Expert: Sherry Wallin - 7/1/2010

Question
How can I get the values of the three variables?

4a + 2b + c = 41
25a - 5b + c = 20
4a - b = 0

Thank You!!

Answer
Hi Christine~
    This is just a system of 3 equations and 3 unknowns. You probably have solved these through substitution or through elimination. Since the 3rd equation only has two variables you could solve for b = 4a and everywhere there is a b put in 4a:

4a + 2(4a) + c = 41  -> 12a + c = 41  Call this equation 1a since it came from the 1st eqn
25a -5(4a) + c = 20  ->  5a + c = 20  Call this equation 2a since it came form the 2nd eqn

subtract eqn 2a from 1a: 12a + c = 41
                                       -5a - c = -20
add them                                7a = 21 -> a = 3

Now substitute a = 3 into 4a - b = 0 getting 4(3) - b = 0 -> 12 - b = 0 -> b = 12
Use either the 1st or 2nd equation to find c: 4(3) + 2(12) + c = 41 -> 12 + 24 +  c = 41 ->
36 + c = 41 -> c = 5

The solution appears to be a = 3, b = 12, and c = 5 but you will want to check that these values work in all three of the ORIGINAL equations.

Math Prof