EXERCISE 5 (a) Solve for \( x \) and \( y \) by using the method of substitution: \( \begin{array}{lll}\text { (1) } x-y=8 \text { and } 2 x+y=10 & \text { (2) } 3 x-y=2 \text { and } 7 x-2 y=8 \\ \text { (3) } 3 x+5 y=8 \text { and } x-2 y=-1 & \text { (4) } 7 x-3 y=41 \text { and } 3 x-y=17\end{array} \) (b) Solve for \( x \) and \( y \) using either the elimination or substitution method: \( \begin{array}{lll}\text { (1) } x+y=1 \text { and } x-2 y=1 & \text { (2) } & \text { (4) } 2 x+2 y=2 \text { and } 5 x-2 y=-18 \\ \text { (3) } x+4 y=14 \text { and } 3 x+2 y=12 & \text { (4) } 2 y-3 x=7 \text { and } 4 y-5 x=21 \\ \text { (5) } 3 x+2 y=6 \text { and } 5 x+3 y=11 & \end{array} \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}7x-3y=41\\3x-y=17\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}7x-3y=41\\y=-17+3x\end{array}\right.\) - step2: Substitute the value of \(y:\) \(7x-3\left(-17+3x\right)=41\) - step3: Simplify: \(-2x+51=41\) - step4: Move the constant to the right side: \(-2x=41-51\) - step5: Subtract the numbers: \(-2x=-10\) - step6: Change the signs: \(2x=10\) - step7: Divide both sides: \(\frac{2x}{2}=\frac{10}{2}\) - step8: Divide the numbers: \(x=5\) - step9: Substitute the value of \(x:\) \(y=-17+3\times 5\) - step10: Calculate: \(y=-2\) - step11: Calculate: \(\left\{ \begin{array}{l}x=5\\y=-2\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=5\\y=-2\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(5,-2\right)\) Solve the system of equations \( 3 x+2 y=6;5 x+3 y=11 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x+2y=6\\5x+3y=11\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{6-2y}{3}\\5x+3y=11\end{array}\right.\) - step2: Substitute the value of \(x:\) \(5\times \frac{6-2y}{3}+3y=11\) - step3: Simplify: \(\frac{5\left(6-2y\right)}{3}+3y=11\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{5\left(6-2y\right)}{3}+3y\right)\times 3=11\times 3\) - step5: Simplify the equation: \(30-y=33\) - step6: Move the constant to the right side: \(-y=33-30\) - step7: Subtract the numbers: \(-y=3\) - step8: Change the signs: \(y=-3\) - step9: Substitute the value of \(y:\) \(x=\frac{6-2\left(-3\right)}{3}\) - step10: Calculate: \(x=4\) - step11: Calculate: \(\left\{ \begin{array}{l}x=4\\y=-3\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=4\\y=-3\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(4,-3\right)\) Solve the system of equations \( 3 x-y=2;7 x-2 y=8 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x-y=2\\7x-2y=8\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=-2+3x\\7x-2y=8\end{array}\right.\) - step2: Substitute the value of \(y:\) \(7x-2\left(-2+3x\right)=8\) - step3: Simplify: \(x+4=8\) - step4: Move the constant to the right side: \(x=8-4\) - step5: Subtract the numbers: \(x=4\) - step6: Substitute the value of \(x:\) \(y=-2+3\times 4\) - step7: Calculate: \(y=10\) - step8: Calculate: \(\left\{ \begin{array}{l}x=4\\y=10\end{array}\right.\) - step9: Check the solution: \(\left\{ \begin{array}{l}x=4\\y=10\end{array}\right.\) - step10: Rewrite: \(\left(x,y\right) = \left(4,10\right)\) Solve the system of equations \( x+y=1;x-2 y=1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y=1\\x-2y=1\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=1-y\\x-2y=1\end{array}\right.