Question

System of Equations
Need help
Trig

Answer 1

Looks like the system is

`\begin{cases}-2x+y+6z=1\\3x+2y+5z=16\\7x+3y-4z=11\end{cases}`

We can eliminate
`y` by taking

`(3x+2y+5z)-2(-2x+y+6z)=16-2(1)`

`\implies3x+2y+5z+4x-2y-12z=16-2`

`\implies7x-7z=14`

`\implies x-z=2`

so that
`z=x-2`, and

`(7x+3y-4z)-3(-2x+y+6z)=11-3(1)`

`\implies7x+3y-4z+6x-3y-18z=11-3`

`\implies13x-22z=8`

Substitute
`z=x-2` into this last equation and solve for
`x`:

`13x-22(x-2)=8`

`\implies13x-22x+44=8`

`\implies-9x=-36`

`\implies x=4`

Then

`z=x-2`

`\implies z=4-2`

`\implies z=2`

Plug these values into any one of the original equation to solve for
`y`:

`-2x+y+6z=1`

`\implies-2(4)+y+6(2)=1`

`\implies-8+y+12=1`

`\implies y=-3`

Hence the solution is x = 4, y = -3, and z = 2.