- A solution of both equations must lie on both lines. - Look for the single point where the two graphs cross. - Read the x-coordinate and y-coordinate from the grid at that point.

1. Locate the point where the two lines cross. 2. Read its coordinates from the grid: the x-coordinate is \(2\) and the y-coordinate is \(3\).

\((2, 3)\)

- One equation already fixes one coordinate. - Use that known coordinate in the other equation. - Report the result as an ordered pair.

1. Substitute \(x=3\) into \(y=2x-5\). 2. This gives \(y=1\). 3. Therefore, the solution is \((3,1)\).

\((3, 1)\)

- A system's solution must make both equations true, not just one. - Put each point's x- and y-coordinates into each equation. - A point that passes one test may still fail the other.

1. For \(A(2,3)\), \(3=2+1\) and \(2\cdot2+3=7\), so both equations hold. 2. For \(B(1,2)\), the first equation holds, but \(2\cdot1+2=4\ne7\). 3. For \(C(3,1)\), \(1\ne3+1\), but \(2\cdot3+1=7\). 4. Only \(A\) solves the whole system.

\(A(2,3)\) solves both. \(B(1,2)\) fails \(2x+y=7\). \(C(3,1)\) fails \(y=x+1\).

- Equal quantities can replace one another in an equation. - Keep the rest of (II) unchanged when replacing one term. - Check the final pair in both original equations.

1. Substitute \(y=3\) into equation (II): \(x+3=8\). 2. Solve: \(x=5\). 3. The ordered pair is \((5,3)\), which makes both equations true.

After substitution: \(x+3=8\). Solution: \((5,3)\).

- Think of substitution as replacing one quantity with an equal expression. - Keep the replacement together with parentheses. - Check that the new equation has only one variable.

1. Equation (I) tells us that \(y\) can be replaced by \(2x+1\). 2. The second equation becomes \(3x+(2x+1)=13\). It now has just one variable.

\(3x+(2x+1)=13\).

- Use an equal expression in place of a variable without changing the value. - The \(x\)-term in (II) is twice \(x\), not just \(x\). - Keep the whole replacement expression in parentheses.

1. Equation (I) lets us replace \(x\) with \(2y-3\). 2. Substituting into equation (II) gives \(2(2y-3)+y=14\).

\(2(2y-3)+y=14\).

- What do you notice about equation (I) when you consider using substitution? - Use parentheses when substituting an expression into equation (II). - How can you use the value of the first variable to find the second variable?

1. Equation (I) is already solved for \(y\), so substitute \(4x - 3\) for \(y\) in equation (II): \(2x + 3(4x - 3) = 19\). 2. Distribute and combine like terms: \(2x + 12x - 9 = 19\), so \(14x - 9 = 19\). 3. Solve for \(x\): \(14x = 28\), so \(x = 2\). 4. Substitute \(x = 2\) into equation (I): \(y = 4 \cdot 2 - 3 = 5\). 5. Therefore, the solution set is \(\{(2, 5)\}\).

The solution set is \(\{(2, 5)\}\).

- What is the same on the left side of both equations? - If two expressions are both equal to \(x\), what must be true about those expressions? - After finding one variable, how can you find the other?

1. Setting the two expressions equal is most direct because both equations are already solved for \(x\). 2. Set the expressions equal: \(2y + 5 = -3y + 15\). 3. Combine like terms: \(5y + 5 = 15\). 4. Solve for \(y\): \(5y = 10\), so \(y = 2\). 5. Substitute into either original equation: \(x = 2 \cdot 2 + 5 = 9\). 6. The solution is \((9, 2)\).

Setting the two expressions for \(x\) equal is most direct because both equations are already solved for \(x\). The solution set is \(\{(9, 2)\}\).

- Which equation lets you isolate one variable with a single step? - Substitute the resulting expression into the other equation. - After finding the ordered pair, replace both variables in each original equation to check it.

1. Solve equation (I) for \(y\): \(y=10-4x\). 2. Substitute into equation (II): \(x-2(10-4x)=7\). 3. Distribute and combine like terms: \(x-20+8x=7\), so \(9x=27\) and \(x=3\). 4. Substitute into \(y=10-4x\): \(y=10-12=-2\). The solution is \((3, -2)\). 5. Check equation (I): \(4(3)+(-2)=10\). Check equation (II): \(3-2(-2)=7\). Both equations are satisfied.

The solution is \((3, -2)\), and it satisfies both original equations.

- Is there a variable that is especially easy to isolate? - Can you express one variable in terms of the other? - Check your result by substituting both values into the original equations.

1. Solve equation (II) for \(y\): \(y = 2x\). 2. Substitute \(2x\) for \(y\) in equation (I): \(x + 3(2x) = 7\). 3. Simplify: \(7x = 7\). 4. Solve for \(x\): \(x = 1\). 5. Substitute \(x = 1\) into \(y = 2x\): \(y = 2\). 6. The solution is \((1, 2)\).

The solution is \((1, 2)\).

- Would it help to solve an equation for a variable whose coefficient is \(1\)? - Pay close attention to decimal place values. - After expressing one variable in terms of the other, substitute that expression into the remaining equation.

1. Solve equation (II) for \(a\): \(a = 6.5 - 0.5b\). 2. Substitute into equation (I): \(0.4(6.5 - 0.5b) - 0.2b = 1\). 3. Distribute: \(2.6 - 0.2b - 0.2b = 1\). 4. Combine like terms: \(2.6 - 0.4b = 1\). 5. Solve for \(b\): \(-0.4b = -1.6\), so \(b = 4\). 6. Substitute into equation (II): \(a = 6.5 - 0.5 \cdot 4 = 4.5\). 7. The solution is \((a, b) = (4.5, 4)\).

The solution is \((a, b) = (4.5, 4)\).

- If two expressions are both equal to \(y\), what must be true about those expressions? - Could multiplying by a common denominator help clear the fractions? - What inverse operations isolate \(x\)?

1. Set the expressions for \(y\) equal: \(\frac{2}{5}x + 3 = \frac{4}{5}x - 1\). 2. Subtract \(\frac{2}{5}x\) from both sides: \(3 = \frac{2}{5}x - 1\). 3. Add \(1\): \(4 = \frac{2}{5}x\). 4. Multiply by \(5\): \(20 = 2x\). 5. Divide by \(2\): \(x = 10\).

\(x = 10\)

- Which variable is easiest to isolate in both equations? - If two expressions are both equal to \(y\), what relationship do they have? - Remember to calculate the second variable after finding the first.

1. Solve both equations for \(y\). Equation (I) becomes \(y = 14 - 3x\). Equation (II) becomes \(y = 2x - 1\). 2. Set the expressions equal: \(14 - 3x = 2x - 1\). 3. Solve for \(x\): \(14 = 5x - 1\), so \(15 = 5x\) and \(x = 3\). 4. Substitute into either expression for \(y\): \(y = 2 \cdot 3 - 1 = 5\). 5. The solution set is \(\{(3, 5)\}\).

The solution is \((3, 5)\).

- Can you replace one variable with an expression involving the other variable? - What happens when you substitute the expression for one variable into the other equation? - Pay attention to signs when distributing. - How can you check your result?

1. Substitute the expression for \(y\) from equation (I) into equation (II): \(3x + 2(4x - 5) = 23\). 2. Distribute: \(3x + 8x - 10 = 23\). 3. Combine like terms: \(11x - 10 = 23\). 4. Solve for \(x\): \(11x = 33\), so \(x = 3\). 5. Substitute into equation (I): \(y = 4 \cdot 3 - 5 = 7\).

The solution is \((3, 7)\).

- Which variable in which equation is easiest to isolate? - Put the entire substituted expression in parentheses. - What inverse operation will isolate the chosen variable?

1. Solve equation (I) for \(y\): \(y = 3x - 7\). 2. Substitute into equation (II): \(2x + 4(3x - 7) = 28\). 3. Distribute: \(2x + 12x - 28 = 28\). 4. Combine like terms and solve: \(14x = 56\), so \(x = 4\). 5. Substitute into \(y = 3x - 7\): \(y = 3 \cdot 4 - 7 = 5\).

The solution is \((4, 5)\).

- When two equations are both solved for \(y\), what must be true at their intersection? - Distribute the decimal factor before collecting the \(x\)-terms. - After finding \(x\), substitute it into either original equation to recover \(y\).

1. Set the expressions for \(y\) equal: \(2.4x-7=0.4(x+12.5)\). 2. Distribute on the right: \(0.4(x+12.5)=0.4x+5\). 3. Solve \(2.4x-7=0.4x+5\): \(2x=12\), so \(x=6\). 4. Substitute into either equation: \(y=2.4(6)-7=7.4\). 5. Therefore, the equations have the same value at \((6, 7.4)\).

\((6, 7.4)\)

- One variable is already isolated in equation (I). - How can you use that expression in equation (II)? - Replace the corresponding variable in equation (II) with the expression from equation (I). - Use parentheses around a substituted expression with more than one term.

1. Substitute the expression for \(y\) from equation (I) into equation (II): \(3x + 2(2x - 4) = 27\). 2. Distribute: \(3x + 4x - 8 = 27\). 3. Combine like terms: \(7x - 8 = 27\). 4. Add \(8\) to both sides: \(7x = 35\). 5. Divide by \(7\): \(x = 5\). 6. Substitute into equation (I): \(y = 2 \cdot 5 - 4 = 6\). 7. The solution set is \(\{(5, 6)\}\).

The solution set is \(\{(5, 6)\}\).

- Can you rearrange one equation so that one variable is alone on one side? - Which choice avoids introducing fractions? - After finding one variable, how can you calculate the other?

1. Solve equation (I) for \(x\): \(x = 4y - 18\). 2. Substitute into equation (II): \(2(4y - 18) + 5y = 16\). 3. Distribute: \(8y - 36 + 5y = 16\). 4. Combine like terms: \(13y - 36 = 16\). 5. Add \(36\) to both sides: \(13y = 52\). 6. Divide by \(13\): \(y = 4\). 7. Substitute into \(x = 4y - 18\): \(x = 4 \cdot 4 - 18 = -2\).

