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

**(a)** We are given \[ x - y = 3 \quad \text{and} \quad 2x + y = 9. \] *Step 1: Add the two equations to eliminate \( y \):* \[ (x - y) + (2x + y) = 3 + 9 \quad \Longrightarrow \quad 3x = 12. \] *Step 2: Solve for \( x \):* \[ x = \frac{12}{3} = 4. \] *Step 3: Substitute \( x = 4 \) into the first equation to solve for \( y \):* \[ 4 - y = 3 \quad \Longrightarrow \quad -y = -1 \quad \Longrightarrow \quad y = 1. \] --- **(b)** We are given \[ x + y = -6 \quad \text{and} \quad 3x + y = -10. \] *Step 1: Subtract the first equation from the second to eliminate \( y \):* \[ (3x + y) - (x + y) = -10 - (-6) \quad \Longrightarrow \quad 2x = -4. \] *Step 2: Solve for \( x \):* \[ x = \frac{-4}{2} = -2. \] *Step 3: Substitute \( x = -2 \) into the first equation to solve for \( y \):* \[ -2 + y = -6 \quad \Longrightarrow \quad y = -6 + 2 = -4. \] --- **(c)** We are given \[ x + 2y = 5 \quad \text{and} \quad x - y = -1. \] *Step 1: Subtract the second equation from the first to eliminate \( x \):* \[ (x + 2y) - (x - y) = 5 - (-1) \quad \Longrightarrow \quad 3y = 6. \] *Step 2: Solve for \( y \):* \[ y = \frac{6}{3} = 2. \] *Step 3: Substitute \( y = 2 \) into the second equation to solve for \( x \):* \[ x - 2 = -1 \quad \Longrightarrow \quad x = -1 + 2 = 1. \] --- **(d)** We are given \[ 3x + 5y = 8 \quad \text{and} \quad x - 2y = -1. \] *Step 1: Multiply the second equation by 3 to make the \( x \)-coefficients the same:* \[ 3(x - 2y) = 3(-1) \quad \Longrightarrow \quad 3x - 6y = -3. \] *Step 2: Subtract the new equation from the first equation to eliminate \( x \):* \[ (3x + 5y) - (3x - 6y) = 8 - (-3) \quad \Longrightarrow \quad 11y = 11. \] *Step 3: Solve for \( y \):* \[ y = \frac{11}{11} = 1. \] *Step 4: Substitute \( y = 1 \) into the second original equation to solve for \( x \):* \[ x - 2(1) = -1 \quad \Longrightarrow \quad x - 2 = -1 \quad \Longrightarrow \quad x = 1. \] --- **(e)** We are given \[ 2x - 3y = 10 \quad \text{and} \quad 4x + 5y = 42. \] *Step 1: Multiply the first equation by 2 to align the \( x \)-coefficients with the second equation:* \[ 2(2x - 3y) = 2(10) \quad \Longrightarrow \quad 4x - 6y = 20. \] *Step 2: Subtract this new equation from the second equation to eliminate \( x \):* \[ (4x + 5y) - (4x - 6y) = 42 - 20 \quad \Longrightarrow \quad 11y = 22. \] *Step 3: Solve for \( y \):* \[ y = \frac{22}{11} = 2. \] *Step 4: Substitute \( y = 2 \) into the first equation to solve for \( x \):* \[ 2x - 3(2) = 10 \quad \Longrightarrow \quad 2x - 6 = 10 \quad \Longrightarrow \quad 2x = 16 \quad \Longrightarrow \quad x = 8. \] --- **(f)** We are given \[ 3y - 4x = 7 \quad \text{and} \quad 2x + 5y = 16. \] *Step 1: Rewrite the first equation in standard form:* \[ -4x + 3y = 7. \] *Step 2: Multiply the second equation by 2 to facilitate eliminating \( x \):* \[ 2(2x + 5y) = 2(16) \quad \Longrightarrow \quad 4x + 10y = 32. \] *Step 3: Add the modified equations to eliminate \( x \):* \[ (-4x + 3y) + (4x + 10y) = 7 + 32 \quad \Longrightarrow \quad 13y = 39. \] *Step 4: Solve for \( y \):* \[ y = \frac{39}{13} = 3. \] *Step 5: Substitute \( y = 3 \) into the second original equation to solve for \( x \):* \[ 2x + 5(3) = 16 \quad \Longrightarrow \quad 2x + 15 = 16 \quad \Longrightarrow \quad 2x = 1 \quad \Longrightarrow \quad x = \frac{1}{2}. \]