Algebra
posted by Samantha
1. I need help solving the system by substitution
xyz=8
4x+4y+5z=7
2x+2z=4

Damon
get rid of y so you can use the third eqn
y = 8xz
so
4x + 4(8xz) + 5z = 7
8 x + z + 32 = 7
8x + z = 25 and we know
+2x +2z = 4
16x +2z =50
+ 2x +2z = 4

18 x = 54
x = 3
now go back 
Reiny
let's clean them up a bit first
xyz=8 > x + y + z = 8 , #1
2x+2z=4 > x + z = 2 , #3
how about subtracting those two:
y = 6 , well that is a good start
let's put that into #1
x + 6 + z = 8
x + z = 2 , but that is what the original #3 said, so we have nothing new.
but let's change that to z = 2x and put that data into #2
4x + 24 + 5(2x) = 7
4x + 24 + 10  5x = 7
9x = 27
x = 3
then z = 23 = 1
x = 3, y = 6, z = 1 
Damon
well, we both used elimination at one point or another, going back a bit we had
8x + z = 25 and we know
+2x +2z = 4
so use z = 8x 25 to substitute in the last one
2x +2(8x25) = 4
18 x 50 = 4
18 x = 54
x = 3 again, whew ! 
Damon
You said to use substitution but there is no practical reason not to use elimination and I did so automatically for the second half of the problem, sorry.

Samantha
Thank you very much Reiny for showing me the proper steps. and Thank you too Damon for trying to help me.
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