## Precalculus (6th Edition) Blitzer

The values of x and y for the given system $3x+5y=-2$ and $2x+3y=0$ are $x=6\,\,\text{ and }\,\,y=-4$.
$3x+5y=-2$ ( $1$ ) $2x+3y=0$ ( $2$ ) And multiply $-2$ on the both sides of the first equation: \begin{align} & 3x+5y=-2 \\ & -2\left( 3x+5y \right)=-2\cdot -2 \end{align} $-6x-10y=4$ (a) And multiply by $3$ on the both sides of the second equation \begin{align} & 2x+3y=0 \\ & 3\left( 2x+3y \right)=0\cdot 3 \end{align} $6x+9y=0$ (b) Add (a) and (b): \begin{align} & -6x-10y+6x+9y=4+0 \\ & -y=4 \\ & y=-4 \end{align} And to obtain the value of x, substitute the value $y=-4$ in (b): \begin{align} & 6x+9\cdot -4=0 \\ & 6x-36=0 \end{align} Add 36 both sides: \begin{align} & 6x-36+36=0+36 \\ & 6x=36 \end{align} And multiply both sides by $\frac{1}{6}$ \begin{align} & \frac{1}{6}\cdot 6x=\frac{1}{6}\cdot 36 \\ & x=6 \\ \end{align} Hence, the values of x and y for the given system $3x+5y=-2$ and $2x+3y=0$ are $x=6\,\,\text{ and }\,\,y=-4$.