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

Solve the system of equations \( 6x+2y=16;2x-2y=32 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}6x+2y=16\\2x-2y=32\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{8-y}{3}\\2x-2y=32\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2\times \frac{8-y}{3}-2y=32\) - step3: Simplify: \(\frac{2\left(8-y\right)}{3}-2y=32\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{2\left(8-y\right)}{3}-2y\right)\times 3=32\times 3\) - step5: Simplify the equation: \(16-8y=96\) - step6: Move the constant to the right side: \(-8y=96-16\) - step7: Subtract the numbers: \(-8y=80\) - step8: Change the signs: \(8y=-80\) - step9: Divide both sides: \(\frac{8y}{8}=\frac{-80}{8}\) - step10: Divide the numbers: \(y=-10\) - step11: Substitute the value of \(y:\) \(x=\frac{8-\left(-10\right)}{3}\) - step12: Substitute back: \(x=\frac{8+10}{3}\) - step13: Calculate: \(x=6\) - step14: Calculate: \(\left\{ \begin{array}{l}x=6\\y=-10\end{array}\right.\) - step15: Check the solution: \(\left\{ \begin{array}{l}x=6\\y=-10\end{array}\right.\) - step16: Rewrite: \(\left(x,y\right) = \left(6,-10\right)\) The solution to the system of equations is \( (x, y) = (6, -10) \).