The solution is \((-2, 4)\).

- The substitution process is unchanged when the coefficients are decimals. - When multiplying by \(2\), distribute to every term inside the parentheses. - Combine the \(x\)-terms carefully before continuing to solve.

1. Substitute the expression for \(y\) from equation (I) into equation (II): \(5x + 2(2.4x - 3) = 33.2\). 2. Distribute: \(5x + 4.8x - 6 = 33.2\). 3. Combine like terms: \(9.8x - 6 = 33.2\). 4. Add \(6\): \(9.8x = 39.2\). 5. Divide by \(9.8\): \(x = 4\). 6. Substitute into equation (I): \(y = 2.4 \cdot 4 - 3 = 6.6\).

The solution is \((4, 6.6)\).

- What must be true of the right sides when both equations are solved for \(y\)? - How can you combine the two equations into one equation in \(x\)? - After finding \(x\), remember to find \(y\). - Check the ordered pair in both original equations.

1. Set the expressions equal: \(3x - 8 = -2x + 7\). 2. Combine like terms: \(5x = 15\). 3. Divide by \(5\): \(x = 3\). 4. Substitute into equation (I): \(y = 3 \cdot 3 - 8 = 1\). 5. The solution is \((3, 1)\).

The solution is \((3, 1)\).

- What must be true of the right sides when both equations are solved for \(y\)? - Bring all variable terms to one side. - How can you check your result?

1. Set the expressions equal: \(3x + 4.5 = -1.5x + 18\). 2. Add \(1.5x\) and subtract \(4.5\): \(4.5x = 13.5\). 3. Divide by \(4.5\): \(x = 3\). 4. Substitute into either original equation: \(y = 3 \cdot 3 + 4.5 = 13.5\).

The solution is \((3, 13.5)\).

- Can you replace one variable with an expression involving the other variable? - What happens when you substitute the expression for \(x\) into equation (II)? - Distribute the factor to every term inside the parentheses. - How can you check your result?

1. Equation (I) is already solved for \(x\), so substitute \(3y + 2\) for \(x\) in equation (II): \(4(3y + 2) - 5y = 29\). 2. Distribute: \(12y + 8 - 5y = 29\). 3. Combine like terms: \(7y + 8 = 29\). 4. Solve for \(y\): \(7y = 21\), so \(y = 3\). 5. Substitute into equation (I): \(x = 3 \cdot 3 + 2 = 11\). 6. The solution is \((11, 3)\).

The solution is \((11, 3)\).

- Which equation is easiest to solve for one variable? - Pay close attention to the negative factor before the parentheses. - After finding one variable, where can you substitute it most easily to find the other? - Check both values in the original equations.

1. Solve equation (I) for \(y\): \(y = 7 - 2x\). 2. Substitute into equation (II): \(3x - 2(7 - 2x) = 21\). 3. Distribute carefully: \(3x - 14 + 4x = 21\). 4. Combine like terms: \(7x - 14 = 21\). 5. Solve for \(x\): \(7x = 35\), so \(x = 5\). 6. Substitute into \(y = 7 - 2x\): \(y = 7 - 2 \cdot 5 = -3\). 7. The solution set is \(\{(5, -3)\}\).

The solution set is \(\{(5, -3)\}\).

- Which variable is already isolated in one equation? - Can you substitute that expression into the other equation? - Pay close attention to the negative factor before the parentheses. - After finding one variable, how can you determine the other?

1. Substitute the expression for \(y\) from the first equation into the second equation: \(5x - 2(2x + 1) = 1\). 2. Distribute: \(5x - 4x - 2 = 1\). 3. Combine like terms and solve: \(x - 2 = 1\), so \(x = 3\). 4. Substitute into the first equation: \(y = 2 \cdot 3 + 1 = 7\). 5. The solution is \((3, 7)\).

The solution is \((3, 7)\).

- Which equation is already solved for one variable? - Put parentheses around the entire expression when substituting it into the second equation. - Remember to find both variable values. - Check your result in both original equations.

1. Substitute the expression for \(y\) from equation (I) into equation (II): \(3x - 4(0.5x + 2) = 2\). 2. Distribute: \(3x - 2x - 8 = 2\). 3. Combine like terms: \(x - 8 = 2\). 4. Add \(8\) to both sides: \(x = 10\). 5. Substitute into equation (I): \(y = 0.5 \cdot 10 + 2\). 6. Calculate \(y\): \(y = 7\). 7. The solution is \((10, 7)\).

The solution is \((10, 7)\).

- Look for a variable with coefficient \(1\) to make isolating it easier. - Pay close attention to signs when distributing. - After finding one variable, substitute it into an original equation to find the other.

1. Solve equation (II) for \(x\), since its coefficient is \(1\): \(x = 3y + 5\). 2. Substitute into equation (I): \(2(3y + 5) + 5y = -1\). 3. Distribute: \(6y + 10 + 5y = -1\). 4. Combine like terms: \(11y + 10 = -1\). 5. Solve for \(y\): \(11y = -11\), so \(y = -1\). 6. Substitute into \(x = 3y + 5\): \(x = 3(-1) + 5 = 2\). 7. The solution is \((2, -1)\).

The solution is \((2, -1)\).

- Which equation is easiest to solve for one variable? - Use parentheses around the expression you substitute into the other equation. - How can you check your result in the original equations?

1. Solve equation (I) for \(x\): \(x = 18 - 4y\). 2. Substitute into equation (II): \(3(18 - 4y) - 2y = -2\). 3. Solve for \(y\): \(54 - 12y - 2y = -2\), so \(-14y = -56\) and \(y = 4\). 4. Substitute into \(x = 18 - 4y\): \(x = 18 - 4 \cdot 4 = 2\). 5. The solution set is \(\{(2, 4)\}\).

The solution set is \(\{(2, 4)\}\).

- Can you rearrange one equation so that one variable is alone on one side? - What happens when you substitute that expression into the other equation? - Put parentheses around the entire substituted expression. - Check your result in both original equations.

1. Solve equation (II) for \(x\): \(x = 2y - 7\). 2. Substitute into equation (I): \(2(2y - 7) + 3y = 7\). 3. Distribute and combine like terms: \(4y - 14 + 3y = 7\), so \(7y - 14 = 7\). 4. Solve for \(y\): \(7y = 21\), so \(y = 3\). 5. Substitute into \(x = 2y - 7\): \(x = 2 \cdot 3 - 7 = -1\). 6. The solution set is \(\{(-1, 3)\}\).

The solution set is \(\{(-1, 3)\}\).

- Which variable can you isolate without introducing fractions? - Substitute the expression into the other equation. - Put parentheses around the entire substituted expression. - Use the first variable value to calculate the second.

1. Solve equation (I) for \(x\): \(x = 4y + 18\). 2. Substitute into equation (II): \(3(4y + 18) + 2y = 12\). 3. Distribute: \(12y + 54 + 2y = 12\). 4. Combine like terms and solve: \(14y + 54 = 12\), so \(14y = -42\) and \(y = -3\). 5. Substitute into \(x = 4y + 18\): \(x = 4(-3) + 18 = 6\). 6. The solution is \((6, -3)\).

The solution is \((6, -3)\).

- Which variable in which equation is easiest to isolate? - Put the entire substituted expression in parentheses. - After finding one variable, substitute it into an original equation to find the other.

1. Solve the first equation for \(y\): \(y = 3x - 5\). 2. Substitute into the second equation: \(2x + 3(3x - 5) = 29\). 3. Solve for \(x\): \(2x + 9x - 15 = 29\), so \(11x = 44\) and \(x = 4\). 4. Substitute into \(y = 3x - 5\): \(y = 3 \cdot 4 - 5 = 7\). 5. The solution is \((4, 7)\).

The solution is \((4, 7)\).

- Can you rearrange one equation so that one variable is alone on one side? - Substitute that expression into the other equation. - Pay close attention to the negative factor before the parentheses. - Check your solution in both original equations.

1. Solve the first equation for \(y\): \(y = 18 - 4x\). 2. Substitute into the second equation: \(3x - 2(18 - 4x) = 8\). 3. Distribute and combine like terms: \(3x - 36 + 8x = 8\), so \(11x - 36 = 8\). 4. Solve for \(x\): \(11x = 44\), so \(x = 4\). 5. Substitute into the expression for \(y\): \(y = 18 - 4 \cdot 4 = 2\).

The solution is \((4, 2)\).

- Which system-solving method is most direct when one equation is already solved for \(y\)? - Replace \(y\) in the other equation with the given expression. - Pay attention to the negative factor before the parentheses. - How can you use the value of \(x\) to find \(y\)?

1. Substitute the expression for \(y\) from line \(h\) into the equation for line \(g\): \(3x - 2(2x - 5) = 4\). 2. Distribute: \(3x - 4x + 10 = 4\). 3. Combine like terms: \(-x + 10 = 4\). 4. Solve for \(x\): \(-x = -6\), so \(x = 6\). 5. Substitute into the equation for line \(h\): \(y = 2 \cdot 6 - 5 = 7\). 6. The intersection is \((6, 7)\).

The intersection is \((6, 7)\).

- Focus on the small grid interval containing the crossing point. - Estimate each coordinate independently before rounding. - Use the requested precision rather than forcing the point onto a grid intersection.

1. The lines cross a little to the right of \(x=2.5\), so the x-coordinate rounds to \(2.5\). 2. The crossing is a little above \(y=2.25\), so the y-coordinate rounds to \(2.3\). 3. The estimated intersection is \((2.5,2.3)\) to the nearest tenth.

Approximately \((2.5, 2.3)\)

- Rewrite the second equation in slope-intercept form before comparing panels. - Check both slope and y-intercept for each line; a panel can match one feature and still be wrong. - A correct panel must represent both equations, not merely have a plausible-looking crossing point.

1. The first equation is already in slope-intercept form: \(y=-x+4\). Its slope is \(-1\) and its y-intercept is \(4\). 2. Rewrite the second equation as \(y=-2x+6\). Its slope is \(-2\) and its y-intercept is \(6\). 3. Panel a) keeps the intersection \((2,2)\) but gives the first line the wrong slope and intercept. Panel b) keeps the two slopes but shifts both y-intercepts. Panel d) keeps the two y-intercepts but uses the wrong slopes. 4. Only panel c) has both required slopes and both required y-intercepts. 5. In panel c), the two lines intersect at \((2,2)\), so that ordered pair is the solution.

