express the rental cost using the contributions and the shortage or surplus. - Use parentheses for the size of the larger group. - Set the two expressions for the rental cost equal.

1. Let \(x\) be the original number of people and \(M\) the rental cost, in dollars. 2. The first situation gives \(M = 20x + 40\). 3. The second situation gives \(M = 18(x + 5) - 20\). 4. Set the expressions equal: \(20x + 40 = 18(x + 5) - 20\). 5. Simplify: \(20x + 40 = 18x + 70\), so \(2x = 30\) and \(x = 15\). 6. The rent is \(M = 20 \cdot 15 + 40 = 340\) dollars. 7. Check: \(18 \cdot (15 + 5) - 20 = 340\).

The original group had \(15\) people, and the cabin rental cost was \(\$340\).

- Use distance equals rate times time. - In each scenario, the cyclists’ distances add to \(60\) miles. - Carefully determine how long each cyclist rides in the second scenario. - Write one equation for each scenario.

1. Let \(v_A\) be Anton’s speed and \(v_B\) Beatrice’s speed, in miles per hour. 2. In the first scenario, their distances after \(2\) hours total \(60\) miles: \(2v_A + 2v_B = 60\). 3. In the second scenario, Anton rides for \(1\) hour and Beatrice rides for \(4\) hours, so \(v_A + 4v_B = 60\). 4. Simplify the first equation: \(v_A + v_B = 30\), so \(v_A = 30 - v_B\). 5. Substitute into the second equation: \((30 - v_B) + 4v_B = 60\). 6. Simplify: \(30 + 3v_B = 60\), so \(v_B = 10\). 7. Then \(v_A = 30 - 10 = 20\).

Anton rides at \(20\,\text{mph}\), and Beatrice rides at \(10\,\text{mph}\).

- Choose variables for the two admission prices. - Write one cost equation for each group. - Simplify the equations before solving the system. - For part b, find the individual-ticket total and compare it with the pass price.

1. Let \(x\) be the adult admission price and \(y\) the child admission price, in dollars. 2. Write the system \(5x + 10y = 90\) and \(3x + 12y = 84\). 3. Simplify the equations to \(x + 2y = 18\) and \(x + 4y = 28\). 4. Subtract: \(2y = 10\), so \(y = 5\). 5. Substitute into \(x + 2y = 18\): \(x + 10 = 18\), so \(x = 8\). 6. Individual tickets for \(2\) adults and \(3\) children cost \(2 \cdot 8 + 3 \cdot 5 = 31\) dollars. 7. Since \(31 < 35\), the family pass costs \(4\) dollars more and does not save money.

a) Adult admission is \(\$8\), and child admission is \(\$5\). b) No. Individual tickets cost \(\$31\), which is \(\$4\) less than the family pass.

- Choose variables for the two prices per pound. - Write one equation for each purchase. - Multiply the equations so one variable can be eliminated. - Check the prices in both purchases.

1. Let \(x\) be the price per pound of apples and \(y\) the price per pound of potatoes, in dollars. 2. Write the system \(2x + 5y = 13.50\) and \(3x + 2y = 12\). 3. Multiply the first equation by \(3\): \(6x + 15y = 40.50\). 4. Multiply the second equation by \(2\): \(6x + 4y = 24\). 5. Subtract: \(11y = 16.50\), so \(y = 1.50\). 6. Substitute into \(3x + 2y = 12\): \(3x + 3 = 12\), so \(x = 3\).

Apples cost \(\$3.00\) per pound, and potatoes cost \(\$1.50\) per pound.

- Express the package size in equal \(4\)-ounce portions. - Write one equation for the total number of portions. - Write another equation for the total value. - Solve for the portions, then convert them to ounces.

1. Let \(x\) be the number of \(4\)-ounce portions of Spring Blend and \(y\) the number of \(4\)-ounce portions of Evening Blend. 2. A \(20\)-ounce package contains \(5\) portions, so \(x + y = 5\). 3. The total value gives \(4.50x + 3.00y = 18\). 4. Solve the first equation for \(y\): \(y = 5 - x\). 5. Substitute: \(4.5x + 3(5 - x) = 18\). 6. Simplify: \(4.5x + 15 - 3x = 18\), so \(1.5x = 3\) and \(x = 2\). 7. Then \(y = 5 - 2 = 3\). 8. Spring Blend contributes \(2 \cdot 4 = 8\) ounces, and Evening Blend contributes \(3 \cdot 4 = 12\) ounces.

The package contains \(8\,\text{oz}\) of Spring Blend and \(12\,\text{oz}\) of Evening Blend.

- Write one equation for total distance. - Use time equals distance divided by speed for each part of the trip. - Add the two travel times to get the total time. - Clear the fractions before solving the system.

1. Let \(x\) be the distance traveled in Eco mode and \(y\) the distance traveled in Sport mode, in miles. 2. The total distance gives \(x + y = 54\). 3. The total time gives \(\frac{x}{18} + \frac{y}{27} = 2.5\). 4. Multiply the time equation by \(54\): \(3x + 2y = 135\). 5. From the distance equation, \(y = 54 - x\). 6. Substitute: \(3x + 2(54 - x) = 135\). 7. Simplify: \(x + 108 = 135\), so \(x = 27\). 8. Then \(y = 54 - 27 = 27\).

Alex travels \(27\,\text{mi}\) in Eco mode and \(27\,\text{mi}\) in Sport mode.

- Identify the desired total volume and concentration. - Choose variables for the two amounts being mixed. - Write one equation for total volume. - Write another equation for the amount of pure acetic acid. - Relate the concentrations of the original solutions to the target concentration.

1. Let \(x\) be the number of liters of the \(25\%\) solution and \(y\) the number of liters of the \(5\%\) solution. 2. The total volume gives \(x + y = 10\). 3. The amount of acetic acid gives \(0.25x + 0.05y = 0.12 \cdot 10\), or \(0.25x + 0.05y = 1.2\). 4. Solve the first equation for \(y\): \(y = 10 - x\). 5. Substitute: \(0.25x + 0.05(10 - x) = 1.2\). 6. Simplify: \(0.25x + 0.5 - 0.05x = 1.2\), so \(0.20x = 0.7\) and \(x = 3.5\). 7. Then \(y = 10 - 3.5 = 6.5\).

The technician should mix \(3.5\,\text{L}\) of the \(25\%\) solution and \(6.5\,\text{L}\) of the \(5\%\) solution.

- Choose variables for the mass per liter of each liquid. - Write one mass equation for each mixture. - Express one variable in terms of the other and substitute. - Check the two masses in both original mixtures.

