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
x2y=1z
3x+y=1+3z

Algebra 2  bobpursley, Saturday, November 13, 2010 at 7:28pm
put them in standard form (x,y,z)
2x+3yz=6
x2y+z=1
3x+y3z=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.
6x5y=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 6x2y=8. And then when combining that with 6x5y=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
6x5y=4 to get
6x2y=10
6x5y=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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