Panel c). The lines must have slopes \(-1\) and \(-2\) with y-intercepts \(4\) and \(6\), respectively. The solution is \((2,2)\).

- Ask which equation supplied the expression for \(y\), and which different equation still needs to be satisfied. - An identity obtained from one equation alone does not classify the whole system. - After replacing \(y\) in the other equation, solve the resulting one-variable equation and then recover the second coordinate.

1. The identity \(x+1=x+1\) appears because Noah substituted an expression from the first equation back into that same equation. It gives no information about whether the second equation is also satisfied. 2. Substitute \(x+1\) for \(y\) in the second equation: \(2x-(x+1)=3\). 3. Simplify: \(x-1=3\), so \(x=4\). 4. Use \(y=x+1\): \(y=5\). The solution is \((4, 5)\).

a) Noah substituted the first equation into itself; the resulting identity does not show that the two equations describe the same line. b) The solution is \((4, 5)\).

- Which variable can you isolate without dividing by a coefficient? - Keep the full expression together when substituting into \(3x\). - Check the resulting pair in both original equations.

1. Equation (I) gives \(x=y+1\). 2. Substitute into equation (II): \(3(y+1)+2y=18\). 3. Simplify to \(5y+3=18\), so \(y=3\). 4. Then \(x=3+1=4\). Both equations hold for \((4,3)\).

Substitution equation: \(3(y+1)+2y=18\). Solution: \((4,3)\).

- First read the coordinates of each marked point from the axes. - What do the changes in x and y tell you about each line's slope? - Once you have the two equations, what must be equal at their crossing?

1. Read \(A=(0,1)\) and \(B=(1,2)\). The slope is \(1\) and the y-intercept is \(1\), so \(g\) is \(y=x+1\). 2. Read \(C=(0,5)\) and \(D=(1,4)\). The slope is \(-1\) and the y-intercept is \(5\), so \(h\) is \(y=-x+5\). 3. At the crossing, \(x+1=-x+5\). Thus \(2x=4\), so \(x=2\) and \(y=3\). 4. The point \((2,3)\) satisfies both line equations.

\(A=(0,1)\), \(B=(1,2)\), \(C=(0,5)\), \(D=(1,4)\). \(g: y=x+1\); \(h: y=-x+5\). The lines would cross at \((2,3)\).

- Try each proposed point in the line's equation. - How can you choose another point while keeping the line vertical? - At the crossing, use the fixed x-coordinate in the second equation.

1. The point \((0,2)\) is wrong because its x-coordinate is \(0\), not \(2\). 2. For example, \((2,4)\) is another point on \(x=2\). All points on this line have x-coordinate \(2\), so it is vertical. 3. At the intersection, \(x=2\). The other equation gives \(y=-2+5=3\). 4. The intersection is \((2,3)\); it satisfies both equations.

\((0,2)\) is wrong: its x-coordinate is not \(2\). Another point on the vertical line is \((2,4)\). Intersection: \((2,3)\).

- Read two points on \(g\) and compare their rise and run. - Substitute the requested x-values in the second equation. - Check the shared point in the new line's equation.

1. The graph of \(g\) passes through \((0,-1)\) and \((2,1)\), so its equation is \(y=x-1\). 2. Points \((0,5)\) and \((3,-1)\) lie on the new line \(y=-2x+5\). 3. Both lines meet at \((2,1)\). 4. The check \(1=-2\cdot2+5\) is true.

Slope of \(g\): \(1\). Points for the new line: \((0,5)\) and \((3,-1)\). Intersection: \((2,1)\). Check: \(1=-2\cdot2+5\).

- Two points with clear grid coordinates can help you find a line's slope. - Once you know the slope, what else do you need to write the line's equation? - At the intersection, both equations give the same y-value for the same x-value.

1. Line \(g\) crosses the y-axis at \(1\) and rises \(1\) for each step right, so \(g: y=x+1\). 2. Line \(h\) crosses the y-axis at \(2\) and falls \(1\) for each step right, so \(h: y=-x+2\). 3. From the graph, the lines appear to meet near \((0.5, 1.5)\). 4. Set the equations equal: \(x+1=-x+2\). Then \(2x=1\), so \(x=\frac{1}{2}\). Substituting gives \(y=\frac{3}{2}\). 5. Check in both equations: \(\frac{3}{2}=\frac{1}{2}+1\) and \(\frac{3}{2}=-\frac{1}{2}+2\). Both are true.

\(g: y=x+1\) \(h: y=-x+2\) Estimated intersection: about \((0.5, 1.5)\). Exact intersection: \(\left(\frac{1}{2}, \frac{3}{2}\right)\). Checks: \(\frac{3}{2}=\frac{1}{2}+1\) and \(\frac{3}{2}=-\frac{1}{2}+2\).

- Trace from the crossing toward each axis to judge its position on the scale. - A crossing between grid lines does not have a whole-number coordinate. - Use the grid spacing rather than rounding the intersection to a marked point.

1. The lines represent \(y=0.5x+1\) and \(y=-x+5\). 2. Their intersection satisfies \(0.5x+1=-x+5\), so \(1.5x=4\) and \(x=\frac{8}{3}\). 3. The y-value is \(\frac{7}{3}\). Both coordinates lie between \(2\) and \(3\), as the graph suggests.

The x-coordinate is between \(2\) and \(3\), and the y-coordinate is between \(2\) and \(3\).

- Read the graphed line's y-intercept and compare two points to find its slope. - At an intersection, both equations give the same y-value. - Substitute the x-value you find into either line equation.

1. The graphed line rises by \(1\) for each step right and crosses the y-axis at \(2\), so its equation is \(y=x+2\). 2. Set the two expressions equal: \(x+2=-x+6\). 3. Then \(2x=4\), so \(x=2\) and \(y=4\).

a) \(y=x+2\) b) \((2,4)\)

- Choose an easy-to-read point on each drawn line and test it in the matching equation. - If a line disagrees with its equation, check whether its height or its rise and run are wrong. - Only use a crossing after checking both lines in that panel.

1. In panel a), the orange line crosses the y-axis at \(3\), not \(5\), so it has the wrong intercept. 2. In panel b), the blue line crosses the y-axis at \(-1\) and rises with slope \(2\). The orange line crosses at \(5\) and falls with slope \(-1\). Both equations match. 3. The lines in panel b) meet at \((2,3)\).

Graph b) is correct. In a), the orange line has y-intercept \(3\) instead of \(5\). The correct intersection is \((2,3)\).

- Which variable has a coefficient of \(1\)? - Why is a coefficient of \(1\) convenient when isolating a variable? - Remember to include both variable values in the solution set.

1. In equation (I), \(x\) has coefficient \(1\), so it can be isolated without division or fractions: \(x = 2y + 8\). 2. Substitute \(x = 2y + 8\) into equation (II): \(3(2y + 8) + 4y = 4\). 3. Solve for \(y\): \(6y + 24 + 4y = 4\), so \(10y = -20\) and \(y = -2\). 4. Substitute \(y = -2\) into \(x = 2y + 8\): \(x = 2(-2) + 8 = 4\). 5. The solution set is \(\{(4, -2)\}\).

a) Isolate \(x\) in equation (I), giving \(x = 2y + 8\), because \(x\) has coefficient \(1\). b) The solution set is \(\{(4, -2)\}\).

- You may isolate either variable in either equation. Which choice looks simplest here? - Pay close attention to the signs when distributing a negative factor. - How can substitution into both original equations verify your result?

1. Solve equation (I) for \(y\): \(y = 1 - 0.5x\). 2. Substitute the expression for \(y\) into equation (II): \(2x - 3(1 - 0.5x) = 11\). 3. Distribute: \(2x - 3 + 1.5x = 11\). 4. Combine like terms and solve: \(3.5x - 3 = 11\), so \(3.5x = 14\) and \(x = 4\). 5. Substitute \(x = 4\) into the rearranged equation (I): \(y = 1 - 0.5 \cdot 4 = -1\). 6. Check equation (I): \(0.5 \cdot 4 + (-1) = 1\). Check equation (II): \(2 \cdot 4 - 3(-1) = 11\). Both equations are true. 7. The solution set is \(\{(4, -1)\}\).

The solution set is \(\{(4, -1)\}\).

- Identify the two facts given about the invested amounts. - Write each interest rate as a decimal. - Translate the total investment and total interest into two equations.

1. Let \(x\) be the amount invested at \(2\%\) and \(y\) the amount invested at \(3\%\). 2. Write the system: \(x + y = 10{,}000\) and \(0.02x + 0.03y = 260\). 3. Solve the first equation for \(x\): \(x = 10{,}000 - y\). 4. Substitute into the second equation: \(0.02(10{,}000 - y) + 0.03y = 260\). 5. Simplify: \(200 - 0.02y + 0.03y = 260\), so \(0.01y = 60\) and \(y = 6000\). 6. Then \(x = 10{,}000 - 6000 = 4000\).

Ms. Miller invested \(\$4000\) in the \(2\%\) account and \(\$6000\) in the \(3\%\) account.

- Find the slope of \(g_1\) from its two points. - Use the known slope and point to find the equation of \(g_2\). - At an intersection point, what must be true about the two y-values?

1. For \(g_1\), the y-intercept is \(2\), and the slope is \(m_1=\frac{6-2}{4-0}=1\). Thus \(g_1: y=x+2\). 2. For \(g_2\), substitute \((6, 2)\) into \(y=-x+b_2\): \(2=-6+b_2\), so \(b_2=8\). Thus \(g_2: y=-x+8\). 3. At the intersection, the y-values are equal: \(x+2=-x+8\). Solving gives \(x=3\). 4. Substitute \(x=3\) into either equation: \(y=5\). The intersection is \((3, 5)\).

\(g_1: y=x+2\), \(g_2: y=-x+8\), and the intersection is \((3, 5)\).