1. Let \(x\) be the mass of one liter of Liquid A and \(y\) the mass of one liter of Liquid B, in kilograms. 2. Write the system \(3x + 2y = 4.4\) and \(x + 4y = 4.8\). 3. Solve the second equation for \(x\): \(x = 4.8 - 4y\). 4. Substitute into the first equation: \(3(4.8 - 4y) + 2y = 4.4\). 5. Simplify: \(14.4 - 12y + 2y = 4.4\), so \(-10y = -10\) and \(y = 1\). 6. Then \(x = 4.8 - 4 \cdot 1 = 0.8\).

One liter of Liquid A has a mass of \(0.8\,\text{kg}\), and one liter of Liquid B has a mass of \(1\,\text{kg}\).

- Write one equation for the amount of nitrogen in each mixture. - Multiply each volume by its unknown concentration. - Express one concentration in terms of the other and substitute. - Convert the decimal solutions back to percentages.

1. Let \(x\) and \(y\) be the nitrogen concentrations of Fertilizers A and B, written as decimals. 2. The first \(5\)-gallon mixture gives \(3x + 2y = 5 \cdot 0.14 = 0.70\). 3. The second \(5\)-gallon mixture gives \(x + 4y = 5 \cdot 0.10 = 0.50\). 4. Solve the second equation for \(x\): \(x = 0.50 - 4y\). 5. Substitute into the first equation: \(3(0.50 - 4y) + 2y = 0.70\). 6. Simplify: \(1.50 - 12y + 2y = 0.70\), so \(-10y = -0.80\) and \(y = 0.08\). 7. Then \(x = 0.50 - 4 \cdot 0.08 = 0.18\). 8. Therefore, Fertilizer A is \(18\%\) nitrogen and Fertilizer B is \(8\%\) nitrogen.

Fertilizer A is \(18\%\) nitrogen, and Fertilizer B is \(8\%\) nitrogen.

- Find the total value of each \(10\)-pound blend. - Write a system using the two pure-bean prices. - Multiply one equation so a variable can be eliminated. - Compare the pure-bean prices and the proportion of each bean in the blends.

1. Let \(a\) be the price per pound of Arabica and \(r\) the price per pound of Robusta, in dollars. 2. Each blend contains \(10\,\text{lb}\). Their total values give \(4a + 6r = 10 \cdot 12.80 = 128\) and \(7a + 3r = 10 \cdot 14.30 = 143\). 3. Multiply the second equation by \(2\): \(14a + 6r = 286\). 4. Subtract the first equation: \(10a = 158\), so \(a = 15.80\). 5. Substitute into \(4a + 6r = 128\): \(4 \cdot 15.80 + 6r = 128\), so \(6r = 64.80\) and \(r = 10.80\). 6. Arabica is more expensive than Robusta. Blend 2 contains \(70\%\) Arabica, while Blend 1 contains only \(40\%\), so Blend 2 has the higher average price.

a) Arabica costs \(\$15.80\) per pound, and Robusta costs \(\$10.80\) per pound. b) Blend 2 costs more because it contains a larger proportion of the more expensive Arabica beans.

- Choose variables that represent prices for the same unit quantity. - Write one total-cost equation for each mix. - Use elimination to solve the system. - Check that your answers use the requested price per \(4\,\text{oz}\).

1. Let \(x\) be the price of \(4\,\text{oz}\) of cashews and \(y\) the price of \(4\,\text{oz}\) of peanuts, in dollars. 2. Mix A gives \(2x + 3y = 5.50\). 3. Mix B gives \(3x + 2y = 7.00\). 4. Multiply the first equation by \(3\): \(6x + 9y = 16.50\). Multiply the second equation by \(-2\): \(-6x - 4y = -14.00\). 5. Add the equations: \(5y = 2.50\), so \(y = 0.50\). 6. Substitute into \(2x + 3y = 5.50\): \(2x + 3 \cdot 0.50 = 5.50\), so \(2x = 4.00\) and \(x = 2.00\).

\(4\,\text{oz}\) of cashews costs \(\$2.00\), and \(4\,\text{oz}\) of peanuts costs \(\$0.50\).

- Find the amount of active ingredient in each \(5\)-liter mixture. - Write one equation for each mixture. - Represent percentages as decimals while solving. - Use variables for the two unknown concentrations.

1. Let \(a\) be the concentration of the first liquid and \(b\) the concentration of the second, written as decimals. 2. Each mixture has a total volume of \(5\,\text{L}\). 3. The first mixture gives \(2a + 3b = 5 \cdot 0.40 = 2\). 4. The second mixture gives \(3a + 2b = 5 \cdot 0.50 = 2.5\). 5. Multiply the first equation by \(3\) and the second by \(2\): \(6a + 9b = 6\) and \(6a + 4b = 5\). 6. Subtract the second equation from the first: \(5b = 1\), so \(b = 0.20\). 7. Substitute into \(2a + 3b = 2\): \(2a + 3 \cdot 0.20 = 2\), so \(2a = 1.40\) and \(a = 0.70\). 8. Convert the decimals to percentages.

The first liquid is \(70\%\) active ingredient, and the second liquid is \(20\%\) active ingredient.

- Translate each person’s statement into an equation. - Choose variables for the two amounts of money. - One equation can be solved for a variable and substituted into the other. - Represent “two-fifths” as a fraction or decimal.

1. Let \(L\) be Lucas’s amount and \(M\) Mia’s amount, in dollars. 2. Translate the statements into the system \(L + \frac{2}{5}M = 30\) and \(M + \frac{1}{2}L = 30\). 3. Solve the second equation for \(M\): \(M = 30 - \frac{1}{2}L\). 4. Substitute into the first equation: \(L + \frac{2}{5}\left(30 - \frac{1}{2}L\right) = 30\). 5. Simplify: \(L + 12 - \frac{1}{5}L = 30\), so \(\frac{4}{5}L = 18\) and \(L = 22.5\). 6. Then \(M = 30 - \frac{1}{2} \cdot 22.5 = 18.75\).

Lucas has \(\$22.50\), and Mia has \(\$18.75\).

- Write one equation for each separate pouring situation. - Include only the amount added to the receiving cup in each equation. - Check that your solution satisfies both conditions. - Add the two original amounts to find the total.

1. Let \(x\) be the original amount in Cup A and \(y\) the original amount in Cup B, in fluid ounces. 2. Write the system \(x + \frac{1}{3}y = 10\) and \(y + \frac{1}{2}x = 12.5\). 3. Multiply the first equation by \(3\): \(3x + y = 30\), so \(y = 30 - 3x\). 4. Substitute into the second equation: \(30 - 3x + \frac{1}{2}x = 12.5\). 5. Simplify: \(-\frac{5}{2}x = -17.5\), so \(x = 7\). 6. Then \(y = 30 - 3 \cdot 7 = 9\). 7. The total amount is \(7 + 9 = 16\,\text{fl oz}\).

a) Cup A originally contained \(7\,\text{fl oz}\), and Cup B contained \(9\,\text{fl oz}\). b) The cups contained \(16\,\text{fl oz}\) altogether.

