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Question 989590: Hi, I have a problem that deals with linear equations with three variables:
SOLVE:
x+y=7
y+z=8
x+2z=2
and no, I cannot simply use matrices to solve this question. I tried eliminating 'z' from the second and third equation, but I got stuck because I can't eliminate 'z' from the first equation. Please help and thank you so much to whoever helps!
Found 2 solutions by stanbon, MathTherapy: Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! SOLVE:
x+y=7
y+z=8
x+2z=2
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express y and x in terms of "z"::
y = 8-z
x = 2-2z
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Substitute for x and y and solve for "z"::
2-2z + 8-z = 7
-3z + 10 = 7
-3z = -3
z = 1
----
y = 8-z = 7
x = 2 - 2z = 2-2 = 0
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Cheers,
Stan H.
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Answer by MathTherapy(10555) (Show Source):
You can put this solution on YOUR website!
Hi, I have a problem that deals with linear equations with three variables:
SOLVE:
x+y=7
y+z=8
x+2z=2
and no, I cannot simply use matrices to solve this question. I tried eliminating 'z' from the second and third equation, but I got stuck because I can't eliminate 'z' from the first equation. Please help and thank you so much to whoever helps!
x + y = 7_____y = 7 - x ------- eq (i)
y + z = 8 ------ eq (ii)
x + 2z = 2 ------ eq (iii)
7 - x + z = 8 ------ Substituting 7 - x for y in eq (ii)
- x + z = 1 ------ eq (iv)
3z = 3 ------ Adding eqs (iv) and (iii)
z = , or 
Complete this by substituting 1 for z in eq (ii) to determine the value of y, and finally, substitute the value of y into eq (i) to get the value of x
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