i need to use the substitution method to find the variables x,y, and z in the equations 3x+2y-4z=20 ; 2x-y+3z=-5 ; and x+y+z=3.
( i solved for y, using the 3rd equation, to form the two new equations of 5x-2z=26 and x=z-8)
You made a mistake with signs.
Solving x+y+z=3 for y gives y=3-x-z
Substituting that in
3x+2y-4z = 20
3x+2(3-x-z)-4z = 20
3x+6-2x-2z-4z = 20
x-6z+6 = 20
x-6z = 14
You had a 26 instead of a 14 because you added 6 to 20 when you
should have subtracted 6 (or added -6 which is the same thing)
Substituting that same thing in the other equation gives:
2x-y+3z=-5
2x-y+3z = -5
2x-(3-x-z)+3z = -5
2x-3+x+z+3z = -5
3x+4z-3 = -5
3x+4z = -2
You got x=z-8. What you did wrong was in this step 2x-(3-x-z)+3z = -5
Now you have this system of equations:
x-6z = 14
3x+4z = -2
I think you will be able to solve that system by solving the first one for
x and substituting in the second one. You seem to be having trouble with
signs. Slow down a bit and concentrate on all the rules of signs.
The solution is x=2, y=3, z=-2
Edwin