Free Math Worksheets

- A solution of a system must satisfy every equation in the system. - The first coordinate is the \(x\)-value and the second is the \(y\)-value. - Satisfying only one equation is not enough.

1. For part a, substitute \((2, 1)\) into the first equation: \(3(2) + 1 = 7\). Substitute into the second equation: \(2 - 2(1) = 0\). Both equations are true, so \((2, 1)\) is a solution. 2. For part b, substitute \((3, 4)\) into the first equation: \(4(3) - 3(4) = 0\), not \(2\). Although the second equation is true because \(2(3) + 4 = 10\), the ordered pair does not satisfy both equations, so it is not a solution of the system.

a) Yes, \((2, 1)\) is a solution. b) No, \((3, 4)\) is not a solution.

- A point lies on a line when its coordinates satisfy the line equation. - An intersection point must lie on both lines. - Test the point in each equation.

1. An intersection point must satisfy both equations. 2. Test \(P(4, 2)\) in line \(g\): \(1.5(4) - 4 = 6 - 4 = 2\). The point lies on \(g\). 3. Test the point in line \(h\): \(-4 + 6 = 2\). The point also lies on \(h\). 4. Therefore, \(P(4, 2)\) is the intersection point, and the claim is true.

The claim is true because \(P(4, 2)\) satisfies both equations.

- Rewrite the equations in comparable forms. - What does it mean graphically when two equations are equivalent? - Identical left sides with different constants create a contradiction. - Can distinct parallel lines intersect?

1. For part a, rewrite the first equation: \(y = 2x - 5\) is equivalent to \(4x - 2y = 10\). The equations represent the same line, so the system has infinitely many solutions. 2. For part b, the left sides are identical, but the constants are different. This would require \(7 = -2\), which is impossible. The equations represent distinct parallel lines, so the system has no solution.

a) Infinitely many solutions b) No solution

- Relate the number of solutions to the positions of the lines. - Rewrite equations in slope-intercept form. - Check whether one equation can be obtained from the other by scaling.

1. For part a, rewrite the second equation: \(3x - 2y = 4\) becomes \(y = 1.5x - 2\). The equations are identical, so there are infinitely many solutions. 2. For part b, the equations have identical left sides but different constants. They represent distinct parallel lines, so there is no solution.

a) Infinitely many solutions b) No solution

- Isolate \(y\) in each equation. - What does it mean when two slope-intercept equations are identical? - How many points do coincident lines share?

1. Rewrite equation (I): \(y = \frac{2}{3}x - 2\). 2. Rewrite equation (II): \(6y = 4x - 12\), so \(y = \frac{2}{3}x - 2\). 3. The slope-intercept forms are identical. Therefore, the equations represent the same line and the system has infinitely many solutions.

a) Both equations become \(y = \frac{2}{3}x - 2\). b) The system has infinitely many solutions because the two lines coincide.

- Compare the slopes of the two lines. What does it mean for the number of solutions when the slopes are equal? - If two lines never intersect, can any ordered pair satisfy both equations?

1. Line \(g\) has \(y\)-intercept \(1\) and slope \(0.5\), so its equation is \(y = 0.5x + 1\). 2. Line \(h\) has \(y\)-intercept \(-2\) and slope \(0.5\), so its equation is \(y = 0.5x - 2\). 3. The lines have the same slope but different \(y\)-intercepts, so they are distinct parallel lines. 4. Because the lines have no point in common, the system has no solution. Its solution set is \(\emptyset\).

(I) \(y = 0.5x + 1\) (II) \(y = 0.5x - 2\) Solution set: \(\emptyset\)

- Which line is higher at \(x = 0\)? - How do the y-values change as you move to the right, based on the slopes? - Which quadrant has positive x-values and negative y-values?

1. At \(x = 0\), the first line has y-value \(-2\) and the second has y-value \(-7\), so the first line is above the second. 2. The first line has negative slope \(-3\), while the second has positive slope \(2\). Moving to the right, the first line decreases and the second increases, so they must meet at a positive x-value. 3. At that positive x-value, the first equation gives \(y = -3x - 2 < 0\). 4. Therefore, the intersection has \(x > 0\) and \(y < 0\), so it lies in Quadrant IV.

The intersection point lies in Quadrant IV.

- Compare the slopes in each system. - If the lower line at \(x = 0\) is going to catch the upper line for positive \(x\), which line must rise faster? - What happens when the lower line has the smaller slope instead?

1. In both systems, at \(x = 0\) the first line is at \(y = 4\) and the second line is at \(y = -1\), so the first line starts above the second. 2. In System A, the second line has slope \(3\), which is greater than \(0.5\). It rises faster, so it catches the first line for a positive x-value. 3. In System B, the second line has slope \(0.2\), which is less than \(0.5\). Moving right, it falls farther behind the first line, so the intersection must occur for a negative x-value. 4. Therefore, System B has an intersection with a negative x-coordinate.

System B

- Rewrite each equation in slope-intercept form. - What must be true about the slopes of distinct parallel lines? - What does parallelism mean for the number of solutions of the system?

1. Rewrite equation (I): \(y = 2x + 4\). Its slope is \(m_1 = 2\). 2. Rewrite equation (II): \(-2y = -4x + 6\), so \(y = 2x - 3\). Its slope is \(m_2 = 2\). 3. The lines have the same slope but different y-intercepts, \(4\) and \(-3\). Therefore, they are distinct parallel lines. 4. Distinct parallel lines do not intersect, so the corresponding system has no solution.

Both lines have slope \(2\). Because their y-intercepts are different, the lines are parallel and have no intersection point. The system has no solution.

- A solution must satisfy both equations. - Substitute the given coordinates into the equation containing \(k\). - Solve the resulting one-variable equation.

