Question

How do I solve this system of equations by elimination?
{-2x + 2y + 3z = 0
{-2x - y + z = -3
{2x + 3y + 3z = 5

Answer 1

add equation 1 and 3, get
\[5y+6z=5\] add equation 2 and 3 get
\[2y+4z=5\] and solve those for \(y\) and \(z\)

Answer 2

How do I solve for them though?

Answer 3

whoa i made a mistake!

Answer 4

Thank you so much for helping me! (by the way)

Answer 5

add 2 and 3 get
\[2y+4z=2\] sorry

Answer 6

no problem, but how do I find the answer? I don't need you to tell me it, I just want to know how to find it.

Answer 7

\[5y+6z=5\]\[2y+4z=2\] or if you prefer
\[5y+6z=5\]\[x+2z=1\]

Answer 8

I'm following so far

Answer 9

now you have a choice,
you can multiply the second equation by \(-3\) then add to eliminate the \(z\) or you can rewrite the second equation as
\[y=1-2z\] the replace in the first equation to get
\[5(1-2z)+6z=5\]

Answer 10

pretty sure you are going to get \(z=0\) then back substitute to find \(y\) and then finally find \(x\)

Answer 11

actually once you have \(z=0\) it is easy,
you know \(y=1-2z\) and since \(z=0\) you get \(y=1\) and so on

Answer 12

Oh ok, thank you so much!