- What must you do to both equations before you can compare their expressions directly? - Solve both equations for the same variable. - After finding \(x\), how can you find \(y\)? - Check the ordered pair in both original equations.

1. Solve both equations for \(y\). Equation (I) becomes \(3y = 6x + 12\), so \(y = 2x + 4\). Equation (II) becomes \(y = -2x + 20\). 2. Set the expressions equal: \(2x + 4 = -2x + 20\). 3. Solve for \(x\): \(4x = 16\), so \(x = 4\). 4. Substitute into either expression for \(y\): \(y = 2 \cdot 4 + 4 = 12\). 5. The solution set is \(\{(4, 12)\}\).

The solution set is \(\{(4, 12)\}\).

- A candle height can be modeled as initial height minus burn rate times time. - Which variable represents time, and which represents height? - What equation represents the moment when the candles have the same height? - Keep track of the units as you interpret the result.

1. Candle A is modeled by \(h = 25 - 1.5t\). Candle B is modeled by \(h = 20 - 0.5t\). 2. Set the expressions for \(h\) equal: \(25 - 1.5t = 20 - 0.5t\). 3. Solve for \(t\): \(5 - 1.5t = -0.5t\), so \(5 = t\). 4. Substitute \(t = 5\): \(h = 25 - 1.5 \cdot 5 = 17.5\). 5. After \(5\) hours, both candles are \(17.5\,\text{cm}\) tall.

a) Candle A: \(h = 25 - 1.5t\) Candle B: \(h = 20 - 0.5t\) b) After \(5\) hours, both candles are \(17.5\,\text{cm}\) tall.

- Which equation is easiest to solve for one variable? - Substitute the resulting expression into the other equation. - After finding one variable, how can you determine the other?

1. Solve equation (II) for \(r\): \(r = \frac{3}{4}s - 1\). 2. Substitute into equation (I): \(\frac{2}{3}(\frac{3}{4}s - 1) + \frac{1}{2}s = 6\). 3. Distribute: \(\frac{1}{2}s - \frac{2}{3} + \frac{1}{2}s = 6\). 4. Combine like terms: \(s - \frac{2}{3} = 6\). 5. Solve for \(s\): \(s = 6 + \frac{2}{3} = \frac{20}{3}\). 6. Substitute into the expression for \(r\): \(r = \frac{3}{4} \cdot \frac{20}{3} - 1 = 4\). 7. The solution is \((r, s) = (4, \frac{20}{3})\).

The solution is \((r, s) = (4, \frac{20}{3})\).

- What value does a function have at a zero? - At an intersection point, what is true about the two function values? - Compare the results of the first two parts with the equation you solve in part c.

1. For part a, set \(f(x) = 0\): \(\frac{2}{3}x - 4 = 0\). Then \(\frac{2}{3}x = 4\), so \(x = 6\). 2. For part b, set \(g(x) = 0\): \(-\frac{1}{3}x + 2 = 0\). Then \(-\frac{1}{3}x = -2\), so \(x = 6\). 3. For part c, set the function values equal: \(\frac{2}{3}x - 4 = -\frac{1}{3}x + 2\). Adding \(\frac{1}{3}x\) gives \(x - 4 = 2\), so \(x = 6\). 4. Both functions equal \(0\) at \(x = 6\), so their intersection is exactly on the x-axis.

a) \(x = 6\) b) \(x = 6\) c) \(x = 6\). The two functions have the same zero, so they intersect on the x-axis.

- Can you simplify the first equation before choosing a method? - Which method is efficient once one variable is isolated? - Pay close attention to the negative factor before the parentheses. - Include values for both variables in your final answer.

1. Simplify equation (I) and solve for \(y\): \(2x - 6 = y + 2\), so \(y = 2x - 8\). 2. Substitute into equation (II): \(4x - 3(2x - 8) = 2\). 3. Distribute and combine like terms: \(4x - 6x + 24 = 2\), so \(-2x = -22\). 4. Solve for \(x\): \(x = 11\). 5. Substitute into \(y = 2x - 8\): \(y = 2 \cdot 11 - 8 = 14\).

The solution is \((11, 14)\).

- Bring all \(x\)-terms to one side of the equation. - How can you clear or simplify the fractions? - Pay close attention to signs when substituting a negative value of \(x\).

1. Set the expressions equal: \(\frac{3}{4}x + 2 = \frac{1}{4}x - 1\). 2. Subtract \(\frac{1}{4}x\): \(\frac{1}{2}x + 2 = -1\). 3. Subtract \(2\): \(\frac{1}{2}x = -3\). 4. Multiply by \(2\): \(x = -6\). 5. Substitute into equation (II): \(y = \frac{1}{4}(-6) - 1 = -2.5\). 6. The intersection is \((-6, -2.5)\).

The intersection is \((-6, -2.5)\).

- What must be true about the slopes of parallel lines? - How can you use the given point to find the y-intercept? - What is true about the two function values at an intersection point?

1. Because \(h\) is parallel to \(g\), its slope is \(-\frac{2}{3}\). Write \(h(x) = -\frac{2}{3}x + b\). Using \(P(6, 2)\), \(2 = -\frac{2}{3} \cdot 6 + b\), so \(b = 6\). Thus \(h(x) = -\frac{2}{3}x + 6\). 2. At the intersection, the function values are equal: \(-\frac{2}{3}x + 6 = x - 3\). 3. Solving gives \(9 = \frac{5}{3}x\), so \(x = \frac{27}{5} = 5.4\). 4. Substitute into \(k\): \(y = 5.4 - 3 = 2.4\). Therefore, \(S(5.4, 2.4)\).

a) \(h(x) = -\frac{2}{3}x + 6\) b) \(S(5.4, 2.4)\)

- How can you find a line's slope from two points? - Can you read the y-intercept of one line directly from one of its points? - What must be true about the function values at an intersection point?

1. For \(f\), \(m_f = \frac{9 - 1}{3 - (-1)} = 2\). Using \(B(3, 9)\), \(9 = 2 \cdot 3 + b\), so \(b = 3\). Thus \(f(x) = 2x + 3\). 2. For \(g\), \(m_g = \frac{0 - 10}{5 - 0} = -2\). Since \(C(0, 10)\) is the y-intercept, \(g(x) = -2x + 10\). 3. At the intersection, \(2x + 3 = -2x + 10\). Then \(4x = 7\), so \(x = \frac{7}{4} = 1.75\). 4. Substitute into \(f\): \(y = 2 \cdot 1.75 + 3 = 6.5\). Therefore, \(S(1.75, 6.5)\).

a) \(f(x) = 2x + 3\) and \(g(x) = -2x + 10\) b) \(S(1.75, 6.5)\)

- For each plan, identify the fixed fee and the rate per minute. - When the plans cost the same, what must be true of the two expressions? - What does \(x\) represent, and what unit should your answer use?

1. Plan A is modeled by \(y = 0.20x + 5\). Plan B is modeled by \(y = 0.45x\). 2. Set the expressions equal: \(0.20x + 5 = 0.45x\). 3. Subtract \(0.20x\): \(5 = 0.25x\). 4. Divide by \(0.25\): \(x = 20\). 5. Check the cost: \(y = 0.45 \cdot 20 = 9.00\), so both plans cost \(\$9.00\).

a) Plan A: \(y = 0.20x + 5\) Plan B: \(y = 0.45x\) b) The plans cost the same after \(20\) minutes. Each costs \(\$9.00\).

- Set the two expressions for \(y\) equal. - Rewrite the fractions with a common denominator. - Substitute the value of \(x\) into either original equation.

1. Set the expressions equal: \(\frac{2}{3}x + 5 = \frac{5}{6}x + 2\). 2. Subtract \(2\) and \(\frac{2}{3}x\): \(3 = \frac{5}{6}x - \frac{4}{6}x\). 3. Simplify: \(3 = \frac{1}{6}x\). 4. Multiply by \(6\): \(x = 18\). 5. Substitute into equation (I): \(y = \frac{2}{3} \cdot 18 + 5 = 17\).

The solution is \((18, 17)\).

- Choose variables for the two planned ticket amounts. - Use the planned total to write the first equation. - Express the percentages as decimal multipliers in the second equation. - After finding the planned amounts, calculate the actual sales.

1. Let \(x\) be the planned number of student tickets and \(y\) the planned number of adult tickets. 2. Write the system \(x + y = 500\) and \(1.05x + 0.90y = 495\). 3. Solve the first equation for \(y\): \(y = 500 - x\). 4. Substitute: \(1.05x + 0.90(500 - x) = 495\). 5. Simplify: \(1.05x + 450 - 0.90x = 495\), so \(0.15x = 45\) and \(x = 300\). 6. Then \(y = 500 - 300 = 200\). 7. Calculate the actual sales: \(1.05 \cdot 300 = 315\) student tickets and \(0.90 \cdot 200 = 180\) adult tickets.

The festival sold \(315\) student tickets and \(180\) adult tickets.

- Determine what the difference between the ending balance and the initial investment represents. - Write one equation for the total principal and another for the total interest. - Convert each percent to a decimal before using it in an equation. - Solve the system using a method such as substitution.

1. Let \(x\) be the amount invested at \(1.5\%\) and \(y\) the amount invested at \(2.5\%\). 2. The total interest earned is \(12{,}230 - 12{,}000 = 230\) dollars. 3. Write the system \(x + y = 12{,}000\) and \(0.015x + 0.025y = 230\). 4. Solve the first equation for \(y\): \(y = 12{,}000 - x\). 5. Substitute: \(0.015x + 0.025(12{,}000 - x) = 230\). 6. Simplify: \(0.015x + 300 - 0.025x = 230\), so \(-0.01x = -70\) and \(x = 7000\). 7. Then \(y = 12{,}000 - 7000 = 5000\).

Ms. Weber invested \(\$7000\) in the \(1.5\%\) account and \(\$5000\) in the \(2.5\%\) account.

- In which equation is one variable easiest to isolate? - Look for a variable with coefficient \(1\). - Substitute the resulting expression into the equation you have not changed. - Check both values in the original equations.

1. Solve the first equation for \(y\), since its coefficient is \(1\): \(y = 10 - 3x\). 2. Substitute into the second equation: \(2x + 3(10 - 3x) = 2\). 3. Distribute: \(2x + 30 - 9x = 2\). 4. Combine like terms and solve: \(-7x + 30 = 2\), so \(-7x = -28\) and \(x = 4\). 5. Substitute into \(y = 10 - 3x\): \(y = 10 - 3 \cdot 4 = -2\). 6. The solution set is \(\{(4, -2)\}\).

The solution set is \(\{(4, -2)\}\).

- Can you isolate a multiple of a variable without introducing fractions? - Can an expression for \(2y\) be used directly to replace \(4y\)? - Remember to determine both \(x\) and \(y\).

1. Solve equation (I) for \(2y\): \(2y = 5x - 19\). 2. Since \(4y = 2(2y)\), substitute \(2(5x - 19)\) for \(4y\) in equation (II): \(3x + 2(5x - 19) = 1\). 3. Solve for \(x\): \(3x + 10x - 38 = 1\), so \(13x = 39\) and \(x = 3\). 4. Substitute into \(2y = 5x - 19\): \(2y = 5 \cdot 3 - 19 = -4\), so \(y = -2\). 5. The solution is \((3, -2)\).

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

- When no coefficient is \(1\), isolating a variable may introduce a fraction. - Which variable is most convenient to isolate? - How can you clear denominators after substituting? - Pay attention to signs when substituting a negative value.

1. Solve equation (I) for \(x\): \(3x = 11 + 5y\), so \(x = \frac{11 + 5y}{3}\). 2. Substitute into equation (II): \(4 \cdot \frac{11 + 5y}{3} + 3y = 5\). 3. Multiply by \(3\) to clear the denominator: \(4(11 + 5y) + 9y = 15\). 4. Distribute and combine like terms: \(44 + 20y + 9y = 15\), so \(44 + 29y = 15\). 5. Solve for \(y\): \(29y = -29\), so \(y = -1\). 6. Substitute into the expression for \(x\): \(x = \frac{11 + 5(-1)}{3} = 2\). 7. The solution set is \(\{(2, -1)\}\).

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

- Solve one equation for one variable. Decide whether fractions or decimals are more convenient. - Substitute the expression into the other equation and distribute carefully. - Check the final ordered pair in both original equations.

1. Solve equation (I) for \(y\): \(2y = 5 - 3x\), so \(y = 2.5 - 1.5x\). 2. Substitute into equation (II): \(5x + 3(2.5 - 1.5x) = 8\). 3. Distribute: \(5x + 7.5 - 4.5x = 8\). 4. Combine like terms and solve: \(0.5x + 7.5 = 8\), so \(0.5x = 0.5\) and \(x = 1\). 5. Substitute into the expression for \(y\): \(y = 2.5 - 1.5 \cdot 1 = 1\). 6. Check equation (I): \(3 \cdot 1 + 2 \cdot 1 = 5\). Check equation (II): \(5 \cdot 1 + 3 \cdot 1 = 8\). Both equations are true. 7. The solution is \((1, 1)\).

The solution is \((1, 1)\).

- Which variable is easiest to isolate? - Treat \(p\) as a constant while solving for \(x\) and \(y\). - Put parentheses around the substituted expression. - Your final ordered pair should be written in terms of \(p\).

1. Solve equation (II) for \(y\): \(y = 2x - 3p\). 2. Substitute into equation (I): \(x + 2(2x - 3p) = 4p\). 3. Solve for \(x\): \(x + 4x - 6p = 4p\), so \(5x = 10p\) and \(x = 2p\). 4. Substitute into the expression for \(y\): \(y = 2(2p) - 3p = p\).

The solution is \((x, y) = (2p, p)\).

- No variable is isolated, so begin by rearranging one equation. - Dividing by a coefficient may produce manageable decimals. - Check your result in both original equations.

1. Solve the second equation for \(y\): \(-2y = 14 - 5x\), so \(y = 2.5x - 7\). 2. Substitute into the first equation: \(4x + 3(2.5x - 7) = 2\). 3. Distribute and solve for \(x\): \(4x + 7.5x - 21 = 2\), so \(11.5x = 23\). 4. Therefore, \(x = 2\). 5. Substitute into the expression for \(y\): \(y = 2.5 \cdot 2 - 7 = -2\). 6. The solution is \((2, -2)\).

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

- How can you clear the fraction in the first equation? - Rewrite both equations in the form \(Aa + Bb = C\). - Which variable is easiest to isolate? - Check both values in the original equations.

1. Simplify the first equation: \(\frac{1}{2}a + 2 = b + 1\), so \(\frac{1}{2}a - b = -1\). Multiply by \(2\) to get \(a - 2b = -2\). 2. Simplify the second equation: \(3a - 3b = 2b + 11\), so \(3a - 5b = 11\). 3. From the first simplified equation, \(a = 2b - 2\). 4. Substitute into the second simplified equation: \(3(2b - 2) - 5b = 11\). 5. Solve for \(b\): \(6b - 6 - 5b = 11\), so \(b = 17\). 6. Find \(a\): \(a = 2 \cdot 17 - 2 = 32\).

The solution is \((a, b) = (32, 17)\).

- Begin by expanding and combining like terms in each equation. - Could the expressions \(x + y\) and \(x - y\) also be treated as temporary variables? - Which system-solving method is most efficient after simplification?

1. Expand and simplify each equation. The first becomes \(2x + 2y + 3x - 3y = 14\), so \(5x - y = 14\). The second becomes \(5x + 5y - 2x + 2y = 16\), so \(3x + 7y = 16\). 2. Solve the first simplified equation for \(y\): \(y = 5x - 14\). 3. Substitute into the second simplified equation: \(3x + 7(5x - 14) = 16\). 4. Distribute and combine like terms: \(3x + 35x - 98 = 16\), so \(38x = 114\). 5. Solve for \(x\): \(x = 3\). 6. Substitute into \(y = 5x - 14\): \(y = 1\). 7. The solution is \((3, 1)\).

The solution is \((3, 1)\).

- Choose variables for last year’s membership in each program. - Use the combined membership from last year to write one equation. - Represent the percentage increase and decrease with decimal multipliers. - Express one variable in terms of the other and substitute.

1. Let \(x\) be last year’s soccer membership and \(y\) last year’s tennis membership. 2. Write the system \(x + y = 200\) and \(1.1x + 0.9y = 202\). 3. Solve the first equation for \(y\): \(y = 200 - x\). 4. Substitute: \(1.1x + 0.9(200 - x) = 202\). 5. Simplify: \(1.1x + 180 - 0.9x = 202\), so \(0.2x = 22\) and \(x = 110\). 6. Then \(y = 200 - 110 = 90\).

Last year, the soccer program had \(110\) members, and the tennis program had \(90\) members.

- Identify whether the unknown variables should represent last month or this month. - Express a \(5\%\) decrease and a \(20\%\) increase with decimal multipliers. - Pay attention to which month the question asks about. - A table comparing last month and this month may help organize the information.

1. Let \(b\) be last month’s number of Basic members and \(p\) last month’s number of Premium members. 2. Write the system \(b + p = 600\) and \(0.95b + 1.2p = 630\). 3. Solve the first equation for \(p\): \(p = 600 - b\). 4. Substitute: \(0.95b + 1.2(600 - b) = 630\). 5. Simplify: \(0.95b + 720 - 1.2b = 630\), so \(-0.25b = -90\) and \(b = 360\). 6. Then \(p = 600 - 360 = 240\). 7. Calculate this month’s membership: \(0.95 \cdot 360 = 342\) Basic members and \(1.2 \cdot 240 = 288\) Premium members.

This month, \(342\) members use the Basic plan and \(288\) members use the Premium plan.

- First determine the technicians’ combined hourly rate. - Express the \(180\) completed components as work done during the individual and joint time periods. - Write a system with the two unknown individual rates. - Use the known combined rate in the second equation.

1. Let \(x\) be Mr. Allen’s rate and \(y\) Ms. Baker’s rate, in components per hour. 2. Their combined work gives \(8(x + y) = 240\), so \(x + y = 30\). 3. The second situation gives \(3x + 4(x + y) = 180\). 4. Substitute \(x + y = 30\): \(3x + 4 \cdot 30 = 180\). 5. Solve: \(3x + 120 = 180\), so \(x = 20\). 6. Then \(20 + y = 30\), so \(y = 10\).

Mr. Allen processes \(20\) components per hour, and Ms. Baker processes \(10\) components per hour.

- Interpret flow rate as the volume moved per unit of time. - Write one equation for each operating schedule. - Once you know both rates, add them to find the combined rate. - Divide the tank’s capacity by the combined rate to find the filling time.

1. Let \(a\) and \(b\) be the flow rates of Pumps A and B, in gallons per minute. 2. Write the system \(3a + 2b = 110\) and \(2a + 4b = 140\). 3. Divide the second equation by \(2\): \(a + 2b = 70\), so \(a = 70 - 2b\). 4. Substitute into the first equation: \(3(70 - 2b) + 2b = 110\). 5. Simplify: \(210 - 6b + 2b = 110\), so \(-4b = -100\) and \(b = 25\). 6. Then \(a = 70 - 2 \cdot 25 = 20\). 7. Together, the pumps move \(20 + 25 = 45\,\text{gal}\) per minute. Filling \(225\,\text{gal}\) takes \(225 \div 45 = 5\) minutes.

a) Pump A moves \(20\,\text{gal}\) per minute, and Pump B moves \(25\,\text{gal}\) per minute. b) Together, the pumps would fill the tank in \(5\) minutes.

- Solve the linear equation for one variable. - Substitute that expression into the squared equation. - Remember both cases when taking a square root.

1. Solve the linear equation for \(x\): \(x = y + 1\). 2. Substitute into the first equation: \((y + 1 + y)^2 = 25\), so \((2y + 1)^2 = 25\). 3. Take both square roots: \(2y + 1 = 5\) or \(2y + 1 = -5\). 4. The first case gives \(y = 2\) and \(x = 3\). The second case gives \(y = -3\) and \(x = -2\). 5. Therefore, the system has exactly two ordered-pair solutions.

