Precalculus (6th Edition) Blitzer

The solution of the system is $\underline{\left( 1,-3 \right)}$.
Let us consider the system of equations: $x=y+4$ $x-y=4$ (I) $3x+7y=-18$ (II) And multiply the equation (I) with $-3$ and then obtain, $-3x+3y=-12$ (III) And add the equations (II) and (III) as given below: \begin{align} & 3x+7y-3x+3y=-18-12 \\ & 7y+3y=-30 \\ & 10y=-30 \\ & y=-3 \end{align} Put the value of y in the equation (I), to compute the value of x. \begin{align} & x-\left( -3 \right)=4 \\ & x+3=4 \\ & x=1 \end{align} Thus, the values are $x=1$ and $y=-3$.