Homework Help: Alg 2

Posted by john on Friday, October 5, 2012 at 9:29pm.

Today we learned how to solve systems of equations with 3 variables, however my teacher didn't go over how to do a problem where not every single equation in the system has the 3 variables. Can someone point me to a place that explains this?
the kind of problem im talking about is like this
solve the system of equations:
6a-2b=18
3b+5c=-34
a+6c=-28

Alg 2 - Reiny, Friday, October 5, 2012 at 11:31pm
notice that each of the equations is missing a different variable.
If we can eliminate the variable "a" from the 1st and 3rd, we have a new equation with b's and c's like the 2nd equation

so. from the 3rd a = -6c - 28
sub that into the 1st
6(-6c - 28) - 2b = 18
-36c - 168 - 2b = 18
36c + 2b = -186
b + 18c = -93
let's multiply that by 3
3b + 54c = -279
subtract the 2nd
49c = -245
c = -5
then a = -6(-5) - 28 = 2
and
3b + 5(-5) = -34
3b = -9
b = -3

a = 2 , b = -3 , c = -5

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