- Choose variables for the two bike prices. - Translate each comparison into an equation that includes the helmet price. - Simplify one equation and use substitution. - Check both original comparisons with your results.

1. Let \(x\) be the price of Jordan’s mountain bike and \(y\) the price of Leo’s mountain bike, in dollars. 2. The first condition gives \(x + 120 = 3(y + 40)\). 3. The second condition gives \(y + 120 = \frac{1}{2}(x + 40)\). 4. Simplify the first equation: \(x + 120 = 3y + 120\), so \(x = 3y\). 5. Substitute into the second equation: \(y + 120 = \frac{1}{2}(3y + 40)\). 6. Multiply by \(2\): \(2y + 240 = 3y + 40\), so \(y = 200\). 7. Then \(x = 3 \cdot 200 = 600\).

Jordan’s mountain bike costs \(\$600.00\), and Leo’s mountain bike costs \(\$200.00\).

- Write one equation for each comparison. - Use parentheses so each extension is added before the lengths are compared. - Solve one equation for a variable and substitute. - Clear the fraction by multiplying the equation by \(4\).

1. Let \(a\) be the original length of Rod A and \(b\) the original length of Rod B, in centimeters. 2. The first condition gives \(a + 50 = 2(b + 10)\). 3. The second condition gives \(b + 50 = \frac{3}{4}(a + 10)\). 4. Solve the first equation for \(a\): \(a = 2b - 30\). 5. Substitute into the second equation: \(b + 50 = \frac{3}{4}(2b - 20)\). 6. Multiply by \(4\): \(4b + 200 = 3(2b - 20)\). 7. Simplify: \(4b + 200 = 6b - 60\), so \(260 = 2b\) and \(b = 130\). 8. Then \(a = 2 \cdot 130 - 30 = 230\).

Rod A is \(230\,\text{cm}\) long, and Rod B is \(130\,\text{cm}\) long.

- Choose a variable for each person’s original number of cards. - Track how both card counts change in each transfer. - Write one equation for each statement. - Translate “twice as many” using multiplication. - Solve the system using an efficient method.

1. Let \(x\) be Lucas’s original number of cards and \(y\) Marie’s original number. 2. If Marie gives Lucas \(10\) cards, \(x + 10 = 2(y - 10)\). 3. If Lucas gives Marie \(10\) cards, \(x - 10 = y + 10\). 4. Solve the second equation for \(x\): \(x = y + 20\). 5. Substitute into the first equation: \((y + 20) + 10 = 2(y - 10)\). 6. Simplify: \(y + 30 = 2y - 20\), so \(y = 50\). 7. Then \(x = 50 + 20 = 70\).

Lucas originally had \(70\) cards, and Marie originally had \(50\) cards.

- Express the touring-bike price using the basic-bike price. - Write two expressions for the racing-bike price. - Set those expressions equal and solve.

1. Let \(b\) be the price of the basic bike and \(r\) the price of the racing bike, in dollars. The touring bike costs \(4b\). 2. The two comparisons for the racing bike give the system \(r = 4b + 1200\) and \(r = b + 4800\). 3. Set the two expressions for \(r\) equal: \(4b + 1200 = b + 4800\). 4. Solve: \(3b = 3600\), so \(b = 1200\). 5. The touring bike costs \(4 \cdot 1200 = 4800\). 6. The racing bike costs \(1200 + 4800 = 6000\).

The basic bike costs \(\$1200.00\), the touring bike costs \(\$4800.00\), and the racing bike costs \(\$6000.00\).

- Calculate the boat’s ground speed in each direction. - Upstream speed is the still-water speed minus the current speed. - Downstream speed is the still-water speed plus the current speed. - A raft with no motor travels at the speed of the current.

1. The upstream ground speed is \(\frac{24}{2} = 12\,\text{mph}\), and the downstream ground speed is \(\frac{24}{1.5} = 16\,\text{mph}\). 2. Let \(v_b\) be the boat’s speed in still water and \(v_c\) the speed of the current. Then \(v_b - v_c = 12\) and \(v_b + v_c = 16\). 3. Add the equations: \(2v_b = 28\), so \(v_b = 14\). 4. Substitute into \(v_b + v_c = 16\): \(14 + v_c = 16\), so \(v_c = 2\). 5. A raft with no motor moves at the current’s speed. Its travel time is \(\frac{24}{2} = 12\) hours.

a) The boat’s speed in still water is \(14\,\text{mph}\), and the current’s speed is \(2\,\text{mph}\). b) The raft would take \(12\) hours.

- Write one equation for the total number of habitats. - Write another equation for the total daily cost. - Solve one equation for a variable and substitute. - Check both the habitat count and the cost.

1. Let \(b\) be the number of bird habitats and \(m\) the number of small-mammal habitats. 2. The total number of habitats gives \(b + m = 42\). 3. The daily costs give \(8b + 12m = 400\). 4. Solve the first equation for \(b\): \(b = 42 - m\). 5. Substitute: \(8(42 - m) + 12m = 400\). 6. Simplify: \(336 - 8m + 12m = 400\), so \(4m = 64\) and \(m = 16\). 7. Then \(b = 42 - 16 = 26\).

There are \(26\) bird habitats and \(16\) small-mammal habitats.

- Write one equation for the total number of items. - Write another equation for the total revenue. - Express one variable in terms of the other and substitute. - Check that both the item count and total revenue match.

1. Let \(x\) be the number of sandwiches and \(y\) the number of fruit cups. 2. The total number of items gives \(x + y = 40\). 3. The total revenue gives \(2.50x + 1.20y = 74\). 4. Solve the first equation for \(y\): \(y = 40 - x\). 5. Substitute: \(2.50x + 1.20(40 - x) = 74\). 6. Simplify: \(2.50x + 48 - 1.20x = 74\), so \(1.30x = 26\) and \(x = 20\). 7. Then \(y = 40 - 20 = 20\).

The stand sold \(20\) sandwiches and \(20\) fruit cups.

- Choose variables for the two individual prices. - Write one cost equation for each purchase. - Multiply an equation so that one variable can be eliminated. - Check both purchase totals with your answers.

1. Let \(x\) be the price of one muffin and \(y\) the price of one bagel, in dollars. 2. The first purchase gives \(4x + 3y = 6.50\). 3. The second purchase gives \(2x + 5y = 7.10\). 4. Multiply the second equation by \(-2\): \(-4x - 10y = -14.20\). 5. Add this equation to the first: \(-7y = -7.70\), so \(y = 1.10\). 6. Substitute into \(2x + 5y = 7.10\): \(2x + 5 \cdot 1.10 = 7.10\). 7. Then \(2x = 1.60\), so \(x = 0.80\).

One muffin costs \(\$0.80\), and one bagel costs \(\$1.10\).

- Write an equation for the mass of copper in each mixture. - Notice that each completed mixture has a mass of \(1000\,\text{g}\). - Simplify the equations before using elimination. - Convert the decimal results to percentages.