1. Because \((5, 2)\) is a solution, it must satisfy equation (II). 2. Substitute \(x = 5\) and \(y = 2\): \(3(5) - 2k = 11\). 3. Simplify: \(15 - 2k = 11\), so \(-2k = -4\). 4. Divide by \(-2\): \(k = 2\).

\(k = 2\)

- Write both equations in slope-intercept form. - When do two equations represent the same line? - What conditions make two lines parallel? - Could these two lines ever intersect once?

1. Solve equation (I) for \(y\): \(6x - 3y = 12\) becomes \(y = 2x - 4\). 2. For infinitely many solutions, equation (II), \(y = 2x + k\), must represent the same line. Therefore, \(k = -4\). 3. For \(k \ne -4\), the two lines have the same slope \(2\) but different \(y\)-intercepts. They are distinct parallel lines, so the system has no solution.

For \(k = -4\), the system has infinitely many solutions. For every \(k \ne -4\), it has no solution.

- When do two equations represent the same line? - What coefficient relationships make two lines parallel? - Scale one equation so that its \(y\)-coefficient matches the other.

1. Multiply equation (I) by \(-2\): \(-10x + 4y = -20\). 2. For infinitely many solutions, equation (II) must be identical to this equation. Therefore, \(a = -10\) and \(b = -20\). 3. For no solution, the variable coefficients must remain proportional while the constant is not. Therefore, \(a = -10\) and \(b \ne -20\).

a) \(a = -10\) and \(b = -20\) b) \(a = -10\) and \(b \ne -20\)

- In \(y = mx + b\), which value gives the slope? - When do two lines fail to intersect? - Which solving method is efficient when both equations are solved for \(y\)?

1. A system of two distinct parallel lines has no solution. Equation (I) has slope \(3\), so equation (II) is parallel when \(k = 3\). Their \(y\)-intercepts, \(-4\) and \(2\), are different, so the lines are distinct. 2. When \(k = 1\), set the expressions for \(y\) equal: \(3x - 4 = x + 2\). 3. Solve: \(2x = 6\), so \(x = 3\). 4. Substitute into equation (I): \(y = 3(3) - 4 = 5\).

a) \(k = 3\) b) \((3, 5)\)

- Write both equations in slope-intercept form. - What does it mean when two lines have the same slope? - Compare their \(y\)-intercepts. - How many points do distinct parallel lines share?

1. Rewrite equation (I) in slope-intercept form: \(6x - 2y = 4\) becomes \(y = 3x - 2\). 2. The two lines have the same slope, \(3\), but different \(y\)-intercepts, \(-2\) and \(5\). 3. Therefore, the lines are distinct and parallel. They have no intersection point, so the system has no solution.

The system has no solution. The lines have \(0\) intersection points.

- Rewrite equations in comparable forms. - What happens when identical expressions are set equal to different constants? - Compare the slopes of the corresponding lines.

1. For system (A), rewrite the second equation as \(y = 2x + 5\). The equations are identical, so there are infinitely many solutions. 2. In system (B), the left sides are identical but the constants are different. The system is inconsistent, so there is no solution. 3. In system (C), the two lines have different slopes. Therefore, they intersect once and the system has exactly one solution.

(A) Infinitely many solutions (B) No solution (C) Exactly one solution

- Multiply the entire first equation by \(2\). - Compare the result with each second equation. - Must the constants also be proportional?

1. In Example 1, multiplying the first equation by \(2\) gives \(2x + 4y = 10\), exactly the second equation. The system has infinitely many solutions. 2. In Example 2, doubling the first equation would give \(2x + 4y = 10\), not \(12\). The equations represent distinct parallel lines, so the system has no solution. 3. The rule is incomplete. The constant must be multiplied by the same factor as the variable coefficients for the equations to be equivalent.

The student’s rule is incomplete. Infinitely many solutions occur only when every term, including the constant, is scaled by the same factor. Otherwise, the system may have no solution.

- Make a small table of values for each equation. - Compare the slopes in the two equations. - What does having the same slope mean for two distinct lines?

1. For \(g\), when \(x = 0\), \(y = -4\), giving \((0, -4)\). When \(x = 2\), \(y = 3 \cdot 2 - 4 = 2\), giving \((2, 2)\). 2. For \(h\), when \(x = 0\), \(y = 2\), giving \((0, 2)\). When \(x = 1\), \(y = 3 \cdot 1 + 2 = 5\), giving \((1, 5)\). 3. Both lines have slope \(3\), but their y-intercepts are different, so the lines are parallel. 4. Parallel distinct lines do not intersect, so there is no ordered pair that satisfies both equations. Algebraically, setting the expressions equal gives \(3x - 4 = 3x + 2\), which simplifies to the contradiction \(-4 = 2\).

a) One possible choice is \((0, -4)\) and \((2, 2)\) for \(g\), and \((0, 2)\) and \((1, 5)\) for \(h\). b) The lines are parallel. c) No. They have the same slope, \(3\), but different y-intercepts, so they never intersect.

- Rewrite the second equation in slope-intercept form. - When do two equations represent the same line? - What happens when slopes match but intercepts differ?

1. Solve equation (II) for \(y\): \(y = 2x - \frac{k}{2}\). 2. For infinitely many solutions, the two equations must be identical. Set the intercepts equal: \(-\frac{k}{2} = -4\), so \(k = 8\). 3. When \(k = 10\), equation (II) becomes \(y = 2x - 5\). The lines have equal slopes but different intercepts, so they are distinct parallel lines and the system has no solution.

a) \(k = 8\) b) No solution

- Compare the slopes of the functions first. - What happens when two distinct lines have the same slope? - When solving the equation with fractions, multiplying both sides by a common denominator may help.

