Use elimination to solve the system \( \begin{array}{l}x+4 y=29 \\ 5 x+6 y=75\end{array} \)

To solve the system of equations using the **elimination method**, follow these steps: \[ \begin{cases} x + 4y = 29 \quad \text{(Equation 1)} \\ 5x + 6y = 75 \quad \text{(Equation 2)} \end{cases} \] ### Step 1: Align the Equations for Elimination First, aim to eliminate one of the variables by making their coefficients equal (or opposites) in both equations. Let's eliminate \( x \) by making the coefficients of \( x \) the same in both equations. - **Multiply Equation 1 by 5** to match the coefficient of \( x \) in Equation 2: \[ 5(x + 4y) = 5 \times 29 \\ 5x + 20y = 145 \quad \text{(Equation 3)} \] Now, the system looks like: \[ \begin{cases} 5x + 20y = 145 \quad \text{(Equation 3)} \\ 5x + 6y = 75 \quad \text{(Equation 2)} \end{cases} \] ### Step 2: Subtract Equations to Eliminate \( x \) Subtract **Equation 2** from **Equation 3** to eliminate \( x \): \[ (5x + 20y) - (5x + 6y) = 145 - 75 \\ 5x - 5x + 20y - 6y = 70 \\ 0x + 14y = 70 \\ 14y = 70 \] ### Step 3: Solve for \( y \) \[ 14y = 70 \\ y = \frac{70}{14} \\ y = 5 \] ### Step 4: Substitute \( y \) Back into One of the Original Equations Use **Equation 1** to find \( x \): \[ x + 4y = 29 \\ x + 4(5) = 29 \\ x + 20 = 29 \\ x = 29 - 20 \\ x = 9 \] ### Solution \[ x = 9 \\ y = 5 \] So, the solution to the system of equations is **\( x = 9 \) and \( y = 5 \)**.