1. Let \(x\) be the copper concentration of Alloy A and \(y\) the copper concentration of Alloy B, written as decimals. 2. The first mixture gives \(400x + 600y = 1000 \cdot 0.52 = 520\). 3. The second mixture gives \(700x + 300y = 1000 \cdot 0.61 = 610\). 4. Divide both equations by \(100\): \(4x + 6y = 5.2\) and \(7x + 3y = 6.1\). 5. Multiply the second equation by \(2\): \(14x + 6y = 12.2\). 6. Subtract the first equation: \(10x = 7\), so \(x = 0.70\). 7. Substitute into \(7x + 3y = 6.1\): \(7 \cdot 0.70 + 3y = 6.1\), so \(3y = 1.2\) and \(y = 0.40\). 8. Convert the decimal concentrations to percentages.

Alloy A is \(70\%\) copper, and Alloy B is \(40\%\) copper.

- Use one equation for the total number of bottles and one for the total volume. - Include each bottle size in the volume equation. - For part b, compare the amount of juice held by a small bottle with the amount held by a large bottle.

1. Let \(x\) be the number of \(16\)-fluid-ounce bottles and \(y\) the number of \(24\)-fluid-ounce bottles. 2. Write the system \(x + y = 200\) and \(16x + 24y = 3840\). 3. Solve the first equation for \(x\): \(x = 200 - y\). 4. Substitute into the volume equation: \(16(200 - y) + 24y = 3840\). 5. Simplify: \(3200 - 16y + 24y = 3840\), so \(8y = 640\) and \(y = 80\). 6. Then \(x = 200 - 80 = 120\). 7. For part b, the larger bottles hold more juice per bottle. Increasing their proportion means fewer bottles are needed for the same total volume.

a) The company fills \(120\) bottles of the \(16\)-fluid-ounce size and \(80\) bottles of the \(24\)-fluid-ounce size. b) The total number of bottles would decrease because more juice would be packaged in each bottle on average.

- Choose variables for the energy in \(100\,\text{g}\) of each nut. - Count the number of \(100\)-gram portions in each mixture. - Write one energy equation for each mixture. - Use substitution or elimination to solve the system.

1. Let \(x\) be the energy in \(100\,\text{g}\) of walnuts and \(y\) the energy in \(100\,\text{g}\) of cashews, in kilocalories. 2. Mix A gives \(x + 2y = 1850\). 3. Mix B gives \(3x + y = 2550\). 4. Solve the second equation for \(y\): \(y = 2550 - 3x\). 5. Substitute into the first equation: \(x + 2(2550 - 3x) = 1850\). 6. Simplify: \(x + 5100 - 6x = 1850\), so \(-5x = -3250\) and \(x = 650\). 7. Then \(y = 2550 - 3 \cdot 650 = 600\).

Walnuts provide \(650\,\text{kcal}\) per \(100\,\text{g}\), and cashews provide \(600\,\text{kcal}\) per \(100\,\text{g}\).

- How can you read the slope and \(y\)-intercept of each line from a graph? - What does the intersection of two lines represent for the solution set of a system? - What does it mean for the solution set when two distinct lines are parallel? - How can you check an equation read from a graph by substituting points on the line?

1. For graph a), line \(g_1\) crosses the \(y\)-axis at \((0, -3)\) and has slope \(2\), so its equation is \(y = 2x - 3\). Line \(g_2\) crosses the \(y\)-axis at \((0, 0)\) and has slope \(-1\), so its equation is \(y = -x\). 2. The lines in graph a) intersect at \((1, -1)\), so the solution set is \(\{(1, -1)\}\). 3. For graph b), line \(h_1\) crosses the \(y\)-axis at \((0, 2)\) and has slope \(-0.5\), so its equation is \(y = -0.5x + 2\). Line \(h_2\) crosses the \(y\)-axis at \((0, -1)\) and has slope \(-0.5\), so its equation is \(y = -0.5x - 1\). 4. The lines in graph b) have the same slope but different \(y\)-intercepts, so they are parallel and have no point in common. The solution set is \(\emptyset\).

a) (I) \(y = 2x - 3\) (II) \(y = -x\) Solution set: \(\{(1, -1)\}\) b) (I) \(y = -0.5x + 2\) (II) \(y = -0.5x - 1\) Solution set: \(\emptyset\)

- What does the intersection of two lines represent for the corresponding system of equations? - Pay attention to the scale on each axis. Is the intersection exactly at a grid point? - Which algebraic methods can you use to solve a system of linear equations? - It may help to rewrite both equations in the form \(y = mx + b\). - How can you check your calculated solution?

1. From the graph, the intersection appears to be near \((1.5, 1.5)\). An exact reading is difficult because the intersection is not at a grid point. 2. Solve both equations for \(y\): equation (I) becomes \(y = -x + 3\), and equation (II) becomes \(y = 1.5x - 1\). 3. Set the expressions equal: \(-x + 3 = 1.5x - 1\). 4. Solve for \(x\): \(4 = 2.5x\), so \(x = 1.6\). 5. Substitute into equation (I): \(y = -1.6 + 3 = 1.4\). 6. The exact intersection is \((1.6, 1.4)\), which is close to the graph-based estimate.

a) A reasonable estimate is about \((1.5, 1.5)\). The intersection is not at a grid point, so the exact coordinates are difficult to read. b) The exact intersection is \((1.6, 1.4)\). c) The algebraic result is close to the estimate from the graph.

- Identify each rider’s starting distance and rate of change. - Use the form \(s = mt + b\) for each equation. - At the intersection, both riders have the same time and distance. - Set the two distance expressions equal to calculate the intersection.

1. Jordan starts at \(0\) miles and travels \(15\) miles each hour, so \(s_J(t) = 15t\). Maya starts \(10\) miles from Jordan’s home and travels \(10\) miles each hour, so \(s_M(t) = 10t + 10\). 2. The graph’s lines intersect at \((2, 30)\). This means that after \(2\) hours, both riders are \(30\) miles from Jordan’s home, so Jordan catches Maya there. 3. Set the distance expressions equal: \(15t = 10t + 10\). Subtract \(10t\) to get \(5t = 10\), so \(t = 2\). Substitute into either equation: \(s = 15 \cdot 2 = 30\).

1. Jordan: \(s_J(t) = 15t\) Maya: \(s_M(t) = 10t + 10\) 2. The intersection is \((2, 30)\). After \(2\) hours, Jordan catches Maya \(30\,\text{mi}\) from Jordan’s home. 3. Solving \(15t = 10t + 10\) gives \(t = 2\), and substitution gives \(s = 30\).

- Read each line's y-intercept and use two grid points to find its slope. - At an intersection point, what must be true about the two function values? - Solve the equation formed by setting the two expressions equal. - To test a point, substitute its x-coordinate and compare the result with its y-coordinate.