1. The functions \(f\) and \(g\) have the same slope, \(\frac{2}{3}\), but different y-intercepts, \(2\) and \(-5\). Therefore, their graphs are distinct parallel lines and do not intersect. 2. To find the intersection of \(g\) and \(h\), set their expressions equal: \(\frac{2}{3}x - 5 = -x + 7\). 3. Add \(x\) and \(5\) to both sides: \(\frac{5}{3}x = 12\). 4. Multiply by \(\frac{3}{5}\): \(x = 12 \cdot \frac{3}{5} = \frac{36}{5} = 7.2\). 5. Substitute into \(h\): \(y = -7.2 + 7 = -0.2\). The intersection is \((7.2, -0.2)\).

a) The graphs are parallel because they have the same slope, \(\frac{2}{3}\), but different y-intercepts. b) The intersection is \((7.2, -0.2)\).

- Connect the number of solutions to the relative positions of two lines. - Compare slopes and intercepts. - Rewrite standard-form equations in slope-intercept form when useful.

1. In system (1), the slopes are equal but the \(y\)-intercepts differ. The lines are distinct and parallel, so there is no solution. 2. In system (2), the slopes are different. The lines intersect once, so there is exactly one solution. 3. In system (3), rewrite the second equation as \(y = 2x + 5\). The equations are identical, so there are infinitely many solutions.

(1) No solution (2) Exactly one solution (3) Infinitely many solutions

- Can you find other pairs whose difference is \(5\)? - A unique solution requires a second line that intersects the first at the desired point. - Can the same expression \(a - b\) equal \(5\) and \(8\) at the same time?

1. The equation \(a - b = 5\) has infinitely many solutions, such as \((10, 5)\), \((6, 1)\), and \((0, -5)\). Thus, \((10, 5)\) is only one solution. 2. A second equation that intersects the first at \((10, 5)\) can make the solution unique. For example, \(a + b = 15\). Together, the two equations have the single solution \((10, 5)\). 3. The equations \(a - b = 5\) and \(a - b = 8\) cannot both be true. Subtracting them gives the contradiction \(0 = 3\), so the system has no solution.

1. The equation has infinitely many solutions, so \((10, 5)\) is only one of them. 2. One possible additional equation is \(a + b = 15\). 3. The system has no solution.

- What do you notice about the slopes of the two lines? - Can two lines be positioned so that they never intersect? - How many common points must two lines have for the system to have a solution? - What does no intersection mean for the solution set?

1. Graph the first line, which has slope \(2\) and \(y\)-intercept \(-1\). 2. Graph the second line, which has slope \(2\) and \(y\)-intercept \(2\). 3. The lines have the same slope but different \(y\)-intercepts, so they are distinct parallel lines. 4. Because the lines never intersect, the system has no solution. 5. Therefore, the solution set is \(\emptyset\).

a) The graph shows two parallel lines with slope \(2\) and \(y\)-intercepts \(-1\) and \(2\). b) The solution set is \(\emptyset\) because the lines do not intersect.

- Multiply the first equation to remove decimals. - Compare the resulting left sides. - What does a contradiction such as \(0 = 1\) mean? - Can the same expression equal two different numbers?

1. Multiply the first equation by \(5\): \(2x - 3y = 6\). 2. Compare it with the second equation, \(2x - 3y = 5\). 3. Subtracting the equations gives \(0 = 1\), a contradiction. 4. Therefore, the system has no solution.

The system has no solution.

- Solve one equation for a variable when substitution is convenient. - Simplify each equation by dividing by a common factor. - How many intersections can two lines have? - Check the solution to part a in both original equations.

1. For system a, solve the second equation for \(y\): \(y = 5x - 3\). Substitute into the first equation: \(3x + 2(5x - 3) = 7\), so \(13x = 13\) and \(x = 1\). Then \(y = 2\). 2. For system b, divide the first equation by \(3\) and the second by \(2\). Both become \(2x - 3y = 4\). The equations represent the same line, so there are infinitely many solutions.

a) \(\{(1, 2)\}\) b) Infinitely many solutions: \(\{(x, y) \mid 2x - 3y = 4\}\)

- Clear fractions or decimals before comparing or solving. - A contradiction means the solution set is empty. - Choose an efficient solving method after simplifying. - Track signs carefully.

1. For system a, multiply the first equation by \(6\): \(3x + 4y = 30\). This contradicts \(3x + 4y = 20\), so the system has no solution. 2. For system b, multiply both equations by \(10\): \(4x - 3y = 11\) and \(5x + 2y = 8\). 3. Multiply the first new equation by \(2\) and the second by \(3\): \(8x - 6y = 22\) and \(15x + 6y = 24\). 4. Add: \(23x = 46\), so \(x = 2\). Substitute into \(5x + 2y = 8\): \(10 + 2y = 8\), so \(y = -1\).

a) No solution; the solution set is empty. b) The solution set is \(\{(2, -1)\}\).

- Relate the number of solutions to the positions of the lines. - Compare slopes first, then intercepts. - Equivalent equations describe the same line. - When both equations are solved for \(y\), set their expressions equal.

1. In part a, the lines have the same slope but different intercepts, so there is no solution. 2. In part b, the slopes are different, so there is exactly one solution. 3. In part c, the second equation is twice the first, so the lines are identical and there are infinitely many solutions. 4. Solve part b by setting the expressions equal: \(-x + 4 = 2x - 2\). Then \(3x = 6\), so \(x = 2\), and \(y = 2\).

a) No solution b) Exactly one solution, \((2, 2)\) c) Infinitely many solutions

- Solve one equation for a variable and substitute. - Be careful with signs when solving equation (II) for \(y\). - Compare the slopes in the two slope-intercept forms. - How do different slopes determine the number of intersections?

1. From equation (I), \(y = -\frac{1}{2}x + 2\). 2. Substitute into equation (II): \(3x - 2(-\frac{1}{2}x + 2) = 12\). This gives \(4x - 4 = 12\), so \(x = 4\). 3. Substitute into equation (I): \(y = -\frac{1}{2}(4) + 2 = 0\). The solution is \((4, 0)\). 4. The slope-intercept forms are \(y = -0.5x + 2\) and \(y = 1.5x - 6\). 5. Their slopes are different, so the lines intersect at exactly one point.

