\( 8 x_{1}-x_{2}=-1 \) \( -7 x_{1}+x_{2}=-2 \) \( \mathbf{x}=\left[\begin{array}{l}\text { EX: } 42 \\ \square\end{array}\right] \)

To solve the given system of equations, we can use substitution or elimination methods. Let’s add the two equations together to see if we can simplify things: 1. \( 8x_1 - x_2 = -1 \) 2. \( -7x_1 + x_2 = -2 \) Adding these gives us: \[ (8x_1 - x_2) + (-7x_1 + x_2) = -1 - 2 \] This simplifies to: \[ x_1 = -3 \] Now, plug \( x_1 = -3 \) back into one of the equations to solve for \( x_2 \). Using the first equation: \[ 8(-3) - x_2 = -1 \implies -24 - x_2 = -1 \implies -x_2 = 23 \implies x_2 = -23 \] Thus, we have: \[ \mathbf{x}=\left[\begin{array}{l} -3 \\ -23 \end{array}\right] \] This works beautifully and is a classic example of solving linear equations! Now go solve some more; it’s great practice.