Here we will learn how to graph linear inequalities and how to graphically represent the solution of a system of linear inequalities.

Linear equations can be written in the form;  y = mx + b
However, liner inequalities come in several different forms. These forms include:
 y \leq mx + b
 y \geq mx + b
 y > mx + b
 y < mx + b
Graphing such inequalities differs a little from graphing regular equations. Graphing the line in an inequality doesn’t change from the original process, but how the line appears on the graph does change. Any line with a  > or < will appear as a dotted line, because  > or < do not include the values on the line, but rather all the values above, or below it. Depending on whether or not the equation is  > or < , you will shade the area above the line ( > greater than) or below the line ( < less than). This process doesn’t change much with  \leq and \geq inequalities. The only difference here is that there is a line below the symbols  > or < . This means that it reads either greater than or equal to, or less than or equal to, and that the equation includes the values of the line, so the line will appear solid instead of dotted.
After graphing and shading the equations, the area where the two equations overlap is the area where the values in it satisfy both equations.

You have 2 inequalities;  y \leq \frac{3}{2}x + 1 and  y > \frac{-2}{3}x + 3
Graphing the first line, you will see that the line is in the same place as if it were written in the form of  y = \frac{3}{2}x + 1 . However, the first line has a  \leq (less than but equal) symbol. So here, the line would be solid, and you would shade the area below the line. For the next equation,  y > \frac{-2}{3}x + 3 , there is a > (greater than) symbol. This means that the line would appear dotted and the shading would be done on the area above the line.

In the end, you have an area where the shaded areas of the two lines overlap. This area is your solution. The values in this area satisfy both  y \leq \frac{3}{2}x + 1 and  y > \frac{-2}{3}x + 3 .

Linear Inequalities

Video-Lesson Transcript

Let’s go over linear inequalities.

Linear equations are equations that have an equal sign and generally has two variables.

An example of a linear equation is y = mx + b which is the equation of a line.

It is a linear equation because it has and equal sign and the variables x and y are to the first degree.

So if we have linear inequality, it’s going be the same form but instead of an equal sign, we’ll have an inequality sign.

We might have greater than, less than, greater than or equal, and less than or equal.

It may also not be in the same format. We need to solve a regular equation.

For example, we have

3x + 2y \textless 7

Now let’s solve for y.

Let’s get y by itself.

3x - 3x + 2y \textless 7 - 3x
2y \textless -3x + 7
\dfrac{2y}{2} \textless \dfrac{-3x}{2} + \dfrac{7}{2}
y \textless -\dfrac{3x}{2} + \dfrac{7}{2}

What does inequality exactly mean?

This will affect the way we graph our solutions.

Let’s say we have

y \leq \dfrac{3}{2}x + 1

This looks like the equation of a line.

Since we know that m is slope in mx + b and b is the y-intercept, we can now graph this.

Using slope is \dfrac{3}{2} and y-intercept is 1, let’s graph as if it’s an equality.

But we don’t have an equals sign here.

It says y is less than or equal to. That means we also want the values of y below the line. But not the values above it.

So let’s shade the area below the line.

Our solution now is the values on the line and all the values below that line.

Let’s have a second one:

y \textgreater -\dfrac{2}{3}x + 3

Our y-intercept is 3 and slope is -\dfrac{2}{3}.

We’ll have a backslash line.

But y is greater than but not equal to.

To represent the “not equal” we draw a dotted line instead of a solid line.

Then shade the area above the dotted line because it is greater than.

Now, if you were given these two systems of functions, the point that satisfies both of these inequalities are represented by the area which is shaded by the two inequalities.

If you were asked to identify a point that satisfies both inequalities, you can pick (6, 2).

But any point in the shaded area is the solution.