Systems of Equations


A system of equations consists of two or more equations considered simultaneously.

A solution to a system of equations is the set of all points that make all of the equations true.

Generally speaking, to solve a system of equations you need as many unique equations as there are variables.

Solution Methods


For simple systems (only two variables) it is often convenient just to graph the equations on a computer or graphing calculator and see where they intercept. By zooming in it is possible to find an answer to several decimal places of precision–more than adequate for most business applications.

Let’s say we are given the following system of equations:

Example of a system of equations

At first glance, this system looks like it will be a little tough to solve algebraically. Luckily, we have our trusty graphing calculator or computer algebra system and we can input them and graph them:

Graph of the Example System of Equations

It is immediately apparent that this system has only one solution that lies on all three curves. Putting our cursor over it and zooming in, we find out that it occurs at (2, 3).

Incidentally, while we are not allowed to use computers on the waiver exam, we are allowed calculators. I assume this means that graphing calculators are OK?


  1. Solve one of the equations for one of the variables.
  2. Use this expression to replace that variable in the next equation.
  3. Solve this equation for the next variable.
  4. Repeat steps 2-3 until you have a solution value for the last variable.
  5. Plug this value into your intermediate equations until you get values for the other variables.

Let’s try solving the system of equations above by substitution. We will work with equations (1) and (3) first because they don’t have any fractions and will be easier to deal with. Equation (3) is already solved in terms of y so we can plug it in on the left side of equation (1):

2x - 1 = x^3 + x^2 - 4x - 1

Subtracting the term 2x – 1 from both sides of the equation gives us:

0 = x^3 + x^2 - 6x

Factoring, we find:

0 = x(x^2 + x - 6) = x(x - 2)(x + 3)

So x = {0, 2, -3}. We substitute these x values back into equation (3) to find the corresponding y values:

Plugging values back into equation (3)

So the solution set for equations (1) and (3) is {(0,-1), (2,3), (-3,-7)}. However, since we are also constrained by equation (2), we need to plug each of these points into it and see if they work:

(0, -1) does not satisfy the equation:

-1 ≠ 5

(2, 3) does, however:

3 = 3

And (-3,-7) does not:

-7 ≠ 0

Therefore, (2,3) is the only point that is a solution to the system of all three equations.

By Elimination

Elimination is the other main algebraic method of solving systems. It is usable as long as the equations in the system are either linear, or can be expressed as linear combinations of similar terms (such as some systems of polynomials). The steps are:

  1. Rewrite all equations in standard form (i.e. terms with variables on the left in descending order by degree, constants on the right).
  2. Pick any two equations in the system. Multiply both sides of one of the equations by a constant such that one of the terms becomes equal to one of the terms of the other equation.
  3. Subtract one of the two equations from the other to cancel out a term.
  4. If the system has more than two equations, repeat the process until you have canceled out all but one variable.
  5. Substitute in the value of this variable to find the values of the other variables.

As an example, let’s consider the following system of two linear equations:

2x + 3y = 12, 3x + 5y = 1

Both of these equations are already in standard form, so the first step has been done for us.

We want the x terms to cancel, so we start out by multiplying both sides of equation (1) by 3/2:

Multiply both sides by 3/2

Now subtract equation (2) from equation (1) to cancel the x terms:

Subtract equation (2) from equation (1)

Solve for y:

Solve for y

And plug the y value of 34/19 into either equation to find a value for x. In This case, we will use equation (1):

Plug in y value to get x value

So the solution to this system is (63/19, 34/19). Sorry about the ugly fractions; I make up these examples on the fly.

When solving larger systems, many people use matrix notation instead of writing out all of the equations. The process, however, is conceptually identical.


Beecher, J., Penna, J., & Bittenger, M. (2008). College Algebra [3rd Ed.]. Boston, MA: Pearson.

Linear EquationsBack to Linear Equations Forward to CalculusCalculus

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: