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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