micahsmith507
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# How do you solve this system of equations using substitution 3x-2y=4 4x-3y=7

$\begin{cases} 3x-2y=4 \\ 4x-3y=7 \end{cases} \\ \begin{cases} -2y=4-3x \\ 4x-3y=7\end{cases} \\ \begin{cases} y= \frac{4-3x}{-2} \\ 4x-3y=7 \end{cases} \\ \begin{cases}y=\frac{4-3x}{-2}\\ 4x-3(\frac{4-3x}{-2})=7\end{cases} \\ \begin{cases}4x+6-4,5x=7\\y=\frac{4-3x}{-2} \end{cases} \\ \begin{cases}6- \frac{1}{2}x=7\\ y=\frac{4-3x}{-2} \end{cases} \\ \begin{cases} - \frac{1}{2}x=1\\y=\frac{4-3x}{-2} \end{cases} \\ \begin{cases}x=-2 \\y=\frac{4-3x}{-2} \end{cases}$ $\begin{cases}x=-2 \\ y= \frac{4-3(-2)}{-2} \end{cases} \\ \begin{cases}x=-2 \\ y=-5 \end{cases}$