Questions: Solve the system of equations (x-7 y=-36) and (-2 x-2 y=8) by c [ beginaligned (x-7 y =-36) (-2 x-2 y =8) x-7 y =-36 -2 x-2 y =8 o x+circ y = endaligned ]

Solution Steps

To solve the system of equations, we can use the method of substitution or elimination. Here, we'll use the elimination method. First, we'll multiply the first equation by 2 to align the coefficients of \(x\) in both equations. Then, we'll add the two equations to eliminate \(x\) and solve for \(y\). Once \(y\) is found, we'll substitute it back into one of the original equations to find \(x\).

We start with the system of equations: \[ \begin{aligned}

1. & \quad x - 7y = -36 \\
2. & \quad -2x - 2y = 8 \end{aligned} \]

To eliminate \(x\), we can manipulate the first equation. Multiplying the first equation by 2 gives: \[ 2x - 14y = -72 \] Now we have: \[ \begin{aligned} 3. & \quad 2x - 14y = -72 \\ 4. & \quad -2x - 2y = 8 \end{aligned} \] Adding equations 3 and 4 results in: \[ -16y = -64 \] Solving for \(y\) gives: \[ y = \frac{-64}{-16} = 4 \]

Now that we have \(y\), we substitute \(y = 4\) back into the first equation: \[ x - 7(4) = -36 \] This simplifies to: \[ x - 28 = -36 \] Solving for \(x\) gives: \[ x = -36 + 28 = -8 \]

Final Answer