1. \((4, 0)\) 2. \(y = -0.5x + 2\) and \(y = 1.5x - 6\) 3. The slopes are different, so the system has exactly one solution.

- Translate each condition into an equation. - Factor the second equation. - Substitute the expression from the first equation. - Does the result produce a true statement or a contradiction?

1. Let \(x\) and \(y\) be the two numbers. The conditions give \(x + y = 12\) and \(4x + 4y = 50\). 2. Factor the second equation: \(4(x + y) = 50\). 3. Substitute \(x + y = 12\): \(4(12) = 50\), which gives the false statement \(48 = 50\). 4. Therefore, the system is inconsistent and no such pair exists.

No such pair exists because the conditions lead to the contradiction \(48 = 50\).

- Write one equation for each price statement. - Compare the coefficients and constants. - What does it mean when one equation is a multiple of the other? - Could more than one pair of ticket prices satisfy both statements?

1. Let \(x\) be the adult ticket price and \(y\) the child ticket price, in dollars. 2. The system is \(3x + 5y = 110\) and \(9x + 15y = 330\). 3. Multiplying the first equation by \(3\) gives the second equation exactly. 4. The equations contain the same information, so the system has infinitely many solutions. Therefore, the adult ticket price cannot be determined uniquely.

No. The second equation is three times the first, so there are infinitely many possible price pairs and the adult ticket price is not unique.

- Write the perimeter formula for a rectangle. - Express both requirements as equations. - Simplify the perimeter equation before comparing. - What does a contradiction mean for the construction?

1. Let \(l\) be the length and \(w\) the width, in centimeters. 2. The perimeter condition gives \(2(l + w) = 80\), so \(l + w = 40\). 3. The second condition gives \(l + w = 45\). 4. The same sum cannot equal both \(40\) and \(45\). The system has no solution, so the frame cannot be built.

No. A perimeter of \(80\,\text{cm}\) requires \(l + w = 40\,\text{cm}\), which contradicts \(l + w = 45\,\text{cm}\).

- Use elimination to compare the equations. - Examine the ratios of corresponding coefficients. - Does the elimination produce an identity or a contradiction? - What does a contradiction mean for the solution set?

1. Multiply the first equation by \(2\): \(24x + 36y = 48\). 2. Multiply the second equation by \(3\): \(24x + 36y = 60\). 3. Subtracting gives \(0 = 12\), a contradiction. 4. Therefore, the system has no solution.

The solution set is empty.

- Compare the \(x\)-coefficients first. - Every term must be scaled by the same factor for equivalent equations. - When do proportional variable coefficients create parallel lines? - When do they create the same line?

1. The \(x\)-coefficient in equation (II) is three times the coefficient in equation (I). For proportional variable coefficients, \(a\) must also be three times \(4\), so \(a = 12\). 2. For infinitely many solutions, the constant must be scaled by the same factor: \(c = 3(12) = 36\). 3. With \(a = 12\), any \(c \ne 36\) gives proportional left sides but nonproportional constants. The lines are distinct and parallel, so there is no solution.

a) \(a = 12\) b) \(c = 36\) c) \(c \ne 36\)

- Rewrite each equation in slope-intercept form. - Compare slopes and intercepts. - Identical equations describe the same line. - Set the two expressions for \(y\) equal in the unique-solution case.

1. In system I, the second equation becomes \(y = 1.5x - 2\). The slopes match but the intercepts differ, so there is no solution. 2. In system II, both equations become \(y = 0.5x - 1.5\). The lines coincide, so there are infinitely many solutions. 3. In system III, the equations become \(y = -2x + 5\) and \(y = x - 1\). The slopes differ, so there is exactly one solution. 4. Set the expressions equal: \(-2x + 5 = x - 1\). Then \(3x = 6\), so \(x = 2\) and \(y = 1\).

I: no solution II: infinitely many solutions III: exactly one solution, \(\{(2, 1)\}\)

- When do two equations describe the same line? - Corresponding coefficients and constants must have the same scale factor. - Find the factor that changes \(4\) into \(6\).

1. The scale factor from the \(x\)-coefficient in equation (I) to equation (II) is \(\frac{6}{4} = 1.5\). 2. Scale the \(y\)-coefficient by the same factor: \(a = 6 \cdot 1.5 = 9\). 3. Scale the constant by the same factor: \(b = 10 \cdot 1.5 = 15\). 4. With these values, the equations are equivalent and the system has infinitely many solutions.

\(a = 9\) and \(b = 15\)

- Multiply equation (I) by \(-2\). - What must happen to the constant for the equations to remain equivalent? - What does it mean when variable coefficients are proportional but constants are not?

1. The left side of equation (II) is \(-2\) times the left side of equation (I). 2. For equivalent equations, the constant must also be multiplied by \(-2\): \(c = -2 \cdot 5 = -10\). 3. If \(c = 10\), the variable coefficients remain proportional but the constant has the wrong scale factor. The lines are distinct and parallel, so the system has no solution.

a) \(c = -10\) b) No solution

- Scale equation (II) so that its \(y\)-coefficient matches equation (I). - When do lines have different slopes? - When are proportional equations identical or inconsistent?

1. Multiply equation (II) by \(2\): \(2ax + 6y = 2c\). 2. If \(a \ne 2\), the \(x\)-coefficients differ, so the lines have different slopes and the system has exactly one solution for any \(c\). 3. If \(a = 2\), the transformed left side matches equation (I). When \(c = 6\), the constants also match and there are infinitely many solutions. 4. If \(a = 2\) and \(c \ne 6\), the lines are distinct and parallel, so there is no solution.

1) \(a \ne 2\), with any \(c\) 2) \(a = 2\), \(c \ne 6\) 3) \(a = 2\), \(c = 6\)

- Rewrite both equations in slope-intercept form. - What must match for two lines to coincide? - What happens when slopes are equal but intercepts differ? - How many intersections do distinct parallel lines have?

