Solving non-Linear Systems of Equations with 2 Unknowns

This lesson will explore how to use the techniques used in the previous method to solve non-linear equations. Specifically, we’ll look at a system of equations whose variable is found in the denominator.


Notice that both equations displayed above consist of terms whose variables (x and y) are positioned in the denominator, and whose coefficients are fractions. While these “look” linear, they are far from it – remember the negative exponent rule, where x-1 is the same as 1 over x+1. If you graph these two equations, you’ll see the non-linear nature of the system.

From the graph, you can also see that there are two points of intersection, whereas with linear systems with two unknowns, there was only 1.

To algebraically find these points of intersection, you can continue using the technique introduced in the previous section where you find the lowest common denominator before eliminating a variable; however, there’s also another technique that might make life easier. This technique involves substituting an arbitrary fraction with a third and fourth variable, say a and b, in for the current variables, x and y, that will make the equation temporarily linear. After finding a and b as you would normally, you reciprocate them and you have your x and y, respectively. If that’s confusing to you, watch the video below:

A more traditional way to solving these systems is by finding the common denominator first, as shown in the video below. Keep in mind that it’s totally up to you which technique you feel more comfortable using. Personally, I prefer the common denominator technique when the system consists of two unknown variables, but not when there are three, which is the focus of the next section.

Sometimes you may find yourself solving a system that doesn’t fit the traditional style of questions. Take, for instance, the system that’s found in the video below: the terms of consist of complex fractions that eventually leads to a quadratic equation that you’ll need to solve using the quadratic formula. In case you’re curious, take a moment to watch how an intricate system like this is solved.