SOLUTION: 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 t

Algebra ->  Expressions-with-variables -> SOLUTION: 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 t      Log On


   



Question 396154: 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)
i dont know what to do next. please help! i would greatly appreciate it

Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
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