1. Rewrite line \(h\): \(y = 1.5x - 0.5b\). 2. The lines always have the same slope, \(1.5\). 3. For identical lines, their intercepts must match: \(4 = -0.5b\), so \(b = -8\). Then they have infinitely many points in common. 4. When \(b = 2\), line \(h\) has intercept \(-1\). The slopes match but the intercepts differ, so the lines are distinct and parallel and have no intersection points.

a) \(b = -8\) b) The lines have \(0\) intersection points.

- Scale one equation so that the variable coefficients match. - Remember to scale the constant as well. - What does a contradiction mean for the number of solutions?

1. Multiply equation (II) by \(-2\): \(2x - 4y = -10\). 2. The transformed equation and equation (I) have identical left sides but different constants. 3. This produces a contradiction, so the system has no solution.

The system has no solution.

- Simplify the first equation. - When are two equations identical? - What does a contradiction imply? - How are lines positioned when slopes match but intercepts differ?

1. Divide equation (I) by \(3\): \(2x - y = 3\). 2. For identical equations, \(k = 3\). Then the system has infinitely many solutions. 3. If \(k = 5\), the equations are \(2x - y = 3\) and \(2x - y = 5\). Subtracting gives \(0 = -2\), a contradiction. 4. The lines have equal slopes and different intercepts, so they are distinct and parallel.

a) \(k = 3\) b) The equations produce the contradiction \(0 = -2\), so there is no solution. c) The lines are distinct and parallel.

- Solve each second equation for \(y\). - Compare the two equations in each system. - When do equal slopes produce distinct parallel lines? - Can equal slopes also describe the same line?

1. In system A, \(x - 2y = -4\) becomes \(y = 0.5x + 2\). 2. In system B, \(x - 2y = 6\) becomes \(y = 0.5x - 3\). 3. In system A, the equations are identical, so there are infinitely many solutions. 4. In system B, the slopes match but the intercepts differ, so there is no solution. 5. The claim is incomplete. Equal slopes give no solution only when the lines have different intercepts. If the intercepts also match, the lines are identical and there are infinitely many solutions.

1. System A: \(y = 0.5x + 2\); System B: \(y = 0.5x - 3\) 2. System A has infinitely many solutions; System B has no solution. 3. Equal slopes give no solution only when the intercepts differ.

- Equivalent equations represent the same line. - Equal variable coefficients with different constants indicate parallel lines. - Compare the ratios of corresponding coefficients. - Does scaling an entire equation change its solution set?

1. In system 1, the first equation is twice the second, including the constant. The equations are equivalent, so there are infinitely many solutions. 2. In system 2, the left sides are identical but the constants differ. The system has no solution. 3. In system 3, the ratios of the \(x\)- and \(y\)-coefficients are not equal. The lines have different slopes and intersect once, so there is exactly one solution.

1) Infinitely many solutions 2) No solution 3) Exactly one solution

- Simplify one equation and compare it with the other. - Equivalent equations have the same solution set. - Add the equations in system B. - Two nonparallel lines can have at most one intersection.

1. In system A, dividing the first equation by \(-2\) gives \(-2x + y = -3\), exactly the second equation. The system has infinitely many solutions. 2. In system B, add the equations: \(4x = 8\), so \(x = 2\). 3. Substitute into \(x - y = 1\): \(2 - y = 1\), so \(y = 1\). The solution set is \(\{(2, 1)\}\).

System A: infinitely many solutions System B: exactly one solution; \(\{(2, 1)\}\)

- What scale factor connects the variable coefficients? - Apply that same factor to the constant for equivalent equations. - What does a contradiction mean? - Can lines with the same slope intersect exactly once?

1. The variable coefficients in equation (II) are twice those in equation (I). 2. For equivalent equations, the constant must also be doubled: \(a = 20\). Then there are infinitely many solutions. 3. If \(a = 15\), equation (II) contradicts the doubled form \(4x + 10y = 20\), so there is no solution. 4. The variable coefficients are always proportional, so the lines always have the same slope. They are either identical or parallel and can never intersect exactly once.

a) \(a = 20\) b) No solution c) The lines always have the same slope, so exactly one solution is impossible.

- Rewrite both equations in slope-intercept form. - What must match for two lines to coincide? - What must match, and what must differ, for distinct parallel lines?

1. Rewrite equation (I): \(y = \frac{2}{3}x - 2\). Rewrite equation (II): \(y = -\frac{a}{9}x + \frac{b}{9}\). 2. For infinitely many solutions, the slopes and intercepts must match. Thus, \(-\frac{a}{9} = \frac{2}{3}\), so \(a = -6\), and \(\frac{b}{9} = -2\), so \(b = -18\). 3. For no solution, the slopes must match while the intercepts differ. Thus, \(a = -6\) and \(b \ne -18\).

a) \(a = -6\), \(b = -18\) b) \(a = -6\), \(b \ne -18\)

- Rewrite both equations in slope-intercept form. - Which values must match for two lines to be identical? - What does it mean graphically when every point is shared?

1. Rewrite equation (II): \(y = 2x - \frac{d}{2}\). 2. For infinitely many solutions, the equations must represent the same line. Therefore, \(k = 2\). 3. Match the intercepts: \(-\frac{d}{2} = 5\), so \(d = -10\). 4. Geometrically, the two graphs coincide and share every point.

\(k = 2\) and \(d = -10\). The two equations represent the same line.

- Rewrite each equation in slope-intercept form. - Compare slopes and \(y\)-intercepts. - Identify parallel, intersecting, and coincident lines in the three graphs. - Relate each graph type to its number of solutions.

