This chapter introduces us to solving a system of equations. In part one of this course (Math 1131), you learned how to solve linear equations (such as, 25 = x + 5), for the unknown variables. This time you’ll learn to solve a system of linear equations with two unknowns, such as:


The reason this is important is because systems of equations appear in almost every branch of science and engineering. Some problems in technology can only be described by two equations. For example, to find the two currents I1 and I2 in the circuit below, you must solve the two equations simultaneously:


In this problem, we have two unknowns (I1 and I2, rather than the typical x and y) found in two separate linear equations. We know they’re both linear equations because both variables are of first degree (having an exponent of 1). We must find values for I1 and I2 that satisfy both equations at the same time. The answer to this problem will be revisited at a later section. Just as some applications need two equations for their description, others need three equations – we also study those in a later section.

What’s interesting about the techniques you will learn in this chapter is that they can be applied to a system of any two questions, that is, the equations don’t have to be linear. Of course, they’re relatively easier to solve if they are of first degree. Many textbooks refer to solving systems of equations as solving simultaneously.

To solve a system of two linear equations means to find the point of intersection when both are graphed on an x-y plane. In other words, by solving a system of two linear equations, you are finding out the coordinates where the two lines intersect. The first technique to solving is by graphing the equations manually, then reading the point where the two lines intersect.

Question:   Find the solution to the following system by graphing. null

Solution: Assuming you already learned how to graph linear equations without a table (see link).

As you can tell, solving by graphing definitely has its limitations: it’s slow, takes up space, and not very accurate if you don’t have the right tools to graph or if the coefficient aren’t integers. A better mousetrap is to solve algebraically instead via two different methods, namely, the method of elimination or the method of substitution. Each method has its pros and its cons, and they will be discussed in the next section.