Alg 2

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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 -

    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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