1. In system (1), the second equation becomes \(y = -x - 1\). The lines are parallel and distinct, so the system matches graph B and has no solution. 2. In system (2), the second equation becomes \(y = -x + 2\). The lines are identical, so the system matches graph A and has infinitely many solutions. 3. In system (3), the second equation becomes \(y = 2x - 1\). The lines intersect once at \((1, 1)\), so the system matches graph C and has exactly one solution.

(1) Graph B; no solution (2) Graph A; infinitely many solutions (3) Graph C; exactly one solution

- A point on a line satisfies that line’s equation. - A solution of the entire system must satisfy every equation. - Count points that lie on all three lines, not pairwise intersections. - Would three pairwise intersections provide one point common to every line?

1. A solution of a system is an ordered pair that satisfies every equation at the same time. Graphically, it must lie on all three lines. 2. The graph shows that lines \(g\), \(h\), and \(j\) all pass through \(S(1, 2)\). 3. Therefore, the system has exactly one solution, \((1, 2)\). If the lines had three different pairwise intersection points, no single point would satisfy all three equations.

The statement is incorrect. The system has exactly one solution, \((1, 2)\), because that is the only point common to all three lines.

- A common point of two lines satisfies two equations. - What must the graph show for one point to satisfy all three equations? - Look for a point lying on every line.

1. A point satisfying all three conditions must lie on all three lines. 2. The graph shows that the lines form a triangle. Their pairwise intersections are \((1, 2)\), \((4, 2)\), and \((2, 3)\). 3. No point lies on all three lines. Therefore, the system has no solution.

There are \(0\) operating points that satisfy all three constraints. The system has no solution.

- A point where two lines intersect represents a common solution. - What does “parallel” imply about common points? - If two lines are exactly the same, which points belong to both lines? - The solution set contains every ordered pair that satisfies both equations.

1. In graph 1, the lines intersect at \((2, -1)\). This point satisfies both equations, so the solution set is \(\{(2, -1)\}\). 2. In graph 2, the lines are distinct and parallel. They have no common point, so the solution set is \(\emptyset\). 3. In graph 3, the lines are identical. Every point on the line satisfies both equations, so the system has infinitely many solutions. The solution set is \(\{(x, y) \mid y = 1.5x + 1\}\).

a) \(\{(2, -1)\}\) b) The lines are parallel and have no intersection, so the solution set is \(\emptyset\). c) The system has infinitely many solutions. The solution set is \(\{(x, y) \mid y = 1.5x + 1\}\).

- What y-coordinate does a point on the x-axis have? - When are two distinct lines parallel? - Recall the signs of x- and y-coordinates in each quadrant.

1. For part a, a point on the x-axis has \(y = 0\). From \(0 = -x + 6\), the intersection must have \(x = 6\). Use this in \(h\): \(0 = 6m + 2\), so \(m = -\frac{1}{3}\). 2. For part b, two distinct lines have no intersection when they are parallel. Since \(g\) has slope \(-1\) and the y-intercepts are different, \(m = -1\). 3. For part c, with \(m = 1\), solve \(-x + 6 = x + 2\). Then \(4 = 2x\), so \(x = 2\). Substitution gives \(y = 4\). Since both coordinates are positive, the intersection lies in Quadrant I.

a) \(m = -\frac{1}{3}\) b) \(m = -1\) c) Quadrant I

- A point lies on a line when its coordinates make the equation true. - Substitute the known coordinate of \(D\) and solve for the missing one. - What relationship between two distinct lines makes a system have no solution?

1. For \(A\), \(0.5 \cdot 2 - 0.25 \cdot 0 = 1\), so \(A\) lies on the line. For \(B\), \(0.5 \cdot 0 - 0.25 \cdot (-4) = 1\), so \(B\) lies on the line. For \(C\), \(0.5 \cdot 4 - 0.25 \cdot 3 = 1.25\), so \(C\) does not lie on the line. 2. For \(D\), \(0.5x_D - 0.25 \cdot 2 = 1\). Thus, \(0.5x_D = 1.5\), so \(x_D = 3\). 3. The original equation is equivalent to \(y = 2x - 4\). A line with the same slope and a different y-intercept will have no intersection with it. For example, \(y = 2x - 8\), equivalently \(0.5x - 0.25y = 2\), gives a system with no solution.

a) \(A\) and \(B\) lie on the line; \(C\) does not. b) \(x_D = 3\) c) One possible equation is \(0.5x - 0.25y = 2\). It represents a distinct parallel line, so the system has no solution.

- What must be true about the slopes of two parallel lines? - How many points do two distinct parallel lines have in common? - For part b), graph both lines carefully in the same coordinate plane.

1. A line parallel to \(g\) must have the same slope, \(\frac{1}{2}\), but a different \(y\)-intercept. One possible equation is \(y = \frac{1}{2}x - 2\). 2. The two lines are distinct and parallel, so the system has no solution. 3. For part b), graph \(y = \frac{1}{2}x + 1\) and \(y = -x + 4\). 4. The lines intersect at \(x = 2\). Substitution gives \(y = \frac{1}{2} \cdot 2 + 1 = 2\), so the solution is \((2, 2)\).

a) One possible equation is \(y = \frac{1}{2}x - 2\). The system has no solution. b) The solution of the new system is \((2, 2)\).

- Rewrite the first equation in slope-intercept form. - When do two lines have no common point? - Can changing only the slope make two different fixed intercepts equal? - What must match for two lines to have infinitely many points in common?

1. Rewrite line \(g\): \(2y = -5x + 10\), so \(y = -2.5x + 5\). Its slope is \(-2.5\), and its \(y\)-intercept is \(5\). 2. For the lines to be parallel, line \(h\) must have the same slope. Thus, \(m = -2.5\). Because its \(y\)-intercept is \(3\), not \(5\), the lines are distinct and the system has no solution. 3. Infinitely many solutions would require equal slopes and equal \(y\)-intercepts. The intercept of line \(h\) is fixed at \(3\), so no value of \(m\) can make the lines identical.

a) \(m = -2.5\) b) No. The fixed \(y\)-intercepts, \(5\) and \(3\), are different.

