EXERCISE 4 Solve for \( x \) and \( y \) by using the method of elimination: \( \begin{array}{ll}\text { (a) } x-y=3 \text { and } 2 x+y=9 & \text { (b) } x+y=-6 \text { and } 3 x+y=-10 \\ \text { (c) } x+2 y=5 \text { and } x-y=-1 & \text { (d) } 3 x+5 y=8 \text { and } x-2 y=-1 \\ \text { (e) } 2 x-3 y=10 \text { and } 4 x+5 y=42 & \text { (f) } 3 y-4 x=7 \text { and } 2 x+5 y=16\end{array} \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+2y=5\\x-y=-1\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=5-2y\\x-y=-1\end{array}\right.\) - step2: Substitute the value of \(x:\) \(5-2y-y=-1\) - step3: Subtract the terms: \(5-3y=-1\) - step4: Move the constant to the right side: \(-3y=-1-5\) - step5: Subtract the numbers: \(-3y=-6\) - step6: Change the signs: \(3y=6\) - step7: Divide both sides: \(\frac{3y}{3}=\frac{6}{3}\) - step8: Divide the numbers: \(y=2\) - step9: Substitute the value of \(y:\) \(x=5-2\times 2\) - step10: Calculate: \(x=1\) - step11: Calculate: \(\left\{ \begin{array}{l}x=1\\y=2\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=1\\y=2\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(1,2\right)\) Solve the system of equations \( 3 y-4 x=7;2 x+5 y=16 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3y-4x=7\\2x+5y=16\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{-7+3y}{4}\\2x+5y=16\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2\times \frac{-7+3y}{4}+5y=16\) - step3: Multiply the terms: \(\frac{-7+3y}{2}+5y=16\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{-7+3y}{2}+5y\right)\times 2=16\times 2\) - step5: Simplify the equation: \(-7+13y=32\) - step6: Move the constant to the right side: \(13y=32+7\) - step7: Add the numbers: \(13y=39\) - step8: Divide both sides: \(\frac{13y}{13}=\frac{39}{13}\) - step9: Divide the numbers: \(y=3\) - step10: Substitute the value of \(y:\) \(x=\frac{-7+3\times 3}{4}\) - step11: Calculate: \(x=\frac{1}{2}\) - step12: Calculate: \(\left\{ \begin{array}{l}x=\frac{1}{2}\\y=3\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=\frac{1}{2}\\y=3\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(\frac{1}{2},3\right)\) Solve the system of equations \( x-y=3;2 x+y=9 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x-y=3\\2x+y=9\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=3+y\\2x+y=9\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2\left(3+y\right)+y=9\) - step3: Simplify: \(6+3y=9\) - step4: Move the constant to the right side: \(3y=9-6\) - step5: Subtract the numbers: \(3y=3\) - step6: Divide both sides: \(\frac{3y}{3}=\frac{3}{3}\) - step7: Divide the numbers: \(y=1\) - step8: Substitute the value of \(y:\) \(x=3+1\) - step9: Calculate: \(x=4\) - step10: Calculate: \(\left\{ \begin{array}{l}x=4\\y=1\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=4\\y=1\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(4,1\right)\) Solve the system of equations \( 3 x+5 y=8;x-2 y=-1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x+5y=8\\x-2y=-1\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}3x+5y=8\\x=-1+2y\end{array}\right.\) - step2: Substitute the value of \(x:\) \(3\left(-1+2y\right)+5y=8\) - step3: Simplify: \(-3+11y=8\) - step4: Move the constant to the right side: \(11y=8+3\) - step5: Add the numbers: \(11y=11\) - step6: Divide both sides: \(\frac{11y}{11}=\frac{11}{11}\) - step7: Divide the numbers: \(y=1\) - step8: Substitute the value of \(y:\) \(x=-1+2\times 1\) - step9: Substitute back: \(x=-1+2\) - step10: Calculate: \(x=1\) - step11: Calculate: \(\left\{ \begin{array}{l}x=1\\y=1\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=1\\y=1\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(1,1\right)\) Solve the system of equations \( x+y=-6;3 x+y=-10 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y=-6\\3x+y=-10\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=-6-y\\3x+y=-10\end{array}\right.\) - step2: Substitute the value of \(x:\) \(3\left(-6-y\right)+y=-10\) - step3: Simplify: \(-18-2y=-10\) - step4: Move the constant to the right side: \(-2y=-10+18\) - step5: Add the numbers: \(-2y=8\) - step6: Change the signs: \(2y=-8\) - step7: Divide both sides: \(\frac{2y}{2}=\frac{-8}{2}\) - step8: Divide the numbers: \(y=-4\) - step9: Substitute the value of \(y:\) \(x=-6-\left(-4\right)\) - step10: Substitute back: \(x=-6+4\) - step11: Calculate: \(x=-2\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-2\\y=-4\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-2\\y=-4\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-2,-4\right)\) Solve the system of equations \( 2 x-3 y=10;4 x+5 y=42 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x-3y=10\\4x+5y=42\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{10+3y}{2}\\4x+5y=42\end{array}\right.\) - step2: Substitute the value of \(x:\) \(4\times \frac{10+3y}{2}+5y=42\) - step3: Simplify: \(20+11y=42\) - step4: Move the constant to the right side: \(11y=42-20\) - step5: Subtract the numbers: \(11y=22\) - step6: Divide both sides: \(\frac{11y}{11}=\frac{22}{11}\) - step7: Divide the numbers: \(y=2\) - step8: Substitute the value of \(y:\) \(x=\frac{10+3\times 2}{2}\) - step9: Calculate: \(x=8\) - step10: Calculate: \(\left\{ \begin{array}{l}x=8\\y=2\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=8\\y=2\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(8,2\right)\) Here are the solutions for each of the systems of equations using the method of elimination: ### (a) Equations: \[ x - y = 3 \] \[ 2x + y = 9 \] **Solution:** \[ (x, y) = (4, 1) \] ### (b) Equations: \[ x + y = -6 \] \[ 3x + y = -10 \] **Solution:** \[ (x, y) = (-2, -4) \] ### (c) Equations: \[ x + 2y = 5 \] \[ x - y = -1 \] **Solution:** \[ (x, y) = (1, 2) \] ### (d) Equations: \[ 3x + 5y = 8 \] \[ x - 2y = -1 \] **Solution:** \[ (x, y) = (1, 1) \] ### (e) Equations: \[ 2x - 3y = 10 \] \[ 4x + 5y = 42 \] **Solution:** \[ (x, y) = (8, 2) \] ### (f) Equations: \[ 3y - 4x = 7 \] \[ 2x + 5y = 16 \] **Solution:** \[ (x, y) = \left(\frac{1}{2}, 3\right) \text{ or } (0.5, 3) \] These results provide the values of \( x \) and \( y \) for each system of equations.