a + b = 27
b + c = 12
c + d = 11
d + a = 26
What are a, b, c & d?
can someone provide a formula for figuring this out, I dont want just the answers, but a way to figure it out over and over 
OK, I don’t think that there can be an equation developed for this that would be helpful. But there is a method. Solve in terms of one of the variables:
b = 27 - a
c = 12 - 27 - a = -15 - a
d = 11 - c = 11 - (-15) - a
d + a = 11 - (-15) - a = 26
26 - a = 26
a = 0
b = 27
c = -15
d = 26
This is algebra, specifically simultaneous equations.
Put two equations together like this:
a + b = 27
b + c = 12
Now subtract one from the other to ‘remove’ b (yes, you are allowed to do this!).
N.B. If there was a multiple of b, eg 5b, you’d multiply both sides of that equation by 5.
a - c = 15
Now try to eliminate c, this time adding:
a - c = 15
c + d = 11
Now this usually provides a unique answer.
Here, alas we get
d + a = 26,
which we already know.
So there must be an infinite number of answers, based on:
a = a
b = 27 - a
c = a + 15
d = 26 - a
(e.g. a = 20, b = 7, c = 5, d = 6)
Were you serious, or was this a test?
There’s a slight but annoying problem with the second line here. It should read:
c = 12 - (27 - a) = -15 + a
I agree with glee that there is no unique solution. Usually, with four equations and four unknowns, you get a solution. Unfortunately, unless you know linear algebra, it’s a bit difficult to test for this without doing it all the way out like glee did. However I can tell you that any time you have a setup like this, there will not be a unique solution (there will either be infinity or zero):
a + b = ??
b + c = ??
c + d = ??
d + a = ??
With simple examples like you’ve given, just solving it like the examples given is probably easiest. If you’re looking for a more mechanical method of solving (ie one that doesn’t require intuition and could be turned into an alogorithm easily) you could put the equations into a matrix and use row reduction to get it into echelon form. I don’t have time to do a comprehensive search for good sides, but to get you started, try this page. (It’s just the first one I got when I googled for “row reduction echelon”.
I, too, get:
a=0
b=27
c= -15
d=26
Thats linear algebra my friend. Just put them into a matrix and it will solve itself.
Oops, just saw robnic said the same thing. Sorry.
Well, this is a linear algebra problem, like robinc mentioned, but it’s an ill-conditioned one, as glee and Achernar pointed out. If you write it as a matrix problem MX = V, with M given by:
1 1 0 0
0 1 1 0
0 0 1 1
1 0 0 1
you’d find that M is not invertible because it has a determinant of zero, and this means that no unique solution exists. For what it’s worth, though, the solution that the other posters have given does work, it’s just not the only one that does.
For the general case with a lot of equations and unknowns, you’re best off doing the row reduction scheme as robinc cited. For small examples like this one, it’s probably easier doing it the way the posters here have been showing you. If it’s really big, I’d just get a calculator or computer to solve it for me because I don’t have the patience to do the row reduction.
Pretty much what gr8guy said. I first noticed that the only useful bit of information I could derive was that b=d+1.
So, arbitrarily assigning different values to a would allow you to figure out b, c, and d.
Therefore, a=25, b=2, c=10. d=1.
Or a=24, b=3, c=9. d=2.
Or a=23, b=4, c=8. d=3.
Or — well, you get the idea.
Are you certain you’ve given us all the information you have?
Here’s another take. Usually, with 4 linar equations and 4 unknowns (a, b, c, d), you expect a unique solution for the unknowns. But this is only true if the 4 equations are linearly independent. Without getting too technical, this means that you cannot derive any one of the equations by adding or subtracting multiples of the other three equations. The four equations: (1) a+b=27, (2) b+c=12, (3) c+d=11, (4) d+a=26 are NOT linearly independent. It’s easy to check that you can derive equation (4) by (1) - (2) + (3). That is, subtract equation (2) from equation (1) and add equation (3). In this sense, equation (4) is redundant. What you really have is a system of 3 linear equations with 4 unknowns. When the number of unknowns exceed the number of equations, you have an infinite number of possible solutions. You can pick any one of the unknowns, give it a value, and solve for the other three. As already pointed out, one way to describe all possible solutions is as follows: b=27-a, c=a-15, d=26-a, a = any number. But you can also use: a=27-b, c=12-b, d=b-1, b=any number. You get the idea. Geometrically, the solution is a line, not a single point. ;j