1. Line \(f\) crosses the y-axis at \((0, 1)\). Using \((0, 1)\) and \((2, 2)\), its slope is \(\frac{1}{2}\). Thus \(f(x) = \frac{1}{2}x + 1\). 2. Line \(g\) crosses the y-axis at \((0, 4)\). Using \((0, 4)\) and \((1, 3)\), its slope is \(-1\). Thus \(g(x) = -x + 4\). 3. At the intersection, \(\frac{1}{2}x + 1 = -x + 4\). Solving gives \(\frac{3}{2}x = 3\), so \(x = 2\). Then \(y = 2\), so \(S(2, 2)\). 4. For \(P(8, 5)\), \(f(8) = \frac{1}{2} \cdot 8 + 1 = 5\), so \(P\) lies on \(f\). 5. For \(Q(-2, 5)\), \(g(-2) = -(-2) + 4 = 6\), so \(Q\) does not lie on \(g\).

a) \(f(x) = \frac{1}{2}x + 1\) and \(g(x) = -x + 4\) b) \(S(2, 2)\) c) \(P(8, 5)\) lies on \(f\); \(Q(-2, 5)\) does not lie on \(g\).

- Represent the decrease and increase as slopes. - Identify the starting amount for each linear equation. - At the intersection of the two graphs, the functions have the same value.

1. Write a linear equation for each tank’s amount after \(t\) hours: \(A(t) = -50t + 500\) and \(B(t) = 150t + 100\). 2. Set the expressions equal: \(-50t + 500 = 150t + 100\). 3. Solve: \(400 = 200t\), so \(t = 2\). 4. Substitute \(t = 2\): \(B(2) = 150 \cdot 2 + 100 = 400\). 5. Both tanks contain \(400\,\text{gal}\) after \(2\) hours.

After \(2\) hours, both tanks contain \(400\,\text{gal}\) of water.

- Use where each graph crosses the vertical axis to find its \(y\)-intercept. - Use the rise and run between two clear points to determine the coefficient of \(x\). - To calculate the intersection, set the two function expressions equal.

1. The graph of \(f\) crosses the \(y\)-axis at \(1\). Using another point such as \((1, 2)\), the slope is \(\frac{2 - 1}{1 - 0} = 1\). Therefore, \(f(x) = x + 1\). 2. The graph of \(g\) crosses the \(y\)-axis at \(5\). Using the point \((2, 1)\), the slope is \(\frac{1 - 5}{2 - 0} = -2\). Therefore, \(g(x) = -2x + 5\). 3. Set the function expressions equal: \(x + 1 = -2x + 5\). 4. Solve for \(x\): \(3x = 4\), so \(x = \frac{4}{3}\). 5. Substitute to find \(y\): \(y = \frac{4}{3} + 1 = \frac{7}{3}\). 6. The exact intersection is \((\frac{4}{3}, \frac{7}{3})\), or approximately \((1.33, 2.33)\).

a) \(f(x) = x + 1\) and \(g(x) = -2x + 5\) b) The intersection is \((\frac{4}{3}, \frac{7}{3})\), or approximately \((1.33, 2.33)\).

- The y-intercept represents a setup fee. - The intersection represents equal cost for the same quantity. - For part c), find which graph lies lower after the intersection. - Evaluate both costs at \(x=20\).

1. Graph 1 has a y-intercept of \(60\), so it represents Vendor A. Graph 2 passes through the origin, so it represents Vendor B. 2. The graphs intersect at \((10, 100)\). Both vendors charge \(\$100\) for \(10\) shirts. 3. For \(x>10\), Vendor A's graph is lower. Therefore, Vendor A is less expensive starting at \(11\) shirts. 4. For \(20\) shirts, Vendor B charges \(10\cdot20=\$200\). Vendor A charges \(4\cdot20+60=\$140\). The savings are \(\$200-\$140=\$60\).

a) Graph 1: Vendor A; Graph 2: Vendor B b) \((10, 100)\); both charge \(\$100\) for \(10\) shirts c) \(11\) shirts d) \(\$60\)

- The y-intercept is the initial volume. - The slope is the inflow rate. - Use the displayed time interval for each domain. - Set the equations equal to find when the volumes match.

1. Tank 1 starts with \(25\) gallons and reaches \(50\) gallons after \(5\) minutes. Its slope is \(m=\frac{50-25}{5}=5\), so \(V_1(x)=5x+25\), with \(0\le x\le8\). Tank 2 starts at \(0\) and reaches \(50\) gallons after \(5\) minutes. Its slope is \(10\), so \(V_2(x)=10x\), with \(0\le x\le8\). 2. Set the equations equal: \(10x=5x+25\). Then \(5x=25\), so \(x=5\). Substitution gives \(y=50\) gallons.

a) \(V_1(x)=5x+25\), with \(0\le x\le8\); \(V_2(x)=10x\), with \(0\le x\le8\) b) After \(5\) minutes; \(50\) gallons

- Rewrite each equation in the form \(y = mx + b\). - Compare the slopes and y-intercepts with the diagrams. - Check the intersection in both original equations.

1. Rewrite each equation in slope-intercept form. Equation (I) becomes \(y = -x + 4\), with slope \(-1\) and y-intercept \(4\). Equation (II) becomes \(y = x - 2\), with slope \(1\) and y-intercept \(-2\). 2. Diagram a shows these two lines. In diagram b, line II has y-intercept \(2\), so it does not represent equation (II). 3. In diagram a, the lines intersect at \((3, 1)\). 4. Check: \(3 + 1 = 4\) and \(3 - 1 = 2\).

Diagram a is correct. The intersection is \((3, 1)\).

- Express the mother’s current age in terms of the daughter’s current age. - Write a second equation for their ages 6 years from now. - Both ages increase by the same amount. - Substitute one age relationship into the other equation.

1. Let \(L\) be Leah’s current age and \(M\) be Ms. Miller’s current age. 2. The current-age relationship is \(M = 4L\). 3. The relationship 6 years from now is \(M + 6 = 3(L + 6)\). 4. Substitute \(M = 4L\) into the second equation: \(4L + 6 = 3(L + 6)\). 5. Distribute: \(4L + 6 = 3L + 18\). 6. Subtract \(3L\) and then subtract \(6\): \(L = 12\). 7. Find Ms. Miller’s age: \(M = 4 \cdot 12 = 48\).

Leah is 12 years old, and Ms. Miller is 48 years old.

- How do you find a point shared by two function graphs? - Which points on the x-axis form the base of the triangle? - Which coordinate of the intersection gives the triangle's height above the x-axis?

