SOLUTION: The whole third variable thing confuses me. When I try and do it by substitution I always end up with this long chain that I don't know how to reduce to find the answer for the sec

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: The whole third variable thing confuses me. When I try and do it by substitution I always end up with this long chain that I don't know how to reduce to find the answer for the sec      Log On


   

Question 6156: The whole third variable thing confuses me. When I try and do it by substitution I always end up with this long chain that I don't know how to reduce to find the answer for the second variable. Thanks in advance.
Solve for X, Y and Z
22X + 5Y + 7Z = 12
10X + 3Y + 2Z = 5
9X + 2Y + 12Z = 14

Answer by glabow(165) About Me  (Show Source):
You can put this solution on YOUR website!
You are looking to find one variable in terms of the others. You do this by rearranging. Rearrange the last equation to find y. (I picked that one because it looked good. Any others will work too.)
9x+%2B+2y%2B12z=14 rearranges to 2y=14-12z-9x and then to y=%2814-12z-9x%29%2F2
Now substitute this value for y in the equations. The first looks like this:
22x%2B%285%2814-12z-9x%29%2F2%29%2B7z=12
22x%2B%28%2870-60z-45x%29%2F2%29%2B7z=12
44x%2B70-60z-45x%2B14z=24
-x-46z=-46
x+%2B46z=46
Do the same substitution with the second equation.
10x%2B%283%2814-12z-9x%29%2F2%29%2B2z=5
10x%2B%28%2842-36z-27x%29%2F2%29%2B2z=5
20x%2B42-36z-27x%2B4z=10
-7x-32z=-32
7x%2B32z=32
Now do the same elimination with the two equations in x and z.
The first equation says
x=46-46z
Substitute x in the second equation.
7%2846-46z%29%2B32z=32
322-322z%2B32z=32 [resist the urge to simplify!]
-322z%2B32z=32-322
32z-322z=32-322
%2832-322%29z=%2832-322%29 [factor out 32-322]
z=1 [divide by (32-322) !]
Now substitute the value for z in x=46-46z
x=0
Now substitute the values for x and z in 9x%2B2y%2B12z=14
2y=14-0-12
y=1
Check in the original equations.