- A specified solution must make both equations true. - Substitute \((3, 2)\) into the general form of equation (II). - What coefficient relationship creates distinct parallel lines?

1. The point \((3, 2)\) satisfies equation (I) because \(2(3) + 3(2) = 12\). 2. To make it satisfy equation (II), substitute the coordinates: \(4(3) + 2d = e\), so \(e = 12 + 2d\). 3. To obtain a unique solution, equation (II) must not be a multiple of equation (I). Since doubling equation (I) gives \(4x + 6y = 24\), choose \(d \ne 6\). For example, \(d = 1\) gives \(e = 14\). 4. For no solution, the variable coefficients must be proportional while the constants are not. Thus, \(d = 6\) and \(e \ne 24\).

a) One possible choice is \(d = 1\) and \(e = 14\). In general, \(e = 12 + 2d\) with \(d \ne 6\). b) \(d = 6\) and \(e \ne 24\)

- What must happen graphically for two lines to have no intersections or infinitely many intersections? - How does multiplying an entire equation by a nonzero number affect its solution set? - What happens if only the constant changes while proportional variable coefficients are kept?

1. For infinitely many solutions, equation (II) must be equivalent to equation (I). Multiplying the entire equation by \(2\) gives \(6x - 4y = 8\). 2. For no solution, the variable coefficients must remain proportional while the constant does not. For example, \(6x - 4y = 10\) represents a distinct parallel line.

a) One possible equation is \(6x - 4y = 8\). b) One possible equation is \(6x - 4y = 10\).

- How must slopes compare for two lines to intersect once? - When are two lines distinct and parallel? - When do two equations represent the same line?

1. Rewrite equation (I) as \(y = \frac{2}{3}x - 2\). 2. For exactly one solution, choose a line with a different slope, such as \(y = x\). 3. For no solution, choose a distinct parallel line, such as \(2x - 3y = 0\). 4. For infinitely many solutions, choose an equivalent equation, such as \(2x - 3y = 6\).

a) One possible equation is \(y = x\). b) One possible equation is \(2x - 3y = 0\). c) One possible equation is \(2x - 3y = 6\).

- When do two lines have no common point? - What happens when two lines have different slopes? - What must match for two equations to represent the same line?

1. For no solution, the lines must be distinct and parallel. Choose \(m = 3\) and any \(n \ne -5\), such as \(n = 0\). 2. For exactly one solution, choose a different slope, so \(m \ne 3\). For example, \(m = 1\) and \(n = -5\). 3. For infinitely many solutions, the equations must be identical. Thus, \(m = 3\) and \(n = -5\).

a) One example is \(m = 3\), \(n = 0\). b) One example is \(m = 1\), \(n = -5\). c) \(m = 3\), \(n = -5\)

- Solve both equations for \(y\). - Compare slopes and intercepts. - Which slope condition guarantees an intersection? - What conditions produce parallel or identical lines?

1. Rewrite equation (I): \(y = -2x + 4\). Rewrite equation (II): \(y = -ax + b\). 2. For infinitely many solutions, the slope and intercept must match: \(a = 2\) and \(b = 4\). 3. For no solution, the slopes must match but the intercepts must differ: \(a = 2\) and \(b \ne 4\). 4. For exactly one solution, the slopes must differ: \(a \ne 2\), with any value of \(b\).

1. \(a = 2\) and \(b = 4\) 2. \(a = 2\) and \(b \ne 4\) 3. \(a \ne 2\), with any \(b\)

- A point of intersection must satisfy both equations. - First find the \(y\)-value when \(x = 2\). - What slope relationship makes two lines parallel?

1. At \(x = 2\), equation (I) gives \(y = 1.5(2) - 3 = 0\). 2. Substitute \((2, 0)\) into equation (II): \(0 = 2m + 2\), so \(m = -1\). 3. For no solution, the lines must be distinct and parallel. Thus, equation (II) must have slope \(1.5\). Because the intercepts \(-3\) and \(2\) differ, \(m = 1.5\) gives no solution.

a) \(m = -1\) b) \(m = 1.5\)

- Compare slopes and \(y\)-intercepts. - What does a shared \(y\)-intercept imply about possible intersections? - Use substitution or elimination for the specific value \(a = 1\).

1. Rewrite equation (I) as \(y = -0.5x + 2\). Rewrite equation (II) as \(y = -\frac{a}{4}x + 2\). 2. For infinitely many solutions, the slopes must match: \(-\frac{a}{4} = -0.5\), so \(a = 2\). The intercepts already match, so the lines are identical. 3. Both lines always have \(y\)-intercept \(2\). Therefore, they can be identical or intersect at \((0, 2)\), but they can never be distinct parallel lines. No value of \(a\) gives no solution. 4. When \(a = 1\), the system is \(3x + 6y = 12\) and \(x + 4y = 8\). Solving gives \(y = 2\) and \(x = 0\).

a) \(a = 2\) b) No value of \(a\) gives no solution. c) \((0, 2)\)

- Simplify the first equation so its \(x\)-coefficient matches the second. - When do matching variable coefficients produce parallel or identical lines? - For part c, use elimination after substituting the parameter values.

1. Divide equation (1) by \(2\): \(2x + 3y = 6\). 2. For part a, the variable coefficients must match while the constants differ. Thus, \(a = 3\). Since \(b = 5 \ne 6\), the system has no solution. 3. For part b, the equations must be identical, so \(a = 3\) and \(b = 6\). 4. For part c, the system is \(4x + 6y = 12\) and \(2x + y = 4\). Multiply the second equation by \(2\): \(4x + 2y = 8\). Subtract to get \(4y = 4\), so \(y = 1\). Then \(2x + 1 = 4\), so \(x = 1.5\).

a) \(a = 3\) b) \(a = 3\), \(b = 6\) c) The solution set is \(\{(1.5, 1)\}\).

