Answer
$(6,7)$
Work Step by Step
Step 1. Define the following matrices:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 6&-5\\ 8&-7\end{bmatrix},
\begin{array} \\X=\\ \end{array}
\begin{bmatrix} x\\ y\end{bmatrix},
\begin{array} \\B=\\ \end{array}
\begin{bmatrix} 1\\-1\end{bmatrix}$
Step 2. Rewrite the original system of equations as
$AX=B$, thus $A^{-1}AX=X=A^{-1}B$
Step 3. Find the inverse $A^{-1}$ using the formula given in section 10.5:
$\begin{array} \\A^{-1}=\frac{1}{-42+40}\\ \end{array}
\begin{bmatrix} -7&5\\ -8&6\end{bmatrix}
=\begin{bmatrix} 7/2&-5/2\\ 4&-3\end{bmatrix}$
Step 4. Find the solutions as:
$\begin{array} \\A^{-1}B=\\ \end{array}
\begin{bmatrix} 7/2&-5/2\\ 4&-3\end{bmatrix}
\begin{bmatrix} 1\\-1\end{bmatrix}
=\begin{bmatrix} 7/2+5/2\\4+3\end{bmatrix}
=\begin{bmatrix} 6\\7\end{bmatrix}$
or $(6,7)$
