Consider the following two linear equations. \[ \begin{array}{l}8 x+2 y=-2 \\ \text { Solve for the point of intersection using elimination }\end{array} \]

It appears that only one linear equation was provided: \[ 8x + 2y = -2 \] To determine the **point of intersection** between two lines using the **elimination method**, we need **two distinct linear equations**. The point of intersection is the solution \((x, y)\) that satisfies both equations simultaneously. **Here's what you need to do:** 1. **Provide a Second Equation:** Please provide the second linear equation so that we can use the elimination method to find the point where the two lines intersect. **Example Scenario:** If you have two equations, such as: 1. \( 8x + 2y = -2 \) 2. \( 4x - y = 3 \) **We can solve them using elimination as follows:** **Step 1: Align the Equations** \[ \begin{align*} 8x + 2y &= -2 \quad \text{(Equation 1)} \\ 4x - y &= 3 \quad \text{(Equation 2)} \end{align*} \] **Step 2: Make the Coefficients of \(y\) Opposite** Multiply Equation 2 by 2 to make the coefficient of \(y\) in Equation 2 equal to \(-2\): \[ \begin{align*} 8x + 2y &= -2 \quad \text{(Equation 1)} \\ 8x - 2y &= 6 \quad \text{(Equation 2 multiplied by 2)} \end{align*} \] **Step 3: Add the Equations to Eliminate \(y\)** \[ (8x + 2y) + (8x - 2y) = -2 + 6 \\ 16x = 4 \\ x = \frac{4}{16} \\ x = \frac{1}{4} \] **Step 4: Substitute \(x\) Back into One of the Original Equations to Find \(y\)** Using Equation 2: \[ 4\left(\frac{1}{4}\right) - y = 3 \\ 1 - y = 3 \\ -y = 2 \\ y = -2 \] **Solution:** The point of intersection is \(\left( \frac{1}{4},\ -2 \right)\). --- **Please provide the second equation**, and I'll be happy to help you find the point of intersection using the elimination method!