# Algebra

posted by Samantha

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

1. Damon

get rid of y so you can use the third eqn

y = 8-x-z
so
-4x + 4(8-x-z) + 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

2. Reiny

let's clean them up a bit first
-x-y-z=-8 ---> x + y + z = 8 , #1
2x+2z=4 ----> x + z = 2 , #3

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 = 2-x and put that data into #2

-4x + 24 + 5(2-x) = 7
-4x + 24 + 10 - 5x = 7
-9x = -27
x = 3
then z = 2-3 = -1

x = 3, y = 6, z = -1

3. 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(8x-25) = 4
18 x -50 = 4
18 x = 54
x = 3 again, whew !

4. 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.

5. 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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