1. Find the intersection by setting the functions equal: \(2x - 4 = -x + 5\). Then \(3x = 9\), so \(x = 3\). Substituting gives \(y = 2\), so \(S = (3, 2)\). 2. For \(f\), solve \(2x - 4 = 0\), giving \(x = 2\). For \(g\), solve \(-x + 5 = 0\), giving \(x = 5\). 3. The base of the triangle lies on the x-axis from \(x = 2\) to \(x = 5\), so its length is \(5 - 2 = 3\). 4. The height is the y-coordinate of \(S\), which is \(2\). 5. The area is \(A = \frac{1}{2} \cdot 3 \cdot 2 = 3\) square units.

a) \(S = (3, 2)\) b) The zero of \(f\) is \(x = 2\), and the zero of \(g\) is \(x = 5\). c) \(3\) square units

- Find each vertex by solving a pair of line equations. - Because one side is horizontal, how can you find its length and the corresponding height? - Use the triangle area formula once you know the base and height.

1. Find the intersection of \(g_1\) and \(g_2\): \(\frac{1}{2}x + 2 = -x + 5\). Then \(1.5x = 3\), so \(x = 2\), and \(y = 3\). Thus \(A = (2, 3)\). 2. Find the intersection of \(g_1\) and \(g_3\): \(-1 = \frac{1}{2}x + 2\). Then \(-3 = \frac{1}{2}x\), so \(x = -6\). Thus \(B = (-6, -1)\). 3. Find the intersection of \(g_2\) and \(g_3\): \(-1 = -x + 5\), so \(x = 6\). Thus \(C = (6, -1)\). 4. Segment \(BC\) is horizontal, so its length is \(6 - (-6) = 12\). The vertical height from \(A\) to \(y = -1\) is \(3 - (-1) = 4\). 5. The area is \(\frac{1}{2} \cdot 12 \cdot 4 = 24\) square units.

a) \(A = (2, 3)\), \(B = (-6, -1)\), \(C = (6, -1)\) b) \(24\) square units

- How do you find a line's slope from two points? - How can you write a line equation from a point and a slope? - Once you have both equations, how do you find their common point?

1. For \(h_1\), the slope is \(m_1 = \frac{9 - 5}{4 - 2} = 2\). Using \(A(2, 5)\), \(5 = 2 \cdot 2 + b_1\), so \(b_1 = 1\). Thus \(h_1(x) = 2x + 1\). 2. For \(h_2\), use \(C(6, 1)\): \(1 = -0.5 \cdot 6 + b_2\), so \(b_2 = 4\). Thus \(h_2(x) = -0.5x + 4\). 3. Set the equations equal: \(2x + 1 = -0.5x + 4\). 4. Then \(2.5x = 3\), so \(x = 1.2\). 5. Substitute into \(h_1\): \(y = 2 \cdot 1.2 + 1 = 3.4\). 6. Therefore, the intersection point is \((1.2, 3.4)\).

The intersection point is \((1.2, 3.4)\).

- Compare the equations of the diagonals in the two configurations. - How does the slope of diagonal \(AC\) change as \(C\) moves farther right? - What equation describes every point on diagonal \(BD\)?

1. With \(C = (6, 6)\), diagonal \(AC\) has equation \(y = x\). Diagonal \(BD\) passes through \((6, 0)\) and \((0, 6)\), so its equation is \(y = -x + 6\). Setting them equal gives \(x = -x + 6\), so \(x = 3\) and \(y = 3\). Thus \(S_1 = (3, 3)\). 2. With \(C' = (12, 6)\), diagonal \(AC'\) has slope \(\frac{6}{12} = 0.5\), so its equation is \(y = 0.5x\). Setting this equal to \(y = -x + 6\) gives \(0.5x = -x + 6\), so \(1.5x = 6\) and \(x = 4\). Then \(y = 2\), so \(S_2 = (4, 2)\). 3. Every intersection lies on segment \(BD\), where \(y = -x + 6\), or equivalently \(x = 6 - y\). The diagonal from \(A\) to any point to the right on \(y = 6\) meets \(BD\) above the x-axis, so \(y > 0\). Therefore \(x = 6 - y < 6\).

a) \(S_1 = (3, 3)\) b) \(S_2 = (4, 2)\) c) The intersection remains above the x-axis on \(BD\). Since \(x = 6 - y\) there and \(y > 0\), its x-coordinate always stays below \(6\).

- Write one speed equation for the tailwind and one for the headwind. - For part b, calculate the time for each \(20\)-mile leg separately. - Compare that sum with the time for \(40\) miles at the still-air speed. - Remember that time equals distance divided by speed.

1. Let \(v_c\) be the cyclist’s speed with no wind and \(v_w\) the wind speed, in miles per hour. 2. The two ground speeds give \(v_c + v_w = 32\) and \(v_c - v_w = 24\). 3. Add the equations: \(2v_c = 56\), so \(v_c = 28\). Then \(v_w = 32 - 28 = 4\). 4. With wind, the outbound time is \(\frac{20}{32} = \frac{5}{8}\) hour, or \(37.5\) minutes. 5. The return time is \(\frac{20}{24} = \frac{5}{6}\) hour, or \(50\) minutes. 6. The total time with wind is \(\frac{5}{8} + \frac{5}{6} = \frac{35}{24}\) hours, or \(87.5\) minutes. 7. With no wind, the \(40\)-mile round trip would take \(\frac{40}{28} = \frac{10}{7}\) hours, or approximately \(85.7\) minutes. 8. Because \(87.5 > 85.7\), the trip with wind takes longer, so the claim is false.

a) The cyclist’s speed with no wind is \(28\,\text{mph}\), and the wind speed is \(4\,\text{mph}\). b) The claim is false. The trip with wind takes \(87.5\) minutes, while the trip with no wind would take approximately \(85.7\) minutes.

- Write the area of the original rectangle. - Express the dimensions after each change. - Compare each new area with the original area. - The \(lw\) terms cancel after each equation is expanded.

1. The original area is \(lw\). 2. The first condition gives \((l + 2)(w - 1) = lw\). 3. Expand and simplify: \(lw - l + 2w - 2 = lw\), so \(-l + 2w = 2\). 4. The second condition gives \((l - 1)(w + 2) = lw + 12\). 5. Expand and simplify: \(lw + 2l - w - 2 = lw + 12\), so \(2l - w = 14\). 6. From \(-l + 2w = 2\), write \(l = 2w - 2\). 7. Substitute into \(2l - w = 14\): \(2(2w - 2) - w = 14\). 8. Simplify: \(3w - 4 = 14\), so \(w = 6\). Then \(l = 2 \cdot 6 - 2 = 10\).

The original rectangle is \(10\,\text{cm}\) long and \(6\,\text{cm}\) wide.

- Use the equal-side property of a square. - Write an area equation for the original rectangle and the adjusted square. - Express one original dimension in terms of the other. - Substitute that relationship into the area equation.

1. Let \(L\) be the original length and \(W\) the original width, in centimeters. 2. Because the adjusted figure is a square, \(L - 3 = W + 2\), so \(L = W + 5\). 3. Equal areas give \(LW = (L - 3)(W + 2)\). 4. Substitute \(L = W + 5\): \((W + 5)W = (W + 2)(W + 2)\). 5. Expand: \(W^2 + 5W = W^2 + 4W + 4\). 6. Simplify: \(W = 4\). 7. Then \(L = 4 + 5 = 9\).

