Solve the system of equations using substitution.
y= 3/4x-7
4x- 6y=50

Y=3/4x-7
4x-6y=50

4x-6y=50
4x-6(3/4x-7)=50
4x-18/4x+42=50
-1/2x=8
x=-16

y=3/4(-16)-7
y=-12-7
y=-19

$$\begin{cases} y= \frac{3 }{4}x-7 \\ 4x- 6y=50 \end{cases}\\ \\\begin{cases} y= \frac{3 }{4}x-7 \\ 4x- 6(\frac{3 }{4}x-7)=50 \end{cases}\\ \\\begin{cases} y= \frac{3 }{4}x-7 \\ 4x- \frac{9 }{2}x+42=50 \end{cases}\\ \\ \begin{cases} y= \frac{3 }{4}x-7 \\ 4x- 4\frac{1 }{2}x =50 -42\end{cases}$$

$$\begin{cases} y= \frac{3 }{4}x-7 \\ - \frac{1 }{2}x =8 \ \ /*(-2)\end{cases}\\ \\ \begin{cases} y= \frac{3 }{4}*(-16)-7 \\ x =-16\end{cases}\\ \\ \begin{cases} y= -12-7 \\ x =-16\end{cases} \\ \\\begin{cases} y= -19 \\ x =-16\end{cases}$$

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