Question

What is the solution of the system? Use elimination.

2x + 2y +z = 7
-X – y +z = -5
x + 3y – 4z = 12

Answer 1

The solutions to the system of the equations by the elimination method will be:

`x=2,\:z=-1,\:y=2`

Explanation:

Given the system of the equations

`2x\:+\:2y\:+z\:=\:7`

`-x-\:y\:+z\:=\:-5`

`x+3y-4z=12`

solving the system of the equations by the elimination method

`\begin{bmatrix}2x+2y+z=7\\ -x-y+z=-5\\ x+3y-4z=12\end{bmatrix}`

`\mathrm{Multiply\:}-x-y+z=-5\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:-2x-2y+2z=-10`

`\begin{bmatrix}2x+2y+z=7\\ -2x-2y+2z=-10\\ x+3y-4z=12\end{bmatrix}`

`-2x-2y+2z=-10`

`+`

`\underline{2x+2y+z=7}`

`3z=-3`

`\begin{bmatrix}2x+2y+z=7\\ 3z=-3\\ x+3y-4z=12\end{bmatrix}`

`2x+6y-8z=24`

`-`

`\underline{2x+2y+z=7}`

`4y-9z=17`

`\begin{bmatrix}2x+2y+z=7\\ 3z=-3\\ 4y-9z=17\end{bmatrix}`

Rearranging the equations

`\begin{bmatrix}2x+2y+z=7\\ 4y-9z=17\\ 3z=-3\end{bmatrix}`

solve
`3z=-3` for z:

`z=-1`

`\mathrm{For\:}4y-9z=17\mathrm{\:plug\:in\:}z=-1`

solve
`4y-9\left(-1\right)=17` for y:

`4y-9\left(-1\right)=17`

`4y+9=17`

`4y=8`

`y=2`

`\mathrm{For\:}2x+2y+z=7\mathrm{\:plug\:in\:}z=-1,\:y=2`

solve
`2x+2\cdot \:2-1=7` for x:

`2x+2\cdot \:2-1=7`

`2x+3=7`

`2x=4`

`x=2`

Therefore, the solutions to the system of the equations by the elimination method will be:

`x=2,\:z=-1,\:y=2`