SOLUTION: please help me with this question.I am having difficulty eliminating variables. solve the linear system without using matrices or determinants x+y+z=2 x-y+3z=12 2x+5y+2z=-2

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Question 117044: please help me with this question.I am having difficulty eliminating variables.
solve the linear system without using matrices or determinants
x+y+z=2
x-y+3z=12
2x+5y+2z=-2

Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
solve the linear system without using matrices or determinants
x + y + z = 2
x - y +3z =12
2x+5y +2z =-2
:
Let's eliminate z, and get it down to two, 2 unknown equations:
Multiply the 2nd equation by -1 and add all three.
x + y + z = 2
-x + y -3z =-12
2x +5y +2z = -2
------------------adding eliminate z
2x +7y +0z = -12
2x + 7y = -12; our 2 unknown equation
:
Here we luck out. Multiply eq1 by -2 and add it to eq3, I should have noticed that at first.
-2x - 2y - 2z = -4
2x + 5y + 2z = -2
--------------------adding eliminates x & z, easy to find y
0x + 3y + 0z = -6
3y = -6
y = -2
:
Use our two unknown equation to find x:
2x + 7y = -12
Substitute -2 for y
2x + 7(-2) = -12
2x -14 = -12
2x = -12 + 14
2x = +2
x = +1
:
Use the 1st equation to find z:
x + y + z = 2
1 - 2 + z = 2
z = 2 + 1
z = +3
:
Check our solution the eq 2
x - y + 3z = 12
1 -(-2) + 3(3) = 12
1 + 2 + 9 = 12 confirms our solutions
:
It pays to study these systems and see if there is a single operation that will eliminate 2 unknowns at once, otherwise just do what is necessary to get two, 2 unknown equations to work with. Then you can use elimination or substitution quite easily.

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