Answer
The values of x and y for the given system $y=3-3x\text{ and}\,\ 3x+4y=6$ are $x=\frac{2}{3}\,\,$ and $y=1$.
Work Step by Step
$y=3-3x$ ( $1$ )
$3x+4y=6$ ( $2$ )
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 $x=\frac{2}{3}\,\,$ , $y=1$.
