Question 570588: How do you solve system of linear equations algebraically that have different coefficients for example
2a+6z=4
3a+7z=6
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 2a + 6z = 4
3a + 7z = 6
:
remember you can multiply both sides of the equation without changing equality
Multiply one or both equations so that adding or subtracting eliminates one of the unknowns
:
Multiply the 1st equation by 3, multiply the 2nd equation by -2, results:
+6a + 18z = 12
-6a - 14z =-12
---------------adding eliminates a, find z
0a + 4z = 0
z = 0
:
Find a using the 1st original equation
2a + 6z = 4
Replace z with 0
2a + 0 = 4
a = 4/2
a = 2
:
The solution: a=2; z=0
;
:
YOu can confirm this in the 2nd original equation
3a + 7z = 6
Replace a with 2, replace z with 0
3(2) + 7(0) = 6
6 + 0 = 6
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