# Linear Inequalities in Two Variables

A. A linear in two variables will look like:

Ax + By ≤ C

Ax + By < C

Ax + By ≥ C

Ax + By > C

B. The boundary line for a linear inequality will be:

1. Dotted for < or > (i.e. – no equality)

2. Solid for ≤ or ≥

C. The solution for a linear inequality is the set of all points that will make the inequality a true statement.

II. Graphing Inequalities

A. Procedure

1. Determine the boundary line and graph it.

2. Pick a checkpoint – a point that is clearly, obviously – not on the boundary line.

3. Substitute the values for x and y into the original
inequality and simplify . If you get:

a. A true statement, shade the side of the line that the checkpoint is on.

b. A false statement, shade the other side of the line.

B. Examples – Graph each inequality .

1. y < −2

First, we see that there is no equality involved, so the boundary line is
dotted .

Next, the boundary line is found by replacing the “<” with “=”, which in this
case will give us a boundary line of:

**Boundary Line is y = −2.
**Now we graph this line. We know from our previous discussions that this is a
horizontal line.

We now pick a checkpoint. The point (0, 0) looks pretty good ! Substituting into the original inequality, we get:

0 < −2

This statement is false, so we shade the side of the line that is opposite of
the point (0, 0).

2. y ≥ 3x – 1

First, we notice that equality is involved here, so our boundary line is solid .

**Boundary Line: y = 3x – 1
**We next notice that this line is in slope-intercept form, so we can use our
knowledge of that to graph this boundary line.

**m = 3, b = −1**

So we will plot the point (0, 1), then rise 3, run 1 to get a second point.

Finally, we need to choose a checkpoint. Again, (0, 0) looks pretty good .

Substituting 0 in for x and 0 in for y in the original inequality, we get:

0 ≥ 3(0) – 1

OR

0 ≥ 0 – 1

OR

0 ≥ −1

This is true, so we shade the side of the line that the checkpoint is on.

3. Now you try one: 3x – 4y < 12

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