\(\{(3, 2), (-2, -3)\}\)

- What is the general shape of a graph whose variable is in the denominator? - Use a graphing tool to display both equations on the same coordinate plane. - Make sure the reciprocal graph is viewed for both positive and negative inputs. - What do common points of the two graphs represent?

1. Enter \(y = \frac{4}{x}\) and \(y = x + 3\) in a graphing tool and display both graphs on the same coordinate plane. 2. The reciprocal graph includes points such as \((1, 4)\), \((2, 2)\), \((4, 1)\), \((-1, -4)\), \((-2, -2)\), and \((-4, -1)\). The line has slope \(1\) and \(y\)-intercept \(3\). 3. The graphs intersect at \((1, 4)\) and \((-4, -1)\). 4. These intersection points are the solutions of the system.

The graphs intersect at \((1, 4)\) and \((-4, -1)\), so these are the solutions.

- Split the equation so that each side becomes a function rule. - Remember that \(x = 0\) is not in the domain of the reciprocal function. - Use a graphing tool to find the intersection points, then identify the requested \(x\)-coordinates. - How can substitution confirm that a proposed value satisfies the original equation?

1. Use \(f(x) = \frac{6}{x}\) and \(g(x) = x + 1\). 2. Enter both functions in a graphing tool and display them on the same coordinate plane. 3. The graphs intersect at \((2, 3)\) and \((-3, -2)\), so the solutions are \(x = 2\) and \(x = -3\). 4. Check \(x = 2\): \(\frac{6}{2} = 3\) and \(2 + 1 = 3\). 5. Check \(x = -3\): \(\frac{6}{-3} = -2\) and \(-3 + 1 = -2\). Both values satisfy the original equation.

a) \(f(x) = \frac{6}{x}\) and \(g(x) = x + 1\) b) The graphs intersect at \((2, 3)\) and \((-3, -2)\), so \(x = 2\) and \(x = -3\). c) \(\frac{6}{2} = 2 + 1 = 3\), and \(\frac{6}{-3} = -3 + 1 = -2\).

- Convert cents per kWh to dollars per kWh before writing the functions. - Equal costs correspond to equal function values. - A percentage increase in a base charge changes the y-intercept. - A lower energy rate produces a smaller slope.

1. In dollars, the cost functions are \(F(x)=0.06x+120\) and \(A(x)=0.045x+180\). Set them equal: \(0.06x+120=0.045x+180\). Then \(0.015x=60\), so \(x=4000\). The plans cost the same at \(4000\,\text{kWh}\). 2. The new Active base charge is \(180\cdot 1.20=\$216\). Solve \(0.06x+120=0.045x+216\): \(0.015x=96\), so \(x=6400\). Active is less expensive for \(x>6400\,\text{kWh}\). 3. Lowering Active's energy rate decreases the slope of its cost graph. Because its base charge remains higher, the graphs intersect at a lower usage value, so Active becomes advantageous sooner.

1. \(4000\,\text{kWh}\) 2. New break-even usage: \(6400\,\text{kWh}\); Active costs less for \(x>6400\,\text{kWh}\). 3. The intersection moves left, to a lower usage value.

- Read the y-intercept and choose another grid point on each line. - Use the two points to calculate each slope. - To get integer coefficients, clear any fractions before rearranging. - At the intersection, the two expressions for \(y\) are equal.

1. Line \(f\) crosses the y-axis at \((0,-2)\) and passes through \((2,1)\). Its slope is \(\frac{1-(-2)}{2-0}=\frac32\), so \(f\) has equation \(y=\frac32x-2\). 2. Line \(g\) crosses the y-axis at \((0,4)\) and passes through \((2,2)\). Its slope is \(-1\), so \(g\) has equation \(y=-x+4\). 3. Rewrite the first equation with integer coefficients: \(2y=3x-4\), so \(3x-2y=4\). The second becomes \(x+y=4\). 4. At the intersection, \(\frac32x-2=-x+4\). Solving gives \(\frac52x=6\), so \(x=\frac{12}{5}\). Then \(y=\frac85\), so \(S\left(\frac{12}{5}, \frac85\right)\).

a) \(f: y=\frac32x-2\); \(g: y=-x+4\) b) \(f: 3x-2y=4\); \(g: x+y=4\) c) \(S\left(\frac{12}{5}, \frac85\right)\)

- Fixed costs are the costs when production is \(0\). - One process is cheaper when its graph lies below the other cost graph. - For each process, find the first intersection of its cost graph with the revenue graph.

1. Fixed cost is the y-intercept of each cost graph. Process A has a fixed cost of \(\$200\), and Process B has a fixed cost of \(\$50\). Therefore, Process A has the greater fixed cost. 2. The cost graphs intersect at \((150, 275)\). For \(x>150\), the graph of \(K_A\) lies below the graph of \(K_B\). Because production quantities are whole numbers, Process A is less expensive starting at \(151\) cases. 3. The graph of \(R\) intersects \(K_B\) at \((50, 125)\), so Process B first reaches revenue equal to cost at \(50\) cases. The graph of \(R\) intersects \(K_A\) at \((100, 250)\), so Process A first reaches revenue equal to cost at \(100\) cases.

a) Process A; Process A: \(\$200\), Process B: \(\$50\) b) For \(x>150\), so starting at \(151\) cases c) Process B: \(50\) cases; Process A: \(100\) cases

- Use whether each displayed line rises or falls and where it crosses the y-axis. - The system's solution is the point shared by the two displayed lines. - Verification means checking the same ordered pair in each original equation.

1. Line \(p\) rises from left to right and crosses the y-axis at \(2\), so \(p\) represents \(y=\frac{1}{2}x+2\). 2. Line \(q\) falls from left to right and crosses the y-axis at \(5\), so \(q\) represents \(y=-x+5\). 3. The lines intersect at \((2, 3)\). 4. In the first equation, \(\frac{1}{2}\cdot2+2=3\). In the second equation, \(-2+5=3\). Thus \((2, 3)\) satisfies both equations.

a) \(p: y=\frac{1}{2}x+2\); \(q: y=-x+5\) b) \((2, 3)\) c) \(\frac{1}{2}\cdot2+2=3\) and \(-2+5=3\), so the point satisfies both equations.

- Separate what the graph can show approximately from what algebra can establish exactly. - Because both equations equal \(y\), what expressions can you set equal for substitution? - Compare the differences between the estimated and exact coordinates after solving.

1. The displayed grid has finite spacing, and the intersection lies between grid lines. A visual reading can estimate the coordinates but cannot establish exact values by itself. 2. Set the two expressions for \(y\) equal: \(\frac{1}{2}x+1=-\frac{1}{3}x+4\). 3. Then \(\frac{5}{6}x=3\), so \(x=\frac{18}{5}\). 4. Substitute into either equation: \(y=\frac{1}{2}\cdot\frac{18}{5}+1=\frac{14}{5}\). 5. The exact solution is \(\left(\frac{18}{5},\frac{14}{5}\right)=(3.6,2.8)\). Jada's estimate \((3.5,3.0)\) is close, so it is reasonable as a graph-based estimate but not as an exact answer.

a) The graph supports only an estimate because the intersection lies between grid lines. b) \(\left(\frac{18}{5},\frac{14}{5}\right)=(3.6,2.8)\) c) Yes. \((3.5,3.0)\) is close to \((3.6,2.8)\), so it is a reasonable estimate but not the exact solution.

- Substitute each requested x-value and solve for y. - Check all four points against the equation for their own line. - At the crossing, one ordered pair must satisfy both equations.

1. For (I), \((0,4)\) and \((4,2)\) work because each has \(x+2y=8\). 2. For (II), \((0,-1)\) and \((3,5)\) work because each has \(2x-y=1\). 3. Solve \(y=2x-1\) in (I): \(x+2(2x-1)=8\), so \(5x=10\) and \(x=2\). 4. Then \(y=3\), so the lines meet at \((2,3)\). 5. Check (I): \(2+2\cdot3=8\). Check (II): \(2\cdot2-3=1\).

For (I): \((0,4)\), \((4,2)\). For (II): \((0,-1)\), \((3,5)\). Intersection: \((2,3)\). Check (I): \(2+2\cdot3=8\). Check (II): \(2\cdot2-3=1\).

- Read the y-intercept and slope of the graphed line. - Does the suggested point satisfy the line equation you found? - At the actual crossing, both line expressions give the same y-value.

1. Line \(g\) crosses the y-axis at \(1\) and rises \(2\) per step right, so \(g\) is \(y=2x+1\). 2. At \(x=1.5\), this line has \(y=4\), so \((1.5,3.5)\) is not on \(g\). 3. Set the two expressions equal: \(2x+1=-x+5\), giving \(3x=4\) and \(x=\frac{4}{3}\). 4. Then \(y=5-\frac{4}{3}=\frac{11}{3}\). The exact intersection is \(\left(\frac{4}{3},\frac{11}{3}\right)\).

\(g: y=2x+1\). The estimate is not exact. The intersection is \(\left(\frac{4}{3},\frac{11}{3}\right)\).

- Each plan has a fixed cost and a cost for every class attended. - At zero classes, only the fixed cost remains. - Where the two cost lines meet, the total charges are equal.

1. If \(x\) is the number of classes and \(y\) the cost in dollars, Plan A is \(y=6+2x\), with points \((0,6)\) and \((2,10)\). 2. Plan B is \(y=2+3x\), with points \((0,2)\) and \((2,8)\). 3. The lines meet where \(6+2x=2+3x\), giving \(x=4\). 4. Both plans cost \(\$14\) for \(4\) classes.

Plan A: \(y=6+2x\), points \((0,6)\), \((2,10)\). Plan B: \(y=2+3x\), points \((0,2)\), \((2,8)\). Both cost \(\$14\) for \(4\) classes.

- Both lines contain the point of intersection. - Which equation lets you calculate the y-coordinate from the known x-coordinate? - Use those two coordinates in the equation with the missing slope.

1. On the second line, \(x=2\) gives \(y=-2+7=5\). The intersection is \((2,5)\). 2. This point also lies on the first line, so \(5=2k-1\). 3. Thus \(2k=6\) and \(k=3\).

