Teaching Linear Inequalities in Two Variables

In Elementary Algebra, I love teaching linear inequalities in two variables at the end of the chapter on graphing lines because it gives me one last chance to go over how to efficiently graph a line. Use the intercepts or use the slope? Is it vertical or horizontal?

It can be a frustrating section for me as well. Some students really have a hard time grasping which half-plane to shade. The way I have always done it:

• Determine if the line is dashed or solid.
• Graph the line using an efficient technique.
• Pick a test point not on the line, preferably the origin (0,0).
• Substitute the coordinates into the original inequality for x and y.
If the inequality is true, then the ordered pair is a solution. Shade the half-plane containing the test point.
If the inequality is false, then the ordered pair is not a solution. Shade the half-plane that does not contain the test point.

Each semester there will be a group that can do everything except figure out which side to shade. It’s frustrating because I think it is pretty clear and straightforward.

Next semester I am changing my approach. Suppose the example is $2x-3y\le 12$.

• I will have them graph the line, using the intercepts of (6,0) & (0,-4).
• We will then label the line $2x-3y=12$ and we will go over the idea that this line represents all of the ordered pairs for which $2x-3y$ is equal to 12.
• I will explain that this line divides the rectangular plane into two half-planes. One of the half-planes contains points for which $2x-3y<12$ and $2x-3y>12$.
• I’ll pick a point below the line to the right, like (6,-5) and substitute the coordinates into the expression $2x-3y$, which will result in 27. Since $27>12$, we will then label that half-plane as $2x-3y>12$.
• I’ll pick the origin, which is above the line to the left, and substitute the coordinates into the expression $2x-3y$, which will result in 0. Since $0<12$, we will then label that half-plane as $2x-3y<12$.
• We will finish by shading the side labeled $2x-3y<12$.

After 2 or 3 examples done this way, I think the students will understand what each half-plane represents and have an easier time determining which half-plane to shade.

-George

I am a math instructor at College of the Sequoias in Visalia, CA. If there’s a particular topic you’d like me to address, or if you have a question or a comment, please let me know. You can reach me through the contact page on my website – http://georgewoodbury.com.

• 1. Whit Ford  |  March 27, 2011 at 11:20 am

My favorite alternative method is:

1) If necessary, convert the equation to slope-intercept form by solving for the dependent variable (do NOT skip this step, as this technique is only reliable when the dependent variable is all by itself on one side and has a coefficient of positive one)

2) Graph the line it represents, using either a solid or dotted line

3) Turn your graph 90 degrees to the right, so that the dependent axis is now horizontal, with positive values to the right and negative values to the left – just like a number line.

4) If the dependent variable is “greater than”, shade to the right of where it crosses the dependent axis (just as you would with a one variable inequality). Otherwise, shade to the left of where the line crosses the axis.