# Graphing Linear Inequalities

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 \(2(-1) + 1\).

Location | Point | True? |

below line | (3,-1) | -1 > 2(3) + 1? no |

below line | (2,1) | 1 > 2(2) + 1? no |

on line | (1,3) | 3 > 2(1) + 1? no |

on line | (0,1) | 1 > 2(0) + 1? no |

above line | (-3,3) | 3 > 2(-3) + 1? yes |

above line | (-1,0) | 0 > 2(-1) + 1? yes |

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!

## Practice Questions

**Question #1:**

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

**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.

**Question #2:**

Which ordered pair shows a possible solution for the inequality \(y≤\frac{1}{2}x-5\)?

**Answer:**

Substitute each point into the inequality as a test point to determine if it is part of the solution set.

\(y≤\frac{1}{2}x-5\)

\((-5)≤\frac{1}{2}(5)-5\)

\(-5≤2\frac{1}{2}-5\)

\(-5≤-2\frac{1}{2}\)✔

Since -5 is less than or equal to \(-2\frac{1}{2}\), the point \((5,-5)\) is part of the solution set. The ordered pairs for Choices B, C, and D are not correct because they do not yield true statements when substituted into the inequality. Therefore, A is the correct answer.

**Question #3:**

Which inequality statement matches the graph shown?

\(y\) < \(3x+2\)

\(y\) < \(3x-2\)

**Answer:**

First, identify the *y*-intercept by looking at where the graph of the line crosses the *y*-axis. The *y*-intercept (*b*) is 2, or \((0,2)\). Since C has -2 as the *y*-intercept (*b*), we can eliminate this answer choice. Because the graph moves in a positive direction, the slope (*m*) is a positive number. Therefore, we can eliminate D as an answer choice, which has -3 for the slope (*m*). The inequality is graphed with a dashed line, which indicates that points on the line are not possible solutions. In other words, the inequality symbol must be either < or >. We can eliminate A because the inequality contains the symbol ≤, which is graphed as a solid line. Therefore, the correct answer is B.

**Question #4:**

Vanessa is at a used bookstore that sells pre-owned books and DVDs. DVDs cost $8 each, and books cost $5 each. Vanessa wants to buy as many books and DVDs as she can afford, but she can’t spend more than $35. Which inequality statement best represents this situation?

\(8x+5y\) < \(35\)

**Answer:**

The inequality statement needs to show that the total amount of money Vanessa spends cannot exceed $35. In other words, Vanessa can spend less than $35 or exactly $35. The inequality sign that matches this scenario is “less than or equal to” (≤).

**Question #5:**

Matthew is at the movie theater with a group of friends. He plans to buy some bags of popcorn and sodas for the group. Each bag of popcorn costs $6, and each soda costs $4. Matthew has $40 to spend and writes the inequality \(6x+4y≤40\) to represent the situation. Which statement below is true?

Matthew can buy 6 bags of popcorn and 2 sodas for his friends.

Matthew can buy 3 bags of popcorn and 6 sodas for his friends.

Matthew can buy 5 bags of popcorn and 3 sodas for his friends.

Matthew can buy 4 bags of popcorn and 4 sodas for his friends.

**Answer:**

Using *x* to represent bags of popcorn and *y* to represent sodas, substitute each set of values into the inequality as a test point. Doing so will determine if it is part of the solution set.

\(6x+4y≤40\)

\(6(4)+4(4)≤40\)

\(24+16≤40\)

\(40≤40\) ✔

Since 40 is less than *or equal to* 40, the set of values in Choice D are part of the solution set. The values given for Choices A, B, and C are not correct because they do not yield true statements when substituted into the inequality. Therefore, D is the correct answer.