(b) Solve for \( x \) and \( y \) using either the elimination or substitution method: (1) \( x+y=1 \) and \( x-2 y=1 \) (2) \( 3 x+2 y=2 \) and \( 5 x-2 y=-18 \) (3) \( x+4 y=14 \) and \( 3 x+2 y=12 \) (4) \( 2 y-3 x=7 \) and \( 4 y-5 x=21 \) (5) \( 3 x+2 y=6 \) and \( 5 x+3 y=11 \) (6) \( \frac{y-2}{3}-\frac{x-1}{2}=1 \) and \( \frac{y-4}{5}-\frac{x-3}{4}=1 \)

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 \( x+y=1;x-2y=1 \). 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 \( (y-2)/3-(x-1)/2=1;(y-4)/5-(x-3)/4=1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}\frac{\left(y-2\right)}{3}-\frac{\left(x-1\right)}{2}=1\\\frac{\left(y-4\right)}{5}-\frac{\left(x-3\right)}{4}=1\end{array}\right.\) - step1: Remove the parentheses: \(\left\{ \begin{array}{l}\frac{y-2}{3}-\frac{x-1}{2}=1\\\frac{y-4}{5}-\frac{x-3}{4}=1\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{-7+2y}{3}\\\frac{y-4}{5}-\frac{x-3}{4}=1\end{array}\right.\) - step3: Substitute the value of \(x:\) \(\frac{y-4}{5}-\frac{\frac{-7+2y}{3}-3}{4}=1\) - step4: Simplify: \(\frac{y-4}{5}+\frac{8-y}{6}=1\) - step5: Multiply both sides of the equation by LCD: \(\left(\frac{y-4}{5}+\frac{8-y}{6}\right)\times 30=1\times 30\) - step6: Simplify the equation: \(y+16=30\) - step7: Move the constant to the right side: \(y=30-16\) - step8: Subtract the numbers: \(y=14\) - step9: Substitute the value of \(y:\) \(x=\frac{-7+2\times 14}{3}\) - step10: Calculate: \(x=7\) - step11: Calculate: \(\left\{ \begin{array}{l}x=7\\y=14\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=7\\y=14\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(7,14\right)\) Solve the system of equations \( 3x+2y=2;5x-2y=-18 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x+2y=2\\5x-2y=-18\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{2-2y}{3}\\5x-2y=-18\end{array}\right.\) - step2: Substitute the value of \(x:\) \(5\times \frac{2-2y}{3}-2y=-18\) - step3: Simplify: \(\frac{5\left(2-2y\right)}{3}-2y=-18\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{5\left(2-2y\right)}{3}-2y\right)\times 3=-18\times 3\) - step5: Simplify the equation: \(10-16y=-54\) - step6: Move the constant to the right side: \(-16y=-54-10\) - step7: Subtract the numbers: \(-16y=-64\) - step8: Change the signs: \(16y=64\) - step9: Divide both sides: \(\frac{16y}{16}=\frac{64}{16}\) - step10: Divide the numbers: \(y=4\) - step11: Substitute the value of \(y:\) \(x=\frac{2-2\times 4}{3}\) - step12: Calculate: \(x=-2\) - step13: Calculate: \(\left\{ \begin{array}{l}x=-2\\y=4\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=-2\\y=4\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(-2,4\right)\) Solve the system of equations \( 2y-3x=7;4y-5x=21 \). 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 \( 3x+2y=6;5x+3y=11 \). 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: \(\left(x,y\right) = \left(4,-3\right)\) Here are the solutions for each system of equations: 1. For the equations \( x + y = 1 \) and \( x - 2y = 1 \): - The solution is \( (x, y) = (1, 0) \). 2. For the equations \( 3x + 2y = 2 \) and \( 5x - 2y = -18 \): - The solution is \( (x, y) = (-2, 4) \). 3. For the equations \( x + 4y = 14 \) and \( 3x + 2y = 12 \): - The solution is \( (x, y) = (2, 3) \). 4. For the equations \( 2y - 3x = 7 \) and \( 4y - 5x = 21 \): - The solution is \( (x, y) = (7, 14) \). 5. For the equations \( 3x + 2y = 6 \) and \( 5x + 3y = 11 \): - The solution is \( (x, y) = (4, -3) \). 6. For the equations \( \frac{y-2}{3} - \frac{x-1}{2} = 1 \) and \( \frac{y-4}{5} - \frac{x-3}{4} = 1 \): - The solution is \( (x, y) = (7, 14) \). If you need further explanations or details on any specific system, feel free to ask!