First off, let me say that graphing linear inequalites is much easier than your book makes it seem.
Think about how you've solved linear inequalites on the number line. Suppose they'd asked you to graph something like x > 2. How did you do that?
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You found your solution by drawing your number line and finding the "equals" part (in this example, x = 2). Then you marked this point with the appropriate notation; namely, an open dot or a parenthesis, indicating that the point x = 2 wasn't included in the solution. Then you'd shade everything to the right, because "greater than" meant "everything off to the right".
The steps for graphing two-variable linear inequalities are very much the same as for graphing the one-variable case.
Just as for number-line inequalities, my first step is to find the "equals" part. For two-variable linear inequalities, the "equals" part is the graph of the straight line; in this case, that means the "equals" part is the line y = 2x + 3:
Okay; the line has given me the boundary between the solution "region" and the not-a-solution region.
Now I'm at the point where the textbook tends to get complicated, with talk of "test points" and such. But back when I did one-variable inequalities (like x < 3), I didn't bother with "test points"; I just shaded one side or the other. I can do the same here. I'll ignore the "test point" stuff, and look at the original inequality, y ≤ 2x + 3.
I've already graphed the "or equal to" part — it's just the line. Now I'm ready to do the "y less than" part. In other words, this is where I need to shade one side of the line or the other. Now think about it: If I need y LESS THAN the line, do I want ABOVE the line, or BELOW?
Naturally, I want below the line. So I shade it in:
And that's all there is to it: the side I shaded is the "solution region" they're wanting.
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This technique (of looking at the original inequality to see which side to shade) worked because we had y alone on one side of the inequality. Just as was the case back when you were graphing plain old straight lines, you always want to solve for y to be alone on one side. Doing so makes everything else simpler and, for inequalities like this, saves the annoyance of doing test points.
First, I'll solve the inequality to get y alone on one side:
2x − 3y < 6
−3y < −2x + 6
y > (2/3)x − 2
[Notice how I the flipped inequality sign in the last line. I mustn't forget to flip the inequality if I multiply or divide through by a negative!]
Now I need to find the "equals" part, which is the line y = (2/3)x − 2. It looks like this:
Okay; the "equals" part has given me the line, which is the boundary between the solution region and the not-a-solution region.
But this exercise is what is called a "strict" inequality. That is, it isn't an "or equal to" inequality; it's only "y greater than". So this affects the "equals" line.
When I had strict inequalities on the number line (such as x < 3), I denoted this by using a parenthesis (instead of a square bracket) or an open [unfilled] dot (instead of a closed [filled] dot).
In the case of these linear inequalities, the notation for a strict inequality is a dashed line. So the border of my solution region actually looks like this:
By using a dashed line, I still know where the border is, but I also know that the border isn't included in the solution. And, since this is a "y greater than" inequality, I want to shade above the line, so my solution looks like this:
Related page: If you need to graph a set of two or more linear inequalities at once, view the lesson on systems of linear inequalities.
You can use the Mathway widget below to practice graphing linear inequalities. Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's.
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