## Student Pointers for Graphing Linear Inequalities

*September 22, 2010 at 12:02 pm* *
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Here is a short video that I put together on this topic: Watch the Video

The overall strategy is:

1) Graph the line – sometimes solid, sometimes dashed.

2) Pick a point not on the line, usually the origin (0,0), as a “test point”.

3) Plug the coordinates of the test point into the original inequality to determine if the “test point” is a solution or not.

4) Shade the side of the graph that contains solutions.

Now for a little more detail.

1) When you graph the line, the first thing you need to determine is whether the line will be solid or dashed.

If the inequality includes “equality” – < or > – then the line is solid because every point on the line is a solution.

If the inequality is a “strict” inequality, “<” or “>”, then we use a dashed line because the points on the line are not solutions of the inequality.

In terms of actually graphing the line:

If it is y = mx + b form, graph the line by plotting the y-intercept (b on the y-axis) and then using the slope (m) to find another point on the line.

If it is in Ax + By = C form, graph the line by finding the x-intercept (set y = 0) and the y-intercept (set x = 0). Put those two points on the graph and draw the line through them.

If it is of the form y = #, like y = 7, then this is a horizontal line.

If it is of the form x = #, like x = 2, then this is a vertical line.

2) Whenever the line does not go through the origin, pick (0,0) as your test point. We do this because plugging in 0 for x and 0 for y is as easy as it can be for us.

If the line goes through the origin, like y = 3x or 2x + 5y = 0, you have to pick another point besides the origin. I’d recommend a point on one of the axes so that at least you are plugging in 0 for one of the variables.

3) Plug the values for x and y into the original inequality.

If this produces a true statement, like 0 < 8 or 0 > -2, then the test point is a solution.

If this produces a false statement, like 0 < -6 or 0 > 5, then the test point is not a solution.

4) The line that you have graphed is like a street on a map, dividing the map into two neighborhoods. One side of the street contains all the solutions, and the other side is “abandoned” with no solutions.

If the test point led to a true inequality, then it is in the “good neighborhood” and we shade that entire neighborhood.

If the test point led to a false inequality, then it is in the “bad neighborhood”. We simply cross the street and shade the other neighborhood.

Do you have any questions? You can reach me through the contact page at my web site – georgewoodbury.com.

-George

*I am a math instructor at College of the Sequoias in Visalia, CA. Each Wednesday I post an article related to General Teaching on my blog. 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.*

Entry filed under: Math. Tags: algebra, amatyc, classroom activities, college, developmental math, education, george woodbury, graph, graphing, Homework, intercepts, linear inequalities, lines, Math, shading, slope intercept form, teaching, test point, woodbury.

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