Add the equations. What value belongs in the green box? \[ \begin{array}{r}-2 x+4 y=15 \\ +12 x-4 y=-8\end{array} \] \[ \]

Solve the system of equations \( -2x+4y=15; 12x-4y=-8 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}-2x+4y=15\\12x-4y=-8\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{-15+4y}{2}\\12x-4y=-8\end{array}\right.\) - step2: Substitute the value of \(x:\) \(12\times \frac{-15+4y}{2}-4y=-8\) - step3: Simplify: \(-90+20y=-8\) - step4: Move the constant to the right side: \(20y=-8+90\) - step5: Add the numbers: \(20y=82\) - step6: Divide both sides: \(\frac{20y}{20}=\frac{82}{20}\) - step7: Divide the numbers: \(y=\frac{41}{10}\) - step8: Substitute the value of \(y:\) \(x=\frac{-15+4\times \frac{41}{10}}{2}\) - step9: Simplify: \(x=\frac{7}{10}\) - step10: Calculate: \(\left\{ \begin{array}{l}x=\frac{7}{10}\\y=\frac{41}{10}\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=\frac{7}{10}\\y=\frac{41}{10}\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(\frac{7}{10},\frac{41}{10}\right)\) The solution to the system of equations is \( x = \frac{7}{10} \) and \( y = \frac{41}{10} \). Therefore, the value that belongs in the green box is \( \frac{7}{10} \).