The original rectangle is \(9\,\text{cm}\) long and \(4\,\text{cm}\) wide.

- A two-digit number with tens digit \(t\) and ones digit \(u\) has value \(10t + u\). - Write the ones digit in terms of the tens digit. - The sum of the digits is \(t + u\). - Substitute one digit relationship into the other equation.

1. Let \(t\) be the tens digit and \(u\) be the ones digit. The number is \(10t + u\). 2. The digit relationship is \(u = t + 3\). 3. The sum of the digits is \(t + u\), so the second equation is \((10t + u) + (t + u) = 45\). 4. Substitute \(u = t + 3\): \((10t + t + 3) + (t + t + 3) = 45\). 5. Combine like terms: \(13t + 6 = 45\). 6. Subtract \(6\): \(13t = 39\). Divide by \(13\): \(t = 3\). 7. Find the ones digit: \(u = 3 + 3 = 6\). 8. The number is \(36\).

The two-digit number is \(36\).

- Represent the original and reversed numbers using their tens and ones digits. - Use the digit sum to express one digit in terms of the other. - Translate “\(36\) less than the original” into an equation comparing the two numbers. - Solve the resulting system by substitution.

1. Let \(t\) be the tens digit and \(u\) be the ones digit. The original number is \(10t + u\). 2. The digit-sum equation is \(t + u = 10\), so \(u = 10 - t\). 3. The reversed number is \(10u + t\). 4. Write the difference equation: \((10t + u) - 36 = 10u + t\). 5. Simplify: \(9t - 9u = 36\), so \(t - u = 4\). 6. Substitute \(u = 10 - t\): \(t - (10 - t) = 4\). 7. Solve: \(2t - 10 = 4\), so \(2t = 14\) and \(t = 7\). 8. Find the ones digit: \(u = 10 - 7 = 3\). 9. The original number is \(73\).

The original number is \(73\).

- Write the number as \(10t + u\) and its digit sum as \(t + u\). - Translate “\(6\) less than seven times the digit sum” carefully. - Use the relationship between the two digits to substitute for one variable. - Check both conditions after finding the number.

1. Let \(t\) be the tens digit and \(u\) be the ones digit. The number is \(10t + u\). 2. The digit relationship is \(t = u + 2\). 3. The value relationship is \(10t + u = 7(t + u) - 6\). 4. Substitute \(t = u + 2\): \(10(u + 2) + u = 7(u + 2 + u) - 6\). 5. Distribute and combine like terms: \(11u + 20 = 14u + 8\). 6. Subtract \(11u\) and then subtract \(8\): \(12 = 3u\), so \(u = 4\). 7. Find the tens digit: \(t = 4 + 2 = 6\). The number is \(64\). 8. Check: the digit sum is \(6 + 4 = 10\), and \(7 \cdot 10 - 6 = 64\). Also, \(6 = 4 + 2\).

The number is \(64\).

- Choose one train’s speed and time as the two unknowns. - Express the other trains’ speeds and times relative to those unknowns. - The product of speed and time is the same route length for all three trains. - Expand the two distance equations; the product of the unknowns cancels.

1. Let \(v\) be the freight train’s speed in miles per hour and \(t\) its travel time in hours. 2. The passenger train’s speed and time are \(v + 20\) and \(t - 3\). The express train’s speed and time are \(v + 50\) and \(t - 5\). 3. Because all three trains travel the same distance, \(vt = (v + 20)(t - 3)\) and \(vt = (v + 50)(t - 5)\). 4. Expand the first equation and simplify: \(20t - 3v = 60\). 5. Expand the second equation and simplify: \(50t - 5v = 250\), or \(10t - v = 50\). 6. From \(10t - v = 50\), write \(v = 10t - 50\). 7. Substitute into \(20t - 3v = 60\): \(20t - 3(10t - 50) = 60\). Then \(-10t + 150 = 60\), so \(t = 9\). 8. The freight train’s speed is \(v = 10 \cdot 9 - 50 = 40\,\text{mph}\). The passenger train travels at \(60\,\text{mph}\), and the express train travels at \(90\,\text{mph}\). 9. The route length is \(40 \cdot 9 = 360\,\text{mi}\).

The freight train travels at \(40\,\text{mph}\), the passenger train at \(60\,\text{mph}\), and the express train at \(90\,\text{mph}\). The route is \(360\,\text{mi}\) long.

- Express the same route length using speed times time in each scenario. - Convert the time changes from minutes to hours. - Expand the equations; the product \(vt\) cancels. - Solve the resulting linear system for the usual speed and time.

1. Let \(v\) be the usual speed in miles per hour and \(t\) the usual travel time in hours. The route length is \(vt\). 2. Convert the time changes: \(4\) minutes is \(\frac{1}{15}\) hour, and \(6\) minutes is \(\frac{1}{10}\) hour. 3. The low-traffic scenario gives \(vt = (v + 10)\left(t - \frac{1}{15}\right)\). 4. Expand and simplify: \(150t - v = 10\). 5. The heavy-rain scenario gives \(vt = (v - 10)\left(t + \frac{1}{10}\right)\). 6. Expand and simplify: \(v - 100t = 10\). 7. Add the equations: \(50t = 20\), so \(t = 0.4\) hour. 8. Substitute into \(v - 100t = 10\): \(v - 40 = 10\), so \(v = 50\). 9. The route length is \(vt = 50 \cdot 0.4 = 20\) miles.

The shuttle’s usual average speed is \(50\,\text{mph}\), and the route is \(20\,\text{mi}\) long.

- Write the area of the original rectangle. - Make one area equation for each proposed change. - Expand and simplify; the product \(lw\) cancels in both equations. - Solve the resulting linear system.

1. Let \(l\) be the original length and \(w\) the original width, in feet. The original area is \(lw\). 2. The first condition gives \((l + 2)(w - 1) = lw\). 3. Expand and simplify: \(lw - l + 2w - 2 = lw\), so \(-l + 2w = 2\). 4. The second condition gives \((l - 3)(w + 3) = lw + 6\). 5. Expand and simplify: \(lw + 3l - 3w - 9 = lw + 6\), so \(l - w = 5\). 6. From \(l - w = 5\), write \(l = w + 5\). 7. Substitute into \(-l + 2w = 2\): \(-(w + 5) + 2w = 2\), so \(w = 7\). 8. Then \(l = 7 + 5 = 12\).

The garden bed was planned to be \(12\,\text{ft}\) long and \(7\,\text{ft}\) wide.

- Convert each copper-to-zinc ratio into a copper percentage. - Determine the total mass of copper required in the final alloy. - Write one equation for total mass and one for copper mass. - Solve the resulting system.