Intersection: \((2,5)\). Missing slope: \(k=3\). At \(x=2\), the second equation gives \(y=5\). Putting \((2,5)\) into the first equation gives \(5=2k-1\), so \(k=3\).

- Start with the given x-coordinate and read the height of the drawn line. - At the crossing, that point lies on both lines. - Substitute the point into the equation with the missing slope.

1. At \(x=2\), the line \(g\) has \(y=3\), so the intersection is \((2,3)\). 2. The unknown line also passes through this point: \(3=2m+1\). 3. Subtract \(1\) and divide by \(2\) to get \(m=1\).

\(m=1\); intersection \((2,3)\).

- Compare how the orange line moves to the right with the sign of the coefficient of \(x\). - Use the two requested x-values to calculate points on the intended line. - Only after correcting the line, look for its crossing with \(g\).

1. The orange line rises to the right, but \(y=-x+3\) must fall by \(1\) when \(x\) increases by \(1\). Its slope is wrong. 2. The correct line \(h\) goes through \((0,3)\) and \((2,1)\). 3. At the intersection, \(x+1=-x+3\), so \(2x=2\) and \(x=1\). 4. Then \(y=2\). The correct lines meet at \((1,2)\).

The drawn orange line rises, but \(h\) should fall to the right. Correct points for \(h\): \((0,3)\) and \((2,1)\). Correct intersection: \((1,2)\).

- Find two clear grid points along each segment. - What are the slopes and y-intercepts of the lines containing those segments? - The segments do not have to touch for their full lines to intersect.

1. Segment \(g\) goes through \((0,1)\) and \((2,3)\), so the line has slope \(1\) and equation \(y=x+1\). 2. Segment \(h\) goes through \((0,6)\) and \((2,4)\), so the line has slope \(-1\) and equation \(y=-x+6\). 3. Set the expressions equal: \(x+1=-x+6\), giving \(2x=5\) and \(x=\frac{5}{2}\). 4. Substitute into either line: \(y=\frac{7}{2}\). The extended lines meet at \(\left(\frac{5}{2},\frac{7}{2}\right)\), beyond the drawn segments.

\(g: y=x+1\). \(h: y=-x+6\). Intersection: \(\left(\frac{5}{2},\frac{7}{2}\right)\).

- What two amounts make up Plan A's monthly cost? - Choose letters for Plan A's fee and charge per gigabyte before writing the two equations. - To compare the plans, write the monthly cost of each as an expression in data use.

1. Let \(b\) be Plan A's base fee in dollars and \(p\) its charge per gigabyte in dollars. Then \(b=p+2\) and \(b+3p=14\). 2. Substitute: \(p+2+3p=14\), so \(4p=12\) and \(p=3\). 3. Plan A's base fee is \(b=3+2=5\). 4. Plan B's base fee is \(5-2=3\), and its charge per gigabyte is \(3+0.50=3.50\). 5. For \(x\) gigabytes, equal costs mean \(5+3x=3+3.5x\). Thus \(2=0.5x\) and \(x=4\).

Plan A: \(\$5\) base fee and \(\$3\) per gigabyte. Plan B: \(\$3\) base fee and \(\$3.50\) per gigabyte. The costs are equal at \(4\,\text{GB}\).

- Represent the smaller and larger numbers with variables. - Translate the ratio into an equation relating the two numbers. - Translate the difference into a second equation. - Substitute one relationship into the other and solve symbolically.

1. Let \(x\) be the smaller number and \(y\) the larger number. 2. The ratio gives \(y = nx\), and the difference gives \(y - x = d\). 3. Substitute \(y = nx\) into the difference equation: \(nx - x = d\). 4. Factor: \(x(n - 1) = d\). 5. Divide by \(n - 1\): \(x = \frac{d}{n - 1}\). 6. Substitute into \(y = nx\): \(y = \frac{nd}{n - 1}\).

The smaller number is \(\frac{d}{n - 1}\), and the larger number is \(\frac{nd}{n - 1}\).

- Treat \(p\) and \(q\) as fixed constants. - Which system-solving method is most efficient here? - Use parentheses when substituting an expression. - Isolate one variable at a time.

1. Solve equation (I) for \(x\): \(x = p - 2y\). 2. Substitute into equation (II): \(2(p - 2y) + y = q\). 3. Distribute and combine like terms: \(2p - 4y + y = q\), so \(2p - 3y = q\). 4. Solve for \(y\): \(3y = 2p - q\), so \(y = \frac{2p - q}{3}\). 5. Substitute into the expression for \(x\): \(x = p - 2 \cdot \frac{2p - q}{3} = \frac{2q - p}{3}\). 6. The solution is \((x, y) = (\frac{2q - p}{3}, \frac{2p - q}{3})\).

The solution is \((x, y) = (\frac{2q - p}{3}, \frac{2p - q}{3})\).

- A minus sign before a fraction affects the entire numerator. - Multiply each equation by the least common multiple of its denominators. - After simplifying, solve the resulting linear system using a method of your choice. - Check your result in the original equations.

1. Multiply the first equation by \(4\): \((3x - 2y + 2) - 2(x + y - 5) = 4\). Expand and simplify to get \(x - 4y = -8\). 2. Multiply the second equation by \(20\): \(4(2x + y - 1) + 5(x - 2y + 6) = 60\). Expand and simplify to get \(13x - 6y = 34\). 3. Solve the first simplified equation for \(x\): \(x = 4y - 8\). 4. Substitute into the second simplified equation: \(13(4y - 8) - 6y = 34\). This gives \(52y - 104 - 6y = 34\), so \(46y = 138\) and \(y = 3\). 5. Substitute into \(x = 4y - 8\): \(x = 4 \cdot 3 - 8 = 4\).

The solution set is \(\{(4, 3)\}\).

- Expand both sides of equation (I). - What happens to the \(xy\)-terms when you simplify equation (I)? - How can you clear the fraction in equation (II)? - Which system-solving method is most efficient after simplification?

1. Expand equation (I): \(xy - 2x + 3y - 6 = xy + 4x - y - 4\). 2. Subtract \(xy\) from both sides: \(-2x + 3y - 6 = 4x - y - 4\). 3. Rearrange equation (I): \(-6x + 4y = 2\), or equivalently \(3x - 2y = -1\). 4. Multiply equation (II) by \(2\): \(x + y = 6\). 5. From equation (II), \(y = 6 - x\). Substitute into the simplified equation (I): \(3x - 2(6 - x) = -1\). Then \(5x = 11\), so \(x = \frac{11}{5}\). 6. Find \(y\): \(y = 6 - \frac{11}{5} = \frac{19}{5}\). 7. The term \(xy\) has the same coefficient on both sides of equation (I), so it cancels. Only first-degree terms in \(x\) and \(y\) remain.

a) The solution set is \(\{(\frac{11}{5}, \frac{19}{5})\}\). b) The \(xy\)-terms cancel during simplification, leaving only linear terms.

- Choose variables for the two original ball counts. - Write one equation for the total purchase cost. - Use each percentage to represent the value of the worn-out part of the inventory. - Simplify the percentage equation before solving the system.

1. Let \(x\) be the number of basketballs and \(y\) the number of volleyballs. 2. The total purchase cost gives \(15x + 20y = 1225\). 3. The value of the worn-out balls gives \(0.20 \cdot 15x + 0.30 \cdot 20y = 285\). 4. Simplify the second equation: \(3x + 6y = 285\), or \(x + 2y = 95\). 5. Solve for \(x\): \(x = 95 - 2y\). 6. Substitute into the cost equation: \(15(95 - 2y) + 20y = 1225\). 7. Simplify: \(1425 - 30y + 20y = 1225\), so \(-10y = -200\) and \(y = 20\). 8. Then \(x = 95 - 2 \cdot 20 = 55\).

The club originally bought \(55\) basketballs and \(20\) volleyballs.

- Notice that the question asks for the inventory remaining after the sale, not the original inventory. - Write one equation for the original inventory value. - Write another equation for the revenue from the percentages sold. - After finding the original counts, calculate the percentages that remain.

1. Let \(x\) be the original number of tablets and \(y\) the original number of laptops. 2. The original inventory value gives \(300x + 800y = 110{,}000\). 3. The sales revenue gives \(0.40 \cdot 300x + 0.60 \cdot 800y = 60{,}000\). 4. Simplify the second equation: \(120x + 480y = 60{,}000\), or \(x + 4y = 500\). 5. Solve for \(x\): \(x = 500 - 4y\). 6. Substitute into the inventory equation: \(300(500 - 4y) + 800y = 110{,}000\). 7. Simplify: \(150{,}000 - 1200y + 800y = 110{,}000\), so \(-400y = -40{,}000\) and \(y = 100\). 8. Then \(x = 500 - 4 \cdot 100 = 100\). 9. The store keeps \(60\%\) of the tablets and \(40\%\) of the laptops, so \(60\) tablets and \(40\) laptops remain.

After the sale, \(60\) tablets and \(40\) laptops remain in stock.

- Interpret worker-hours as the number of workers multiplied by the hours per worker. - Choose variables for the two unknown planned quantities. - Express the same total work for each staffing scenario. - Expand the equations so the common product cancels. - Verify that both alternative staffing plans produce the same total worker-hours.

1. Let \(n\) be the planned number of workers and \(t\) the planned hours per worker. The total work is \(nt\) worker-hours. 2. Equal total work gives \(nt = (n + 4)(t - 1)\) and \(nt = (n - 3)(t + 1.5)\). 3. Expand and simplify the first equation: \(nt = nt - n + 4t - 4\), so \(-n + 4t = 4\). 4. Expand and simplify the second equation: \(nt = nt + 1.5n - 3t - 4.5\), so \(1.5n - 3t = 4.5\). 5. From the first equation, \(n = 4t - 4\). Substitute into the second: \(1.5(4t - 4) - 3t = 4.5\). 6. Simplify: \(6t - 6 - 3t = 4.5\), so \(3t = 10.5\) and \(t = 3.5\). 7. Then \(n = 4 \cdot 3.5 - 4 = 10\). 8. The total work is \(nt = 10 \cdot 3.5 = 35\) worker-hours.

