Posted by **MaKenzie Smith** on Saturday, November 13, 2010 at 7:16pm.

Solving systems of linear equations in three variables.

Ok so i don't really totally get these problems. Can you help me solve this one? and explain everything please?

2x+3y=6+z

x-2y=-1-z

3x+y=-1+3z

- Algebra 2 -
**bobpursley**, Saturday, November 13, 2010 at 7:28pm
put them in standard form (x,y,z)

2x+3y-z=6

x-2y+z=-1

3x+y-3z=-1

You can solve these a number of ways, I will show you one.

add 1st and second equations

3x+y=5

now multiply equation 2 by three, then add equations 2 and 3.

6x-5y=-4 check that.

now on these two remaining equations, multiply the first by 2, then subtract.

7y=14

y=2 work back up to find x, then z.

- Algebra 2 -
**MaKenzie Smith**, Saturday, November 13, 2010 at 7:39pm
Ok, this helps a lot, but hold on, i don't get how you got 7y=14? because when i multiply 3x+y=5 by (-2) i get -6x-2y=8. And then when combining that with 6x-5y=-4 i get -7y=4? how did you get 7y=14???

- Algebra 2 -
**bobpursley**, Saturday, November 13, 2010 at 7:43pm
3x+y=5 multiply -2, your way

6x-5y=-4 to get

-6x-2y=-10

6x-5y=-4 add to get

-7y=-14

y=2

- Algebra 2 -
**MaKenzie Smith**, Saturday, November 13, 2010 at 7:46pm
Oh wait, i forgot to multiply the 5 by -2! (idk how i got 8, but watever its not that important) so right you would get y=2.

ok i think i get it now! THANKS!

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