Question Find the solution of the system of equations. \[ \begin{array}{r}6 x+2 y=16 \\ 2 x-2 y=32\end{array} \] Answer Attempt 1 out of 2

To solve the system of equations, let's start by simplifying the second equation. Divide the entire equation \(2x - 2y = 32\) by 2: \[ x - y = 16 \quad \text{(Equation 2')} \] Now, we can use substitution or elimination. From Equation 2', we can express \(y\) in terms of \(x\): \[ y = x - 16 \] Substituting this into the first equation \(6x + 2y = 16\): \[ 6x + 2(x - 16) = 16 \] Simplifying: \[ 6x + 2x - 32 = 16 \] \[ 8x - 32 = 16 \] \[ 8x = 48 \] So, \[ x = 6 \] Now, substitute \(x = 6\) back into \(y = x - 16\): \[ y = 6 - 16 = -10 \] Thus, the solution of the system of equations is \(x = 6\) and \(y = -10\). To check your work, always plug your solution back into both original equations to ensure they hold true!