Question

HELP PLS this is my last question

Answer 1

The solutions to the system of the equations are:

`y=1,\:x=3`

Explanation:

Given the equations

`y=(2)/(3)x-1;\:y=-x+4`

solving the system of equation

`\begin{bmatrix}y=(2)/(3)x-1\\ y=-x+4\end{bmatrix}`

Arrange equation variables for elimination

`\begin{bmatrix}y-(2)/(3)x=-1\\ y+x=4\end{bmatrix}`

`y+x=4`

`-`

`\underline{y-(2)/(3)x=-1}`

`(5)/(3)x=5`

`\begin{bmatrix}y-(2)/(3)x=-1\\ (5)/(3)x=5\end{bmatrix}`

solving for x

`(5)/(3)x=5`

`5x=15`

Divide both sides by 15

`(5x)/(5)=(15)/(5)`

`x=3`

`\mathrm{For\:}y-(2)/(3)x=-1\mathrm{\:plug\:in\:}x=3`

`y-(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`