Consider the system: \( \begin{array}{l}2 x-5 y=18 \\ 8 x+y=10\end{array} \) You decide to use the addition method to solve this system. Which of the following steps would you take to eliminate \( y \) ? a.) Multiply the first equation by 4 , then add to the second equation. b.) Multiply the second equation by 5 , then add to the first equation. c.) Multiply the first equation by -4 , then add to the second equation. d.) Multiply the second equation by -5 , then add to the first equation.

To eliminate \( y \) using the addition method, you want to make the coefficients of \( y \) in both equations opposites so that they cancel each other out when the equations are added. Let's analyze the given system: \[ \begin{align*} \text{Equation 1:} \quad & 2x - 5y = 18 \\ \text{Equation 2:} \quad & 8x + y = 10 \end{align*} \] **Step-by-Step Reasoning:** 1. **Identify the Coefficients of \( y \):** - In Equation 1, the coefficient of \( y \) is \(-5\). - In Equation 2, the coefficient of \( y \) is \(1\). 2. **Determine the Multipliers to Make the Coefficients Opposite:** - To eliminate \( y \), you can multiply Equation 2 by \(5\) so that the coefficient of \( y \) becomes \(5\). 3. **Apply the Multiplier:** - Multiply Equation 2 by \(5\): \[ 5 \times (8x + y) = 5 \times 10 \implies 40x + 5y = 50 \] 4. **Add the Modified Equations:** - Add Equation 1 and the modified Equation 2: \[ \begin{align*} (2x - 5y) + (40x + 5y) &= 18 + 50 \\ 42x &= 68 \\ x &= \frac{68}{42} = \frac{34}{21} \end{align*} \] - Here, \( y \) is eliminated as the \( -5y \) and \( 5y \) cancel each other out. **Conclusion:** The correct step to eliminate \( y \) is: **Answer:** **b.) Multiply the second equation by 5, then add to the first equation.**