Answer:
The solutions to the system of the equations are:
[tex]y=1,\:x=3[/tex]
Step-by-step explanation:
Given the equations
[tex]y=\frac{2}{3}x-1;\:y=-x+4[/tex]
solving the system of equation
[tex]\begin{bmatrix}y=\frac{2}{3}x-1\\ y=-x+4\end{bmatrix}[/tex]
Arrange equation variables for elimination
[tex]\begin{bmatrix}y-\frac{2}{3}x=-1\\ y+x=4\end{bmatrix}[/tex]
[tex]y+x=4[/tex]
[tex]-[/tex]
[tex]\underline{y-\frac{2}{3}x=-1}[/tex]
[tex]\frac{5}{3}x=5[/tex]
[tex]\begin{bmatrix}y-\frac{2}{3}x=-1\\ \frac{5}{3}x=5\end{bmatrix}[/tex]
solving for x
[tex]\frac{5}{3}x=5[/tex]
[tex]5x=15[/tex]
Divide both sides by 15
[tex]\frac{5x}{5}=\frac{15}{5}[/tex]
[tex]x=3[/tex]
[tex]\mathrm{For\:}y-\frac{2}{3}x=-1\mathrm{\:plug\:in\:}x=3[/tex]
[tex]y-\frac{2}{3}\cdot \:3=-1[/tex]
[tex]y-2=-1[/tex]
Add 2 to both sides
[tex]y-2+2=-1+2[/tex]
[tex]y=1[/tex]
Therefore, the solutions to the system of the equations are:
[tex]y=1,\:x=3[/tex]
