These are called "simultaneous equations". The way to figure such things out is to solve for one variable in terms of the others in one equation. Then you use that to substitute into the second and third equations (which equation is "first" or "second" is entirely up to you). That eliminates one equation and one variable. Now you are left with just two equations and two variables. Here's an example:
(i) a + 2b + 3c = 52
(ii) 7a - 5b - c = -11
(iii) a + b - c = 1
Let's start by solving equation 1 for a:
(iv) a = 52 - 2b - 3c
Now substitute that into equation (ii) and solve for b:
7(52 - 2b - 3c) - 5b - c = -11
364 - 14b - 21c - 5b - c = -11
19b = 375 - 22c
(iv) b = (375 - 22c) / 19
Do the same for equation (iii):
(52 - 2b - 3c) + b - c = 1
51 - b - 4c = 0
(v) b = 51 - 4c
So now to solve what's left, we substitute the value for b in (iv) into the equation (v):
(375 - 22c) / 19 = 51 - 4c
375 - 22c = 969 - 76c
54c = 969 - 375 = 594
c = 594 / 54 = 11.
Now that we know c, we put it into equation (v) or (iv) to get b:
b = 51 - 4c = 51 - 4(11) = 51 - 44 = 7
And then put both values into any of the original three equations:
a + b - c = 1
a = 1 - b - c = 1 - 7 - (-11) = 5.
Now that you can see how it is done, try this procedure with the numbers in your own question.
Hope this helps,
Stephen La Rocque.