Question: 3) \(3y - x + 10 = 0\) \(3x + 4y + 22 = 0\)

Solution

To solve the system of equations: 1) Rearrange each equation to express \( x \) in terms of \( y \). Starting with the first equation: \[ 3y - x + 10 = 0 \] Add \( x \) to both sides: \[ 3y + 10 = x \] 2) Substitute this expression for \( x \) from the first equation into the second equation: \[ 3x + 4y + 22 = 0 \] Substitute \( x = 3y + 10 \): \[ 3(3y + 10) + 4y + 22 = 0 \] 3) Solve for \( y \): \[ 9y + 30 + 4y + 22 = 0 \] Combine like terms: \[ 13y + 52 = 0 \] Subtract 52 from both sides: \[ 13y = -52 \] Divide by 13: \[ y = -4 \] 4) Substitute \( y = -4 \) back into the expression for \( x \): \[ x = 3(-4) + 10 \] Calculate: \[ x = -12 + 10 \] \[ x = -2 \] 5) The solution to the system is \( x = -2 \) and \( y = -4 \).