SOLUTION: i am working on solving problems wit h3 variables. it has been over an hour and can't get a correct answer. it seems easy, but i am having no luck. if I could see the work to arri

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Question 234909: i am working on solving problems wit h3 variables. it has been over an hour and can't get a correct answer. it seems easy, but i am having no luck.
if I could see the work to arrive the right answers that would be great. i have the answers in the back of my book.
2x-3y+z=5
x+3y+8z=22
3x-y+2z=12

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
2x-3y+z=5
x+3y+8z=22
3x-y+2z=12
-----------------------
Let the 2nd equation become the 1st:
x+3y+8z=22
2x-3y+z=5
3x-y+2z=12
-----------------
Subtract 2 times the 1st from the 2nd;
Subtract 3 times the 1st from the 3rd.
x+3y+8z=22
0-9y-15z=-39
0-10y-22z=-54
--------------------
Subtract 3rd from 2nd to get:
x+3y+8z=22
0+y+7z=15
0-10y-22z=-54
-------------------
Add 10 times 2nd to 3rd to get:
x+3y+8z=22
0+y+7z=15
0+0+48z=96
-------------------
Divide thru 3rd by 48 to get z = 2
x+3y+8z=22
0+y+7z=15
0+0+z=2
-----------------
Substitute z=2 into the 2nd and into the 1st lines:
x+3y+16=22
0+y+14=15
0+0+z=2
---------------
Solve the 2nd for y: y = 1
---------------------------------
Substitute that into the 1st to get:
x + 3*1+16=22
0 + y + 0 = 1
0 + 0 + z = 2
----------------------
solve the 1st for x: x = 3
--------------------------------
x = 3
y = 1
x = 2
=========================
Cheers,
Stan H.




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