- When do two equations represent the same line? - Which terms must be proportional for equivalent equations? - How can you keep a line parallel but move it to a different position?

1. Divide equation (I) by \(2\): \(x + 2y = 4\). 2. For infinitely many solutions, equation (II) must be identical, so \(a = 4\). 3. For no solution, choose a distinct parallel line. For example, \(x + 2y = 5\) has the same variable coefficients but a different constant.

a) \(a = 4\) b) One possible equation is \(x + 2y = 5\).

- Rewrite the equations in comparable forms. - Which coefficients determine whether the slopes match? - How do constants distinguish identical from parallel lines? - What happens when the slopes differ?

1. One solution means the lines intersect once. No solution means they are distinct and parallel. Infinitely many solutions means they are the same line. 2. Rewrite line \(g\) in standard form: \(3x - 2y = -4\). 3. For line \(h\) to be distinct and parallel, its variable coefficients must match those of \(g\), while the constant differs. Thus, \(a = 2\) and \(b \ne -4\). 4. If \(a \ne 2\), the lines are not parallel and intersect once. This includes \(a = 0\), when \(h\) is the vertical line \(x = \frac{b}{3}\). The value of \(b\) changes the location of the intersection but not its existence.

a) One solution: one intersection; no solution: distinct parallel lines; infinitely many solutions: identical lines. b) \(a = 2\) and \(b \ne -4\) c) \(a \ne 2\); \(b\) changes the intersection’s location but not whether it exists.

- When do two equations describe the same line? - How can a line remain parallel but move to a different position? - A required solution must satisfy the new equation. - Choose a new line that is not parallel to equation (I).

1. For infinitely many solutions, choose an equivalent equation, such as \(2x - 3y = 4\). 2. For no solution, choose a distinct parallel line, such as \(6x - 9y = 10\). The lines have the same slope but different intercepts. 3. For the unique solution \((2, 0)\), choose a nonparallel line through that point, such as \(x + y = 2\). The point satisfies both equations, and the lines intersect only once.

a) One possible equation is \(2x - 3y = 4\). b) One possible equation is \(6x - 9y = 10\). c) One possible equation is \(x + y = 2\).

- Compare the quantities and totals in the first two orders. - A scaled copy of an equation does not create a new independent condition. - Combine one original equation with the third-order equation. - Choose substitution or elimination to solve.

1. The second order is exactly half of the first order, including the total cost. Therefore, the two equations are dependent and provide only one independent condition. 2. Let \(a\) be the price of one case of apple juice and \(w\) the price of one case of sparkling water. 3. Divide the first-order equation by \(2\) to get \(5a + 4w = 92\). The third order gives \(3a + 5w = 76\). 4. Multiply the first equation by \(3\) and the second by \(5\): \(15a + 12w = 276\) and \(15a + 25w = 380\). 5. Subtract to get \(13w = 104\), so \(w = 8\). Then \(5a + 32 = 92\), so \(a = 12\). 6. Therefore, apple juice costs \(\$12.00\) per case and sparkling water costs \(\$8.00\) per case.

a) The second statement is only half of the first, so it adds no independent information. b) Apple juice costs \(\$12.00\) per case, and sparkling water costs \(\$8.00\) per case.

- Define variables for the two item counts. - Multiply each unit price by its quantity. - Try several nonnegative integer values for one variable. - What kind of second condition would create another independent equation?

1. Let \(x\) be the number of apples and \(y\) the number of granola bars. The equation is \(0.5x + y = 4.5\). 2. Nonnegative integer solutions include \((1, 4)\), \((3, 3)\), \((5, 2)\), \((7, 1)\), and \((9, 0)\). 3. One equation with two unknowns is underdetermined. In this context, several nonnegative integer pairs satisfy the equation. 4. A second independent condition is needed. For example, if the student bought \(6\) items total, then \(x + y = 6\), and the unique solution is \((3, 3)\).

1. \(0.5x + y = 4.5\) 2. Examples: \((1, 4)\), \((3, 3)\), and \((5, 2)\) 3. One equation in two unknowns has multiple possible solutions. 4. Example: “The student bought \(6\) items total.”

- Scale one equation so the variable coefficients can be compared. - When do two lines share every point? - How can the constant change while the slope stays the same? - What coefficient change will make the slopes different?

1. Multiply equation (II) by \(-3\): \(3x - 6y = 12\), which is exactly equation (I). Therefore, the system has infinitely many solutions. 2. The equations represent the same line. 3. Keep the left side of equation (II) but change the constant, for example to \(-x + 2y = 0\). This creates a distinct parallel line and no solution. 4. Change a variable coefficient so the lines are not parallel, for example \(-x + 3y = -4\). Then the lines intersect once.

1. Infinitely many solutions 2. The lines are identical. 3. One example is \(-x + 2y = 0\). 4. One example is \(-x + 3y = -4\).

- If two lines are identical, how are their equations related? - How many points do two coincident lines have in common? - You can write one equation in the form \(y = mx + b\) and obtain an equivalent second equation by rearranging or multiplying it.

1. The visible line passes through \((0, -1)\) and \((2, 2)\). Its slope is \(\frac{2 - (-1)}{2 - 0} = \frac{3}{2}\), and its \(y\)-intercept is \(-1\). 2. One equation for the line is \(y = \frac{3}{2}x - 1\). 3. An equivalent second equation is \(3x - 2y = 2\). 4. Because both equations describe the same line, every point on the line satisfies both equations. The system has infinitely many solutions. 5. The solution set can be written as \(\{(t, \frac{3}{2}t - 1) \mid t \in \mathbb{R}\}\).

(I) \(y = \frac{3}{2}x - 1\) (II) \(3x - 2y = 2\) Solution set: \(\{(t, \frac{3}{2}t - 1) \mid t \in \mathbb{R}\}\)