Question

Hi! I need to find the values of x, y, and z. I'm not really looking for just the correct answer, but HOW to solve it.
x-2y+3z=3
2x+y+5z=8
3x-y-3z=-22

Answer 1

So we are solving a system of 3 equations...same techniques used for a system of 2 equations

\[\large x-2y + 3z = 3\]
\[\large 2x + y + 5z = 8\]
\[\large 3x - y - 3z = -22\]

Now we can either eliminate one of these equations...or we can do a BUNCH of substitutions...your call :D

Answer 2

Ehh, I'll show the substitution...either way is fine though

\[\left(\begin{matrix} x - 2y + 3z = 3 \\ 2x + y + 5z = 8 \\ 3x - y - 3z = -22 \end{matrix}\right)\]

So lets solve that first equation for 'x'

\[\large x = 3 + 2y - 3z\]

Sub that into the second and third equations

so now all 3 we have are

\[\left(\begin{matrix}x = 3 + 2y - 3z \\ 5y - z = 2 \\5y - 12z = -31 \end{matrix}\right)\]

Now solve that second equation for 'y'

\[\large y = \frac{2 + z}{5}\]

Plug that into the third equation for 'y'

The 3 equations we have now are

\[\left(\begin{matrix}x = 3 + 2y - 3z \\ y = \frac{2 + z}{5} \\-11z = -33 \end{matrix}\right)\]

It can now be seen, that last equation we have...can be solved...we would find z = 3

You can then use that in equation 2...to find y = 1

And finally knowing both we can use equation 1 to find x = -4

Answer 3

This was super helpful :) Thank you!