Question

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

Answer 1

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

Answer 2

let's clean them up a bit first
-x-y-z=-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 = 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

Answer 3

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 !

Answer 4

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.

Answer 5

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