The plan called for \(10\) workers at \(3.5\) hours each, for a total of \(35\) worker-hours.

- Represent each pump’s hourly rate as a fraction of the full reservoir. - Translate each operating schedule into an equation. - Write \(40\%\) as a decimal or fraction. - Once you know a pump’s hourly rate, use its reciprocal to find the total filling time.

1. Let \(x\) and \(y\) be the fractions of the reservoir filled per hour by Pumps 1 and 2. 2. Write the system \(4(x + y) = \frac{1}{3}\) and \(6x + 3y = 0.4\). 3. Simplify the first equation: \(x + y = \frac{1}{12}\). 4. Solve for \(y\): \(y = \frac{1}{12} - x\). Substitute into the second equation: \(6x + 3\left(\frac{1}{12} - x\right) = \frac{2}{5}\). 5. Simplify: \(3x + \frac{1}{4} = \frac{2}{5}\), so \(3x = \frac{3}{20}\) and \(x = \frac{1}{20}\). 6. Then \(y = \frac{1}{12} - \frac{1}{20} = \frac{1}{30}\). 7. Time is the reciprocal of rate, so Pump 1 takes \(20\) hours and Pump 2 takes \(30\) hours.

Pump 1 would take \(20\) hours, and Pump 2 would take \(30\) hours.

- Write a system using each crew’s fraction of the job completed per hour. - Translate one-third and one-half into the equations. - For part b, determine the time required at the combined rate. - Compare the combined time with the time of the faster crew.

1. Let \(x\) and \(y\) be the fractions of the job completed per hour by Crews 1 and 2. 2. Write the system \(2(x + y) = \frac{1}{3}\) and \(4x + y = \frac{1}{2}\). 3. The first equation gives \(x + y = \frac{1}{6}\), so \(y = \frac{1}{6} - x\). 4. Substitute into the second equation: \(4x + \frac{1}{6} - x = \frac{1}{2}\). 5. Simplify: \(3x = \frac{1}{3}\), so \(x = \frac{1}{9}\). Then \(y = \frac{1}{18}\). 6. Crew 1 takes \(9\) hours alone, and Crew 2 takes \(18\) hours alone. 7. Their combined rate is \(\frac{1}{9} + \frac{1}{18} = \frac{1}{6}\), so together they take \(6\) hours. 8. Compared with the faster crew’s \(9\) hours, the time saved is \(9 - 6 = 3\) hours.

a) Crew 1 would take \(9\) hours, and Crew 2 would take \(18\) hours. b) Working together saves \(3\) hours.

- Write an equation for the salt in the initial mixture. - Write a second salt-balance equation for the mixture after the additional solution is added. - For part b, compare the concentration being added with the concentration of the initial mixture.

1. Let \(x\) be the volume of the \(5\%\) solution and \(y\) the volume of the \(20\%\) solution, in liters. 2. The initial mixture gives \(0.05x + 0.20y = 0.10(x + y)\). Simplifying gives \(0.10y = 0.05x\), so \(x = 2y\). 3. After adding \(3\,\text{L}\) of the \(20\%\) solution, the salt balance is \(0.10(x + y) + 0.20 \cdot 3 = 0.15(x + y + 3)\). 4. Substitute \(x = 2y\): \(0.10(3y) + 0.60 = 0.15(3y + 3)\). 5. Simplify: \(0.30y + 0.60 = 0.45y + 0.45\), so \(0.15 = 0.15y\) and \(y = 1\). 6. Then \(x = 2y = 2\). 7. For part b, the added solution is less concentrated than the \(10\%\) mixture. Adding it dilutes the mixture, so the concentration falls below \(10\%\).

a) The initial mixture used \(2\,\text{L}\) of the \(5\%\) solution and \(1\,\text{L}\) of the \(20\%\) solution. b) The concentration would decrease below \(10\%\).

- Choose variables for the original amount of each drink. - Write an equation for the amount of apple juice in the original mixture. - Use the first equation to relate the two amounts. - Write a second equation after increasing both amounts by \(3\,\text{gal}\).

1. Let \(x\) be the volume of Drink A and \(y\) the volume of Drink B, in gallons. 2. The original mixture gives \(0.10x + 0.40y = 0.20(x + y)\). Simplifying gives \(0.20y = 0.10x\), so \(x = 2y\). 3. The larger mixture gives \(0.10(x + 3) + 0.40(y + 3) = 0.225(x + y + 6)\). 4. Substitute \(x = 2y\): \(0.10(2y + 3) + 0.40(y + 3) = 0.225(3y + 6)\). 5. Simplify: \(0.60y + 1.50 = 0.675y + 1.35\). 6. Then \(0.15 = 0.075y\), so \(y = 2\). 7. Therefore, \(x = 2y = 4\).

The original plan used \(4\,\text{gal}\) of Drink A and \(2\,\text{gal}\) of Drink B.

- Determine what percent of each full dispenser is removed to fill the pitcher. - Express \(A\) and \(B\) in terms of \(C\). - Translate “\(20\) more than four times” into an equation. - Substitute and solve.

1. Filling \(C\) from \(A\) removes \(40\%\) of \(A\), so \(C = 0.40A\), or \(A = 2.5C\). 2. Filling \(C\) from \(B\) removes \(25\%\) of \(B\), so \(C = 0.25B\), or \(B = 4C\). 3. The total-capacity condition is \(A + B = 4C + 20\). 4. Substitute the expressions for \(A\) and \(B\): \(2.5C + 4C = 4C + 20\). 5. Solving gives \(2.5C = 20\), so \(C = 8\). 6. Then \(A = 2.5(8) = 20\) and \(B = 4(8) = 32\).

Dispenser \(A\) holds \(20\) gallons, dispenser \(B\) holds \(32\) gallons, and pitcher \(C\) holds \(8\) gallons.

- Solve the system while treating \(k\) as a constant. - Write the conditions \(x > 0\) and \(y > 0\) as inequalities in \(k\). - Find the overlap of the two resulting intervals.

1. Solve equation (II) for \(x\): \(x = k - 2y\). 2. Substitute into equation (I): \(4(k - 2y) + 3y = 24\), so \(y = \frac{4k - 24}{5}\). 3. Then \(x = k - 2y = \frac{48 - 3k}{5}\). 4. Require \(y > 0\): \(4k - 24 > 0\), so \(k > 6\). 5. Require \(x > 0\): \(48 - 3k > 0\), so \(k < 16\). 6. Both conditions hold when \(6 < k < 16\).

\(6 < k < 16\)

- Write one equation for the number of vehicles and one for the number of wheels. - Treat \(n\) and \(r\) as fixed quantities while solving for \(m\) and \(a\). - Isolate one of the vehicle variables and substitute. - Consider the minimum and maximum possible number of wheels for \(n\) vehicles, and determine when the formulas are integers.

1. Write the system \(m + a = n\) and \(2m + 4a = r\). 2. Solve the first equation for \(m\): \(m = n - a\). 3. Substitute into the second equation: \(2(n - a) + 4a = r\). 4. Simplify: \(2n + 2a = r\), so \(a = \frac{r - 2n}{2}\). 5. Substitute into \(m = n - a\): \(m = n - \frac{r - 2n}{2} = \frac{4n - r}{2}\). 6. For nonnegative counts, \(a \ge 0\) requires \(r \ge 2n\), and \(m \ge 0\) requires \(r \le 4n\). 7. For whole-number counts, \(n\) and \(r\) must be nonnegative integers and \(r\) must be even. Thus the conditions are \(n, r \in \mathbb{Z}_{\ge 0}\), \(2n \le r \le 4n\), and even \(r\).

The formulas are \(a = \frac{r - 2n}{2}\) and \(m = \frac{4n - r}{2}\). They give nonnegative whole numbers when \(n, r \in \mathbb{Z}_{\ge 0}\), \(2n \le r \le 4n\), and \(r\) is even.

- Write one equation for total mass and one for the mass of salt. - Convert each percent concentration into a fraction over \(100\). - Use the total-mass equation to replace one unknown. - Treat \(a\), \(b\), \(k\), and \(M\) as constants while solving for \(x\) and \(y\).

1. Let \(x\) be the mass of Solution A and \(y\) the mass of Solution B, in grams. 2. The total-mass equation is \(x + y = M\). 3. The salt-mass equation is \(\frac{a}{100}x + \frac{b}{100}y = \frac{k}{100}M\). Multiplying by \(100\) gives \(ax + by = kM\). 4. Solve the first equation for \(y\): \(y = M - x\). 5. Substitute into the salt equation: \(ax + b(M - x) = kM\). 6. Simplify: \((a - b)x = M(k - b)\), so \(x = \frac{M(k - b)}{a - b}\). 7. Use \(y = M - x\): \(y = M - \frac{M(k - b)}{a - b} = \frac{M(a - k)}{a - b}\). 8. Because \(a > k > b\), both calculated masses are positive.

The required masses are \(x = \frac{M(k - b)}{a - b}\,\text{g}\) of Solution A and \(y = \frac{M(a - k)}{a - b}\,\text{g}\) of Solution B.

- Express \(x\) and \(y\) in terms of \(a\). - A Quadrant II point must have a negative x-coordinate and a positive y-coordinate. - Determine when each rational expression has the required sign. - Remember to exclude parameter values that make a denominator zero.

1. Solve the first equation for \(x\): \(x=5-y\). 2. Substitute into the second equation: \(2(5-y)+ay=12\). 3. Simplify: \(10-2y+ay=12\), so \((a-2)y=2\). Thus \(y=\frac{2}{a-2}\), where \(a\neq2\). 4. Substitute into \(x=5-y\): \(x=5-\frac{2}{a-2}=\frac{5a-12}{a-2}\). 5. Since \(y>0\), \(\frac{2}{a-2}>0\), which requires \(a>2\). 6. With \(a>2\), the denominator of \(x\) is positive. Therefore, \(x<0\) requires \(5a-12<0\), so \(a<\frac{12}{5}\). 7. Combining the conditions gives \(2<a<\frac{12}{5}\).

\(2<a<\frac{12}{5}\)