\( 1 \& \) Solve the system of equations by the addition method. \[ \begin{array}{r}20 x-2 y=-30 \\ 5 x+7 y=-45\end{array} \] Select the correct choice below and fill in any answer boxes present in your choice. A. The solution set is \{ ? (Simplity your answer. Type an ordered pair.) B. There are infinitely many solutions.

Solve the system of equations \( 20x-2y=-30;5x+7y=-45 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}20x-2y=-30\\5x+7y=-45\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{-15+y}{10}\\5x+7y=-45\end{array}\right.\) - step2: Substitute the value of \(x:\) \(5\times \frac{-15+y}{10}+7y=-45\) - step3: Multiply the terms: \(\frac{-15+y}{2}+7y=-45\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{-15+y}{2}+7y\right)\times 2=-45\times 2\) - step5: Simplify the equation: \(-15+15y=-90\) - step6: Move the constant to the right side: \(15y=-90+15\) - step7: Add the numbers: \(15y=-75\) - step8: Divide both sides: \(\frac{15y}{15}=\frac{-75}{15}\) - step9: Divide the numbers: \(y=-5\) - step10: Substitute the value of \(y:\) \(x=\frac{-15-5}{10}\) - step11: Substitute back: \(x=-\frac{15+5}{10}\) - step12: Calculate: \(x=-2\) - step13: Calculate: \(\left\{ \begin{array}{l}x=-2\\y=-5\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=-2\\y=-5\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(-2,-5\right)\) The solution to the system of equations by the addition method is \( (x,y) = (-2,-5) \). Therefore, the correct choice is: A. The solution set is \((-2,-5)\).