Question

HELP PLS this is my last question

Answer 1

Answer:

The solutions to the system of the equations are:

$y=1,\:x=3$

Step-by-step explanation:

Given the equations

$y=\frac{2}{3}x-1;\:y=-x+4$

solving the system of equation

$\begin{bmatrix}y=\frac{2}{3}x-1\\ y=-x+4\end{bmatrix}$

Arrange equation variables for elimination

$\begin{bmatrix}y-\frac{2}{3}x=-1\\ y+x=4\end{bmatrix}$

$y+x=4$

$-$

$\underline{y-\frac{2}{3}x=-1}$

$\frac{5}{3}x=5$

$\begin{bmatrix}y-\frac{2}{3}x=-1\\ \frac{5}{3}x=5\end{bmatrix}$

solving for x

$\frac{5}{3}x=5$

$5x=15$

Divide both sides by 15

$\frac{5x}{5}=\frac{15}{5}$

$x=3$

$\mathrm{For\:}y-\frac{2}{3}x=-1\mathrm{\:plug\:in\:}x=3$

$y-\frac{2}{3}\cdot \:3=-1$

$y-2=-1$

Add 2 to both sides

$y-2+2=-1+2$

$y=1$

Therefore, the solutions to the system of the equations are:

$y=1,\:x=3$