SOLUTION: 2x+3y=-5 4y-5z=-32 3x+2z=14 Please show me how to solve for X, Y , and Z using substitution. Thank You

Algebra ->  Linear-equations -> SOLUTION: 2x+3y=-5 4y-5z=-32 3x+2z=14 Please show me how to solve for X, Y , and Z using substitution. Thank You      Log On


   



Question 908468: 2x+3y=-5
4y-5z=-32
3x+2z=14
Please show me how to solve for X, Y , and Z using substitution. Thank You

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Substitution isn't the easiest way. Elimination is easier,
But you specified substitution, so here goes:

(1)  2x%2B3y=-5
(2)  4y-5z=-32
(3)  3x%2B2z=14

Pick any equation and solve for any letter:

I'll arbitrarily pick equation 2 and arbitrarily solve for y

4y-5z=-32
4y=5z-32
(4)  y=%285z-32%29%2F4

Substitute for y in (1)

(1)  2x%2B3y=-5
     2x%2B3%28%285z-32%29%2F4%29=-5

Multiply through by 4 to clear the fraction:

     8x%2B3%285z-32%29=-20
     8x%2B15z-96=-20
(5)  8x%2B15z=76

We notice that (3) contains the same letters as (5)

Solve (3) for one of its letters.

I'll arbitrarily solve (3) for z

(3)  3x%2B2z=14
     2z=14-3x
(6)  z=%2814-3x%29%2F2

Substitute in (5)

(5)   8x%2B15z=76
      8x%2B15%2814-3x%29%2F2=76

Multiply by 2 to clear the fraction:

      16x%2B15%2814-3x%29=152
      16x%2B210-45x=152
      -29x%2B210=152
      -29x=-58
      x=%28-58%29%2F%28-29%29
      x=2

Substitute in (6)

(6)  z=%2814-3x%29%2F2
     z=%2814-3%282%29%29%2F2
     z=%2814-6%29%2F2
     z=8%2F2
     z=4

(4)  y=%285z-32%29%2F4  
     y=%285%284%29-32%29%2F4
     y=%2820-32%29%2F4
     y=%28-12%29%2F4
     y=-3   

(x,y,z) = (2,-3,4)

Edwin