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A general system of simultaneous linear equations is a group of n linear equations that relate n variables, where n is any natural number.  Ideally, the number of equations in the system equals the number of variables, and the system is written as:

Here x, y, z, ... are the variables, and a₁ through a_n, b₁ through b_n, c₁ through c_n, ..., and k₁ through k_n are constants.

Each practical problem has its own properties and relationships and may involve any number of equations and any number of variables. The variety determines whether or not the system can be solved. If the system consists of linear equations, it is usually solvable when the number of equations is at least equal to the number of variables.  But the equations have to be compatible and not redundant.

To be compatible, equations have to have a common value among them. When plotted on a graph, for instance, the lines representing the respective equations have to cross one another at a common point.

An equation in a system is redundant if it expresses the same relationship among the variables as another equation in the system. For example, 3x + 6 y = 12 and x + 2y = 4 are redundant.

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In a two-variable application of the system of linear equations, all but two of the variables have zero coefficients. For the variables x and y, for instance, we might have the equations:

2x - y = 4

and

x - 2y = 5.

Solving this system of equations is more difficult than solving a one-variable equation. The approach is to find the value of one of the variables at a time. This can be done by eliminating one of them, and can be attained either (1) by subtracting one equation from the other or (2) expressing one variable in terms of the other and adding or substituting.

Using subtraction to eliminate x, we subtract one equation from the other, which means we subtract the left side of one of them from the left side of the other and the right side from the right side. So let's subtract the second from the first.

As the equations stand, straight subtraction wouldn't eliminate any variable at all. To make subtraction work, we have to change the form of one of them to an equivalent form. To this end, say to eliminate x, we can multiply the second equation by 2, to get:

2x - 4y = 10.

If we now subtract the two sides, we get:

2x - 2x - y + 4y = 4 - 10,

which is the same as:

3y = 6,

or

y = 2.

Now, by substituting for y in either one of the equations, we can find x. Plugging 2 in for y in the first equation, for instance, we get:

2x - 2 = 4

or

x = 3..

If these are the correct values for x and y, they should satisfy both equations. So let's see. We first substitute in the first equation, and get:

6 - 2 = 4,

which is OK.

Substituting in the second equation, we get:

3 - 4 = 5.

Did I make a mistake? Or are the two equations incompatible? Let's try solving them by substitution to see if we get the same wrong result.

To use substitution, we have to rearrange one of the equations to express one of the variables in terms of the other one and then substitute that expression in the other equation. So let's start with the second equation and use it to express x as:

x = 5 + 2y.

If we now substitute this expression for x in the first equation we get:

2(5+ 2y) -y = 4,

or

10 + 4y - y = 4,

which is the same as:

3y = - 6,

or

y = - 2.

Ah ha! We have a negative value for y this time! Before, we had y = +2.

But let's carry on to see what we now get for x. Substituting 2 for y in the second equation, we get

x = 5 - 4,

or

x = 1.

Now let's test our new results in each of our original equations.

Substituting for x and y in 2x - y = 4, we get:

2 -(- 2) = 4,

which is OK.

And substituting in x - 2y = 5, we get

1 - 2(-2) = 1 + 4 = 5,

which is also OK.

So now we've solved our system of two simultaneous linear equations properly and learned that y = -2, not +2.

The mistake, as you can see for yourself, was made in the interchange of values between the two sides. This mistake led to a value of 3 for x, when the proper result should have been 1.

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In a 3-variable system, the coefficients of all but three of the variables in the general system are zero, as you will have guessed.

As an example of such a system, we might have:

x + y + z = 0

3x - 2y = 0

2x + 5y - 3z = 15.

In this example, notice that z doesn't appear in the second equation, so the coefficient for z has to be zero. That is, 0z = 0.

Solving for x, y, and z in such a system of equations uses the same methods as when solving systems of two variables, but the process gets more involved because you have to work through two different levels. I'll leave the details to you.

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