# Solving and Graphing Two-Variable Inequalities

Instructor: Glenda Boozer
With an inequality like y < 2x - 3, where do we start? We can look at y = 2x - 3 and graph it. We can use the graph to define all of the possible solutions.

## Two-Variable Inequalities Defined

When we have an inequality with two variables, we can't just write the solution in a short mathematical statement like x < 1, and we can't just show it on a number line. Two variables means two dimensions, so we need to graph it on a grid, graph paper, or something similar.

## Solving Inequalities versus Solving Equations

Just as we solve an inequality like 3x + 2 > 4 by using almost all of the same techniques as when we solve the equation 3x + 2 = 4, we start the process of solving y < 2x - 3 by working with y = 2x - 3.

## Graphing the Equation

If we make a graph of y = 2x - 3, we can take a few values of x, like 1, 2, 3 and 4, and for each one, we can find the value of y that goes with it:

If x = 1, then y = 2(1) - 3 = 2 - 3 = -1, so the ordered pair (x,y) = (1,-1) goes on the graph.

If x = 2, then y = 2(2) - 3 = 4 - 3 = 1, so the ordered pair (2,1) goes on the graph.

If x = 3, then y = 2(3) - 3 = 6 - 3 = 3, so (3,3) goes on the graph.

If x = 4, then y = 2(4) - 3 = 8 - 3 = 5, so (4,5) goes on the graph.

Now that we have some points, we can draw a line, like this:

## The Inequality

Now, this is nice, but what about y < 2x - 3? Take a look at this: if we take the (1,-1) pair for (x,y) and put them into the inequality, we get

-1 < 2(1) - 3

-1 < 2 - 3

-1 < -1.

That's not true, since -1 can't be less than itself. This means that all of our points are ruled out (try them!) and that every point on the line is ruled out (you might choose a point on the line and try it, too). In that case, are we wasting our time? No - you'll soon see why not. Right now, though, let's change the solid line to a dotted line in order to show that the exact points are not part of the solution set; that is, the set of all possible solutions.

If it were -1 ? -1, that would be fine, because -1 is equal to itself, so if our inequality uses ? or ?, then we don't have to change the line into a dotted line.

We still need some points that do make the inequality true, so now what? Let's try the ordered pair (0,0) and see if it makes the inequality true:

0 < 2(0) - 3

0 < -3: but that isn't true!

The point (0,0) is not part of the solution set. In fact, every point on the left/upper side of the dotted line makes the inequality false. How about a point on the other side? Let's try (2,-1), since we can see that point on our graph:

-1 < 2(2) - 3

-1 < 4 - 3

-1 < 1.

That's true, so (2,-1) is part of the solution set. In fact, it turns out that every point on the right/lower side of the line is part of the solution set. We can show this by shading the entire side of the graph, like this:

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