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
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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