## Precalculus (6th Edition) Blitzer

The solution of the system is $\underline{\left( 4,-2 \right)}$
Let us consider the system of the given equations: $2x+5y=-2$ (I) $3x-4y=20$ (II) Multiply equation (I) with 4 and thus obtain $8x+20y=-8$ (III) Multiply equation (II) with 5 and thus obtain $15x-20y=100$ (IV) Add equations (III) and (IV) as follows: \begin{align} & 8x+20y+15x-20y=-8+100 \\ & 23x=92 \\ & x=\frac{92}{23} \\ & x=4 \end{align} Put the value of x in equation (I), to obtain the value of y as shown below: \begin{align} & 2\left( 4 \right)+5y=-2 \\ & 8+5y=-2 \\ & 5y=-10 \\ & y=-2 \end{align} Thus, the values are $x=4$ and $y=-2$.