Simultaneous Equations

[David Ellis]

It's a long time since I was at school, and I've forgotten a lot. But I'm
trying to solve three equations. They look like they're solvable, but I
can't do it. It's possible that they contain some absurdity that makes them
a=10+b+c
2b=10+a+c
3c=10+a+b
Is there as solution?

This is a fairly simple system of three simultaneous linear equations
in three variables. Here is one way to solve the system. The above
equations can be rewritten as:

(1) a - b - c = 10
(2) -a + 2b - c = 10
(3) -a -b +3c = 10

We can easily combine them to eliminate the variable a.

Adding (1) and (2) yields:

(4) b -2c = 20

Adding (1) and (3) yields:

(5) -2b + 2c = 20

Equations (4) and (5) are a system of two linear equations in two
variables. Adding (4) and (5) yields b = -40, and substituting this
value into either (4) or (5) yields c = -30.

Substituting these values of b and c into any of the equations (1),
(2) or (3) yields a = -60.

It is easy to verify that the assignment (a,b,c) = (-60,-40,-30)
satisfies all three of the original equations.

--
David J Ellis
7 Hampton Road / Natick, MA 01760
***@verizon.net

[Barry Schwarz]

It's a long time since I was at school, and I've forgotten a lot. But I'm
trying to solve three equations. They look like they're solvable, but I
can't do it. It's possible that they contain some absurdity that makes them
a=10+b+c
2b=10+a+c
3c=10+a+b
Is there as solution?

The general method is to pick an equation and rearrange it so one
variable is expressed in terms of the others. (In this case, eq 1
already gives you a in terms of b and c.) You then substitute for
that variable in the other equations. In this case, you get

2b = 10 + (10 + b + c) + c = 20 + b + 2c
3c = 10 + (10 + b + c) + b = 20 + 2b + c

You now have the same problem but one less expression. Rearranging
the first of these results in b = 20 + 2c and substituting in the
second produces

3c = 20 + 2(20 + 2c) + c = 60 + 5c

Now that you have only one variable, solve normally. Substitute this
value in the expression for b and solve. Then substitute the values
for both b and c in the expression for a and solve.

To make sure you understand, go back to the first step and use eq 2 to
solve for either a or c and repeat the process.

When there is a unique solution, this process will work regardless of
the number of variables or the values of any exponents they are raised
to. You also need to be aware of two exceptional conditions:

It is possible for the equations to be inconsistent. This
means there is no solution, as in:

a = b+6
2b = 2a -50

It is possible for the equations to not be independent. This
means there are multiple solutions, as in:

2a = b+6
3b = 6a-18


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