\( \left.\begin{array}{l}\text { Find the solution of the system of equations. } \\ \qquad \begin{array}{r}5 x+10 y=-5 \\ -5 x-y=32\end{array} \\ \text { Answer Attempt 1 out of 2 }\end{array}+\square . \square\right) \) Submit Answer \( \quad \).

Let’s solve the system of equations step by step! First, we have the equations: 1. \( 5x + 10y = -5 \) 2. \( -5x - y = 32 \) From the second equation, we can express \( y \) in terms of \( x \): \[ -y = 32 + 5x \] \[ y = -5x - 32 \] Now, substitute \( y \) into the first equation: \[ 5x + 10(-5x - 32) = -5 \] This simplifies to: \[ 5x - 50x - 320 = -5 \] \[ -45x - 320 = -5 \] Adding 320 to both sides gives: \[ -45x = 315 \] Now, divide by -45 to find \( x \): \[ x = -7 \] Now that we have \( x \), substitute it back to find \( y \): \[ y = -5(-7) - 32 \] \[ y = 35 - 32 = 3 \] So, the solution to the system of equations is \( x = -7 \) and \( y = 3 \). Therefore, the solution is \( (-7, 3) \)! Feel free to celebrate your quadratic abilities with a little victory dance!