\( \begin{array}{l}40 \text { Listen } \\ \text { What is the solution to the system of equations? } \\ \begin{array}{l}3 x+5 y=-3 \\ 4 x+10 y=-14\end{array}\end{array} \)

To solve the system of equations: \[ \begin{cases} 3x + 5y = -3 \\ 4x + 10y = -14 \end{cases} \] **Step 1: Elimination Method** 1. **Multiply the first equation by 2** to make the coefficients of \( y \) the same: \[ 2(3x + 5y) = 2(-3) \\ 6x + 10y = -6 \quad \text{(Equation 1a)} \] 2. **Write down the second equation as is:** \[ 4x + 10y = -14 \quad \text{(Equation 2)} \] 3. **Subtract Equation 2 from Equation 1a:** \[ (6x + 10y) - (4x + 10y) = -6 - (-14) \\ 2x = 8 \\ x = 4 \] **Step 2: Substitute \( x = 4 \) into the first original equation to find \( y \):** \[ 3(4) + 5y = -3 \\ 12 + 5y = -3 \\ 5y = -15 \\ y = -3 \] **Solution:** \[ x = 4 \\ y = -3 \] **Verification:** Plugging \( x = 4 \) and \( y = -3 \) into the second original equation: \[ 4(4) + 10(-3) = 16 - 30 = -14 \] Which matches the given equation, confirming that the solution is correct. **Final Answer:** The solution is \( x = 4 \) and \( y = -3 \).