Alg 2
posted by john .
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:
6a2b=18
3b+5c=34
a+6c=28

Alg 2 
Reiny
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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