Graphing Linear Inequalities

An inequality is a mathematical expression where the two sides are not equal.

There are 4 inequality symbols:

We read the inequality, \(y≤2x+6\), as “\(y\) is less than or equal to six more than two times \(x\)”. To graph a linear inequality with two variables, we use the same process as when graphing a linear equation plus a few extra steps. When graphing the line, we either make our line solid, which includes the values on the line as part of the solution set, or we use a dotted line, which does not include the values on the line as part of the solution set. Here is how we decide if the line should be solid or dotted:

The last step in the process is to decide which side of the line to shade. One method is to pick one point on each side of the line and substitute into the equation and simplify. Whichever results in a true statement will be the side that is shaded. Another method is to shade below the line if the inequality symbol is < (less than) or ≤ (less than or equal to), and shade above the line if the inequality symbol is > (greater than) or ≥ (greater than or equal to).

Example: Graph the inequality \(y≤2x+6\).

Start by graphing the line as if it is an equation, \(y=2x+6\). Since it is already in slope-intercept form, we know that the \(y\)-intercept is \(6\) and the slope is \(2\). Since the inequality is \(≤\), we will make the line solid.

As a last step, we pick one point on each side of the line and plug it into the inequality. For ease, I am going to pick \((0,0)\), on the right side and \((-4,4)\) on the left side.

We will shade the side that gave us the true statement because this means that the ordered pair is part of the solution set of the inequality.

Hi, and welcome to this video about linear inequalities! Today, we’ll explore what linear inequalities are, and see how to express their solutions.

You probably have some experience with linear equations. They can be written in point-slope form \(y-y_{1}=m(x-x_{1})\), slope-intercept form \((y=mx + b)\), or standard form \((Ax + By = C)\) and their graphs are straight lines. For an example, let’s use the equation y equals 2x plus 1 \(y=2x + 1\).

There are an infinite number of solutions to the equation. Solutions, in this case, are coordinate pairs that lie on the line—algebraically, they make the equation true. So (0,1) is a solution because 1 equals 2 times 0, plus 1. \(1 = 2(0) + 1\). (negative 1, negative 1) is also a solution since negative 1 equals 2 times negative 1, plus 1. (-1, -1) is also a solution since -1 = 2(-1) + 1. On the other hand, (3,2) is not a solution because does not equal 2 times 3, plus 1. \(2 ≠ 2(3) + 1\).

What if, instead of the equation y equals 2x plus 1 \(y = 2x + 1\), we replaced the equals sign with, say, this sign: > to get y is greater than 2x plus 1 \(y > 2x + 1\)? Well, for starters, that would give us a linear inequality! Linear inequalities are simply inequalities that involve linear equations.

The solution set is still an infinite set of points, but let’s see where they lie. We’ll start by graphing the line y equals 2x plus 1 \(y = 2x + 1\) as before

Okay, now let’s make a quick table so we can test some points on this graph. We’re interested in points on each side of the line, as well as on the line.

Let’s test (3, negative 1) first. We can see this is below the line, and if we plug these numbers into our inequality, we see that it does not work. Or, we would say it isn’t “true”. Let’s test another point, (2, 1). This is also below the line, and plugging the values into our inequality tells us it is not true. Let’s try some that are on the line. (1, 3) and (0, 1) are both points on the line, but plugging in those values still doesn’t work. Now let’s try above the line. (Negative 3,3) is above the line, and if we plug those values into our inequality, we find that it is true! 3 is greater than 2 times negative 3, plus 1. \(2(-3) + 1\). (Negative 1, 0) is also above the line, and if we plug those values into our inequality, we find that it is true as well! 0 is greater than 2 times negative 1, plus 1 \(2(-1) + 1\).

We can test as many points as we’d like, but this tells us more than enough information. All the points in the solution set are above the line, not on it. To show that, we shade that entire half of the graph.

The reason all of the points in the solution set are above the line is because of our inequality symbol. Our solution is “greater than”, so it will be above the line. If we used <, our solution would be below the line. So, that’s really the whole concept of linear inequalities. In practice, we’re talking about 3 key steps: Graph the line as normal
Test 1 or 2 points; (0,0) is always the easiest. If it lies on the line, try something like (1,1) or (negative 1,1).
Dash the line if needed. If the inequality contains a less than or greater than, use a dashed line. If it contains less than or equal to or greater than or equal to, keep it solid.

Let’s use these steps for the inequality 3x plus 6y is less than or equal to negative 12! \(3x + 6y ≤ -12\)!

Here’s the graph.

Let’s test (0,0): 3 times 0 plus, 6 times 0 is less than or equal to negative \(3(0) + 6(0) ≤ -12\) is not true. Therefore, we shade in the direction that does not include (0,0) and keep the line solid.

Before we go, let’s have a little practice!

Write the inequality represented by this graph. Pause the video if you need more time:

The correct answer is y is greater than or equal to 4x minus 8. \(y ≥ 4x – 8\).

I hope this review was helpful! Thanks for watching, and happy studying!

Graphing Linear Inequalities Practice Questions

Question #1:

 
Which graph best represents the inequality \(y>-2x+4\)?

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Answer:

First, graph the line for \(y>-2x+4\) on the coordinate plane. Since the y-intercept (b) is 4, the line will intersect the y-axis at 4, or \((0,4)\). The slope (m) is -2 or \(-\frac{2}{1}\). Starting at \((0,4)\), use the slope to find another point on the line. Since slope is rise over run, move down two units and right one unit to the point \((1,2)\).

Next, draw a line to connect the points. Since the inequality sign is “greater than”, connect the ordered pairs with a dashed line. This indicates that points on this line are not possible solutions.

Then, substitute origin point \((0,0)\) into the inequality as a test point to determine which portion of the coordinate plane to shade:
\(y>-2x+4\)
\((0)>-2(0)+4\)
\(0>0+4\)
\(0>4\) ✘

Since 0 is not greater than 4, the point \((0,0)\) cannot be part of the solution set. That means the values below the graphed line are not solutions to the inequality. Instead, shade the area above the line. The finished graph has a dashed line with a shaded region above it. All possible solutions are points above the line graphed.

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Question #6:

 
Which shows the graph of the inequality \(y≤-\frac{2}{3}x+4\)?

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Answer:

To graph an inequality, we start by graphing it as a linear equation. Since \(y≤-\frac{2}{3}x+4\) is in slope-intercept form, we know the y-intercept is \(4\) and the slope is \(-\frac{2}{3}\). Once the line is graphed, we will determine if the line should be solid or dotted. In this case since it is \(≤\) (less than or equal to), the line will be solid, which means the values on the line are part of the solution set for the inequality. The last step is to determine which side of the line should be shaded. One method is to pick a point one and use it to evaluate the inequality. If the point that results in a true statement, it will be on the side that is shaded. Start with \((0,0)\), \(0≤-\frac{2}{3}(0)+4\), after simplifying we are left with \(0≤4\), which is a true statement, so we will shade below the line.

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