1. Alloy A is \(\frac{4}{4 + 1} = 0.80\), or \(80\%\), copper. Alloy B is \(\frac{1}{1 + 3} = 0.25\), or \(25\%\), copper. 2. Let \(x\) be the pounds of Alloy A and \(y\) the pounds of Alloy B. 3. The total mass gives \(x + y = 50\). 4. The copper mass gives \(0.80x + 0.25y = 0.47 \cdot 50\), or \(0.8x + 0.25y = 23.5\). 5. Solve the first equation for \(y\): \(y = 50 - x\). 6. Substitute: \(0.8x + 0.25(50 - x) = 23.5\). 7. Simplify: \(0.8x + 12.5 - 0.25x = 23.5\), so \(0.55x = 11\) and \(x = 20\). 8. Then \(y = 50 - 20 = 30\).

The foundry needs \(20\,\text{lb}\) of Alloy A and \(30\,\text{lb}\) of Alloy B.

- Determine the total mass of salt needed in the final solution. - Account for the \(1\,\text{kg}\) of water when writing the total-mass equation. - Decide how much salt the pure water contributes. - Write one equation for total mass and one for salt mass.

1. Let \(x\) be the mass of Solution A and \(y\) the mass of Solution B, in kilograms. 2. The final mixture has a mass of \(10\,\text{kg}\), including the \(1\,\text{kg}\) of water, so \(x + y + 1 = 10\), or \(x + y = 9\). 3. The final mixture must contain \(0.12 \cdot 10 = 1.2\,\text{kg}\) of salt. Pure water contributes no salt, so \(0.05x + 0.20y = 1.2\). 4. From \(x + y = 9\), write \(x = 9 - y\). 5. Substitute: \(0.05(9 - y) + 0.20y = 1.2\). 6. Simplify: \(0.45 - 0.05y + 0.20y = 1.2\), so \(0.15y = 0.75\) and \(y = 5\). 7. Then \(x = 9 - 5 = 4\).

Add \(4\,\text{kg}\) of Solution A and \(5\,\text{kg}\) of Solution B.

- Determine what happens to the total mass when the same amount is removed from one alloy and added to the other. - Use the original mixture to find a relationship between the two masses. - Write a silver-mass equation for each scenario.

1. Let \(x\) be the planned mass of Alloy A and \(y\) the planned mass of Alloy B, in grams. 2. The original mixture gives \(0.60x + 0.90y = 0.70(x + y)\). 3. Simplify the first equation: \(0.20y = 0.10x\), so \(x = 2y\). 4. The adjusted mixture has the same total mass, so \(0.60(x - 100) + 0.90(y + 100) = 0.80(x + y)\). 5. Substitute \(x = 2y\): \(0.60(2y - 100) + 0.90(y + 100) = 0.80(3y)\). 6. Simplify: \(1.20y - 60 + 0.90y + 90 = 2.40y\), so \(2.10y + 30 = 2.40y\). 7. Therefore, \(30 = 0.30y\), so \(y = 100\), and \(x = 2 \cdot 100 = 200\).

The original plan used \(200\,\text{g}\) of Alloy A and \(100\,\text{g}\) of Alloy B.

- Write a cost rule for Company A. - Represent Company B’s unknown hourly rate and base fee with variables. - Use the \(4\)-hour and \(8\)-hour information to write two equations. - After finding Company B’s rule, compare the two cost expressions with an inequality.

1. Company A’s cost is \(A(t) = 10t + 40\), where \(t\) is the number of hours. 2. Let Company B’s cost be \(B(t) = mt + b\), where \(m\) is the hourly rate and \(b\) is the base fee. 3. At \(4\) hours, the costs are equal: \(4m + b = 10 \cdot 4 + 40 = 80\). 4. At \(8\) hours, Company B costs \(\$20.00\) more: \(8m + b = 10 \cdot 8 + 40 + 20 = 140\). 5. Subtract the first equation from the second: \(4m = 60\), so \(m = 15\). 6. Substitute into \(4m + b = 80\): \(4 \cdot 15 + b = 80\), so \(b = 20\). 7. Compare the costs: Company A is cheaper when \(10t + 40 < 15t + 20\). 8. Solve the inequality: \(20 < 5t\), so \(t > 4\).

Company B charges a \(\$20.00\) base fee and \(\$15.00\) per hour. Company A is less expensive for rentals longer than \(4\) hours.

- Represent the current capacity as rows times seats per row. - Write an equation for each renovation plan. - Expand carefully and cancel the original capacity term. - Solve the resulting linear system and check both plans.

1. Let \(x\) be the current number of rows and \(y\) the current number of seats per row. 2. Plan A gives \((x + 2)(y + 5) = xy + 200\). 3. Plan B gives \((x - 5)(y - 2) = xy - 150\). 4. Expand the first equation: \(xy + 5x + 2y + 10 = xy + 200\), so \(5x + 2y = 190\). 5. Expand the second equation: \(xy - 2x - 5y + 10 = xy - 150\), so \(2x + 5y = 160\). 6. Multiply \(5x + 2y = 190\) by \(5\): \(25x + 10y = 950\). Multiply \(2x + 5y = 160\) by \(-2\): \(-4x - 10y = -320\). 7. Add the equations: \(21x = 630\), so \(x = 30\). 8. Substitute into \(2x + 5y = 160\): \(2 \cdot 30 + 5y = 160\), so \(y = 20\).

The theater currently has \(30\) rows with \(20\) seats in each row.

- Sketch an isosceles trapezoid and label the equal sides. - Write one equation for the perimeter and one for the side-length relationship. - After finding the side lengths, split the trapezoid into a rectangle and two right triangles. - Check whether the right triangles can have a positive height.

1. Let \(x\) be the common length of the two legs and the shorter base, and let \(y\) be the length of the longer base, in centimeters. 2. The perimeter gives \(y + 3x = 24\). 3. The side-length relationship gives \(y = 3x\). 4. Substitute: \(3x + 3x = 24\), so \(6x = 24\) and \(x = 4\). 5. Then \(y = 3 \cdot 4 = 12\). The proposed side lengths are \(4\,\text{cm}\), \(4\,\text{cm}\), \(4\,\text{cm}\), and \(12\,\text{cm}\). 6. In an isosceles trapezoid with longer base \(a\), shorter base \(c\), leg \(s\), and height \(h\), \(h^2 = s^2 - \left(\frac{a-c}{2}\right)^2\). 7. Substitute the proposed lengths: \(h^2 = 4^2 - \left(\frac{12-4}{2}\right)^2 = 16 - 16 = 0\). 8. Therefore, \(h = 0\). The figure collapses into a line segment, so no nondegenerate trapezoid with positive area exists.

The equations give a longer base of \(12\,\text{cm}\) and three other sides of \(4\,\text{cm}\) each. A nondegenerate trapezoid with these side lengths does not exist because its height would be \(0\,\text{cm}\).