## Precalculus (6th Edition) Blitzer

The values of x and y for the given system $y=3-3x\text{ and}\,\ 3x+4y=6$ are $x=\frac{2}{3}\,\,$ and $y=1$.
$y=3-3x$ ( $1$ ) $3x+4y=6$ ( $2$ ) Put the value of y from equation ( $1$ ) in equation ( $2$ ): \begin{align} & 3x+4y=6 \\ & 3x+4\cdot \left( 3-3x \right)=6 \\ & 3x+12-12x=6 \end{align} And subtract like terms: $12-9x=6$ And add $-12$ to both sides: \begin{align} & 12-9x-12=6-12 \\ & -9x=-6 \\ & x=\frac{2}{3} \end{align} And to obtain the value of y, substitute the value $x=\frac{2}{3}$ in equation ( $1$ ) \begin{align} & y=3-3x \\ & y=3-3\cdot \frac{2}{3} \\ & y=1 \\ \end{align} Thus, the values of x and y for the given system $y=3-3x\text{ and}\,\ 3x+4y=6$ are $x=\frac{2}{3}\,\,$ , $y=1$.