## Precalculus (6th Edition) Blitzer

The solution is, $\left( 4,-3 \right)$.
Let us consider the equation: $2x+4y=-4$ …… (I) $3x+5y=-3$ …… (II) And for the addition method, the first equation (I) is multiplied by 3 and equation (II) is multiplied by $-2$ so that: \begin{align} & 6x+12y=-12 \\ & -6x-10y=+6 \\ & \overline{\begin{align} & +2y=-6 \\ & \text{ }y=-3 \\ \end{align}} \\ \end{align} Put $y$ into equation (I): \begin{align} & 2x+4\left( -3 \right)=-4 \\ & 2x-12=-4 \\ & 2x=-4+12 \\ & 2x=8 \end{align} After that, simplify, $x=4$ Hence, the values are: $x=4,y=-3$