## Precalculus (6th Edition) Blitzer

The required value is $x=3,y=1$
We have to multiply the second equation with 2 and then subtract this from the first equation: \begin{align} & 4x+3y-2\left( 2x-5y \right)=15-2\left( 1 \right) \\ & 4x+3y-4x+10y=15-2 \\ & 13y=13 \\ & y=1 \end{align} Substitute the value of $y=1$ in the second equation: \begin{align} & 2x-5\left( 1 \right)=1 \\ & 2x=1+5 \\ & 2x=6 \\ & x=3 \end{align} Thus, the value of $x=3$ and $y=1$.