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

Solution

We have the following system of equations: \[ \begin{cases} 3y - x + 10 = 0 & \\ 3x + 4y + 22 = 0 & \end{cases} \] First, solve the first equation for \( x \): \[ 3y - x + 10 = 0 \] Rearrange it to solve for \( x \): \[ -x = -3y - 10 \] \[ x = 3y + 10 \] Substitute \( x = 3y + 10 \) into the second equation: \[ 3(3y + 10) + 4y + 22 = 0 \] Expand and simplify: \[ 9y + 30 + 4y + 22 = 0 \] Combine like terms: \[ 13y + 52 = 0 \] Solve for \( y \): \[ 13y = -52 \] \[ y = -4 \] Substitute \( y = -4 \) back into \( x = 3y + 10 \): \[ x = 3(-4) + 10 \] \[ x = -12 + 10 \] \[ x = -2 \] The solution is \( x = -2 \) and \( y = -4 \).