\) - step2: Substitute the value of \(x:\) \(1-y-2y=1\) - step3: Subtract the terms: \(1-3y=1\) - step4: Move the constant to the right side: \(-3y=1-1\) - step5: Subtract the terms: \(-3y=0\) - step6: Change the signs: \(3y=0\) - step7: Rewrite the expression: \(y=0\) - step8: Substitute the value of \(y:\) \(x=1-0\) - step9: Substitute back: \(x=1\) - step10: Calculate: \(\left\{ \begin{array}{l}x=1\\y=0\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=1\\y=0\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(1,0\right)\) Solve the system of equations \( 2 x+2 y=2;5 x-2 y=-18 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x+2y=2\\5x-2y=-18\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=1-y\\5x-2y=-18\end{array}\right.\) - step2: Substitute the value of \(x:\) \(5\left(1-y\right)-2y=-18\) - step3: Simplify: \(5-7y=-18\) - step4: Move the constant to the right side: \(-7y=-18-5\) - step5: Subtract the numbers: \(-7y=-23\) - step6: Change the signs: \(7y=23\) - step7: Divide both sides: \(\frac{7y}{7}=\frac{23}{7}\) - step8: Divide the numbers: \(y=\frac{23}{7}\) - step9: Substitute the value of \(y:\) \(x=1-\frac{23}{7}\) - step10: Calculate: \(x=-\frac{16}{7}\) - step11: Calculate: \(\left\{ \begin{array}{l}x=-\frac{16}{7}\\y=\frac{23}{7}\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{16}{7}\\y=\frac{23}{7}\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(-\frac{16}{7},\frac{23}{7}\right)\) Solve the system of equations \( x-y=8;2 x+y=10 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x-y=8\\2x+y=10\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=8+y\\2x+y=10\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2\left(8+y\right)+y=10\) - step3: Simplify: \(16+3y=10\) - step4: Move the constant to the right side: \(3y=10-16\) - step5: Subtract the numbers: \(3y=-6\) - step6: Divide both sides: \(\frac{3y}{3}=\frac{-6}{3}\) - step7: Divide the numbers: \(y=-2\) - step8: Substitute the value of \(y:\) \(x=8-2\) - step9: Calculate: \(x=6\) - step10: Calculate: \(\left\{ \begin{array}{l}x=6\\y=-2\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=6\\y=-2\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(6,-2\right)\) Solve the system of equations \( x+4 y=14;3 x+2 y=12 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+4y=14\\3x+2y=12\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=14-4y\\3x+2y=12\end{array}\right.\) - step2: Substitute the value of \(x:\) \(3\left(14-4y\right)+2y=12\) - step3: Simplify: \(42-10y=12\) - step4: Move the constant to the right side: \(-10y=12-42\) - step5: Subtract the numbers: \(-10y=-30\) - step6: Change the signs: \(10y=30\) - step7: Divide both sides: \(\frac{10y}{10}=\frac{30}{10}\) - step8: Divide the numbers: \(y=3\) - step9: Substitute the value of \(y:\) \(x=14-4\times 3\) - step10: Calculate: \(x=2\) - step11: Calculate: \(\left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(2,3\right)\) Solve the system of equations \( 2 y-3 x=7;4 y-5 x=21 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2y-3x=7\\4y-5x=21\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{-7+2y}{3}\\4y-5x=21\end{array}\right.