Solve any way you can (system of linear equations)

[Guest]

Okay I don't even know where to start on this one. Any suggestions would be awesome!!

2x – y + z = 8
x + 2y + z = -3
x - 2y – z = 7

Thanks
Andrea

 

[TchrWill]

Re: Solve any way you can

Okay I don't even know where to start on this one. Any suggestions would be awesome!!

2x – y + z = 8
x + 2y + z = -3
x - 2y – z = 7

Try subtracting the 3rd equation from the 2nd and twice the 2nd from the first to eliminate x. Perform a similar operation on the resulting equations and solve for either y or z.

You can do it.

 

[soroban]

Re: Solve any way you can

Hello, Andrea!

[1]\(\displaystyle \;\;2x\)\(\displaystyle \,-\,\)\(\displaystyle y\ +\ z\:=\:8\)
[2]\(\displaystyle \;\;x\ +\ 2y\ +\ z\:=\:-3\)
[3]\(\displaystyle \;\;x\ -\ 2y\)\(\displaystyle \,-\,\)\(\displaystyle z\:=\:7\)

Add [1] and [3]: \(\displaystyle \,3x\,-\,3y\:=\:15\;\;\Rightarrow\;\;x\,-\,y\:=\:5\) [4]

Add [2] and [3]: \(\displaystyle \,2x\:=\:4\;\;\Rightarrow\;\;\L x\,=\,2\)

Substitute into [4]: \(\displaystyle \,2\,-\,y\:=\:5\;\;\Rightarrow\;\;\L y\,=\,-3\)

Substitute into [2]: \(\displaystyle \,2\,+\,2(-3)\,+\,z\:=\:-3\;\;\Rightarrow\;\;\L z\,=\,1\)


Solution: \(\displaystyle \L x\,=\,2,\;y\,=\,-3,\;z\,=\,1\)