SOLUTION: solve the given system of equations
x+y+8z=40
x+y+6z=32
x+7y+3z=50
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Question 802534: solve the given system of equations
x+y+8z=40
x+y+6z=32
x+7y+3z=50
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
solve the given system of equations
x+y+8z=40
x+y+6z=32
x+7y+3z=50
------------
All coefficients of x are 1, so eliminate x first.
x+y+8z=40
x+y+6z=32
------------ Subtract
2z = 8
z = 4 even better
=============
x+y+6z=32
x+7y+3z=50
-------------- Subtract
-6y + 3z = -18
-6y + 12 = -18
y = 5
==================
Sub for y & z in any equation to find x
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