\) - step2: Substitute the value of \(x:\) \(4y-5\times \frac{-7+2y}{3}=21\) - step3: Simplify: \(4y-\frac{5\left(-7+2y\right)}{3}=21\) - step4: Multiply both sides of the equation by LCD: \(\left(4y-\frac{5\left(-7+2y\right)}{3}\right)\times 3=21\times 3\) - step5: Simplify the equation: \(2y+35=63\) - step6: Move the constant to the right side: \(2y=63-35\) - step7: Subtract the numbers: \(2y=28\) - step8: Divide both sides: \(\frac{2y}{2}=\frac{28}{2}\) - step9: Divide the numbers: \(y=14\) - step10: Substitute the value of \(y:\) \(x=\frac{-7+2\times 14}{3}\) - step11: Calculate: \(x=7\) - step12: Calculate: \(\left\{ \begin{array}{l}x=7\\y=14\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=7\\y=14\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(7,14\right)\) 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: \(\left(x,y\right) = \left(1,1\right)\) ### Part (a): Solve for \( x \) and \( y \) using the method of substitution 1. **Equations (1)**: \[ x - y = 8 \quad (1) \\ 2x + y = 10 \quad (2) \] From equation (1), we can express \( y \) in terms of \( x \): \[ y = x - 8 \] Substituting this into equation (2): \[ 2x + (x - 8) = 10 \\ 3x - 8 = 10 \\ 3x = 18 \\ x = 6 \] Now substituting \( x = 6 \) back into \( y = x - 8 \): \[ y = 6 - 8 = -2 \] **Solution**: \( (x, y) = (6, -2) \) 2. **Equations (2)**: \[ 3x - y = 2 \quad (1) \\ 7x - 2y = 8 \quad (2) \] From equation (1), express \( y \): \[ y = 3x - 2 \] Substitute into equation (2): \[ 7x - 2(3x - 2) = 8 \\ 7x - 6x + 4 = 8 \\ x + 4 = 8 \\ x = 4 \] Substitute \( x = 4 \) back into \( y = 3x - 2 \): \[ y = 3(4) - 2 = 12 - 2 = 10 \] **Solution**: \( (x, y) = (4, 10) \) 3. **Equations (3)**: \[ 3x + 5y = 8 \quad (1) \\ x - 2y = -1 \quad (2) \] From equation (2), express \( x \): \[ x = 2y - 1 \] Substitute into equation (1): \[ 3(2y - 1) + 5y = 8 \\ 6y - 3 + 5y = 8 \\ 11y - 3 = 8 \\ 11y = 11 \\ y = 1 \] Substitute \( y = 1 \) back into \( x = 2y - 1 \): \[ x = 2(1) - 1 = 1 \] **Solution**: \( (x, y) = (1, 1) \) 4. **Equations (4)**: \[ 7x - 3y = 41 \quad (1) \\ 3x - y = 17 \quad (2) \] From equation (2), express \( y \): \[ y = 3x - 17 \] Substitute into equation (1): \[ 7x - 3(3x - 17) = 41 \\ 7x - 9x + 51 = 41 \\ -2x + 51 = 41 \\ -2x = -10 \\ x = 5 \] Substitute \( x = 5 \) back into \( y = 3x - 17 \): \[ y = 3(5) - 17 = 15 - 17 = -2 \] **Solution**: \( (x, y) = (5, -2) \) ### Part (b): Solve for \( x \) and \( y \) using either the elimination or substitution method 1. **Equations (1)**: \[ x + y = 1 \quad (1) \\ x - 2y = 1 \quad (2) \] From equation (1), express \( x \): \[ x = 1 - y \] Substitute into equation (2): \[ (1 - y) - 2y = 1 \\ 1 - 3y = 1 \\ -3y = 0 \\ y = 0 \] Substitute \( y = 0 \) back into \( x = 1 - y \): \[ x = 1 - 0 = 1 \] **Solution**: \( (x, y) = (1, 0) \) 2. **Equations (2)**: \[ 2x + 2y = 2 \quad (1) \\ 5x - 2y = -18 \quad (2) \] Simplifying equation (1): \[ x + y = 1 \quad (1) \] From equation (1), express \( y \): \[ y = 1 - x \] Substitute into equation (2): \[ 5x - 2(1 - x) = -18 \\ 5x - 2 + 2x = -18 \\ 7x - 2 = -18 \\ 7x = -16 \\ x = -\frac{16}{7} \] Substitute \( x = -\frac{16}{7} \) back into \( y = 1 - x \): \[ y = 1 - \left(-\frac{16}{7}\right) = 1 + \frac{16}{7} = \frac{23}{7} \] **Solution**: \( (x, y) = \left(-\frac{16}{7}, \frac{23}{7}\right) \) 3. **Equations (3)**: \[ x + 4y = 14 \quad (1) \\ 3x + 2y = 12 \quad (2) \] From equation (1), express \( x \): \[ x = 14 - 4y \] Substitute into equation (2): \[ 3(14 - 4y) + 2y = 12 \\ 42 - 12y + 2y = 12 \\ 42 - 10y = 12 \\ -10y = -30 \\ y = 3 \] Substitute \( y