SOLUTION: Solve. 2x - 3y + z = 5 x + 3y + 8z = 22 3x - y + 2z = 12 I've tried it by: Multiplying the second equation by 2 and the third equation by 2; as well as multiplying the second

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Question 639177: Solve.
2x - 3y + z = 5
x + 3y + 8z = 22
3x - y + 2z = 12
I've tried it by: Multiplying the second equation by 2 and the third equation by 2; as well as multiplying the second equation by 4 and the third equation by 2 but neither seemed to work. Please help.
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!


Multiply the 2nd equation by -2, then add the first and second equations.

Then multiply the original 2nd equation by -3, then add the second and third equations.

Write back and tell me what you have so far.

John

My calculator said it, I believe it, that settles it
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Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
(1)   2x - 3y +  z =  5
(2)    x + 3y + 8z = 22
(3)   3x -  y + 2z = 12

The smart thing to notice is that if you add (1) and (2)
the -3y and the +3y will eliminate the y's

(1)   2x - 3y +  z =  5
(2)    x + 3y + 8z = 22
-----------------------
      3x      + 9z = 27

and that can be divided through by 3

(4)         x + 3z = 9

Now we need another equation that does not contain y.

We multiply (3) by 3 and add it to (2)

(3)   3x -  y + 2z = 12

Multiply it by 3:

      9x - 3y + 6z = 36
(2)    x + 3y + 8z = 22
-----------------------
     10x      +14z = 58

We can divide that through by 2 and get

(5)        5x + 7z = 29

Now we put (4) and (5) together and we now
have a system of only 2 equations in only
2 unknowns.

(4)         x + 3z =  9
(5)        5x + 7z = 29

Multiply (4) through by -5 and add to (5)

         -5x - 15z = -45
(5)       5x +  7z =  29
-------------------------
               -8z = -16
(6)              z = 2

Substitute z = 2 into (4)

(4)         x + 3z = 9
          x + 3(2) = 9
             x + 6 = 9
(7)              x = 3

Substitute x=3 and z=2 into (3)

(3)   3x -  y + 2z = 12
    3(3)- y + 2(2) = 12
         9 - y + 4 = 12
            13 - y = 12
                -y = -1
(8)              y = 1

From (7), (8), and (6),

Solution = (x,y,z) = (3,1,2)

Edwin
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