# More with Compound Inequalities

In this Lesson we will be doing more work with compound inequalities.

The solutions in the above animation are repeated with a more detailed explanation below.

## Question 1.1

Graph these “or” statements on a number line.

x ≤ 2 or x > 5

Just to remind ourselves what we are doing, we are plotting the solution set of values to the statement x ≤ 2 or x > 5 on a number line. The lines represent all the values of x that satisfy the statement.

We will have two lines, one starting at 2 and one starting at 5.

Write a solid dot over 2 and a hollow dot over 5 with arrowed lines extending over the values that are less than or equal to 2 or greater than 5. x > 8 or x < -1

Write a hollow dot over -1 and 8 and then an arrowed line extending over all the values that are either less than -1 or greater than 8. ## Question 1.2

Graph these “and” statements on number line, then look at the graphed number lines to help you write one single inequality.

x > -3 and x < 3,

Before we write our number line, remember what the statement means, x > -3 and x < 3.

The solution set that we graph will be the values that satisfy both parts in this statement, as it is an “and” statement.

So, the solution set is all the values that are greater than -3 and less than 3. First lets try graphing x > -3 and x < 3 The solution set to the statement x > -3 and x < 3 are the values that satisfy both parts. Visually this is the part of the graph that is covered by both lines that we plotted in the line graph above.

Now, to properly graph the solution set on a line graph for x > -3 and x < 3, we write the line that is overlapped by the line we drew for x > - 3 and the line we drew for x < 3. Now we can use this visual to help us write a singular inequality.

We want to write a singular statement that says x is greater than - 3 and less than 3.

This will look like -3 < x < 3,

Compare this to the line we have written and you may be able to see the similarities.

x ≥ 1 and x ≤ 6

Again, to help us write the solution set of the statement, lets first write out an individual line for x ≥ 1 and then one for x ≤ 6.

Then, when we have graphed those two lines, graph the line that will be the solution set to the statement above that (it will be the line where the two other lines have overlapped). Now, we want to write a singular inequality that represents the values greater than or equal to 1 and less than or equal to 6.

Similar to the previous question, except this time we have ≤ / ≥ rather than < / >,

1 ≤ x ≤ 6

Now, we are going to practice doing the opposite of what we have just done.

When we talk about Singular Inequalities, note that Inequalities that involve “and” are almost always written as Singular Inequalities.

The solutions in the above animation are repeated with a more detailed explanation below.

## Question 2

Combining all the skills we have been developing, solve the given inequality and graph its solution set on a number line.

2x + 3 > 7 and -2x + 6 > -10

Here we are combining all the skills we have developed so far. We are asked to graph this inequality, but in order to do that, we must know the solution set, which means we have to solve it!

Let’s first solve each part of the inequality individually.

Let’s solve 2x + 3 > 7

Remember we can solve inequalities almost the exact same way we solve equations, except when we multiply or divide both sides by a negative. When we do that, we have to switch the inequality (i.e. switch < to > or ≤ to ≥)

So, we start by subtracting both sides by 3.

2x + 3 - 3 > 7 - 3

2x > 4,

Now, divide both sides by 2, (not dividing by negative, so don't have to switch the inequality)

2x ÷ 2 > 4 ÷ 2

x > 2

The we have our first part of the inequality. We now know part of our solution set will be the values that are greater than or equal to 2.

Now solve, -2x + 6 > -10,

Any time we are solving an inequality and see a negative multiplying our x, take a note that we will most likely have to divide by a negative at some point when solving, and have to switch the inequality.

First, subtract both sides by 6

-2x + 6 - 6 > -10 - 6

-2x > -16

Now we divide both sides by -2, and switch the inequality

-2x ÷ -2 < -16 ÷ -2

x < 8

There we have our two parts of the inequality,

x > 2 and x < 8

Or, as this is an “and” inequality, we can more commonly write it as:

2 < x < 8

Now we can graph this.

First, graph our line for,

x > 2

A hollow dot over 2,

Our x values are greater than 2, so a line extending to the right of 2 Now on that same line graph,

Graph the line for,

x < 8

A hollow dot over 8,

Our x values are less than 8, so a line extending to the left of 8 Now we see our two lines. This is an “and” inequality, so we want a line that is true for both parts of the inequality. So, we are looking for the line where these two parts overlap.

This will look like: ½ (x + 2) < 4 or -3(x - 3) ≤ 12

First, note we have an “or” inequality which means a value is a solution to the inequality if it makes at least one part of the inequality true.

Let’s solve each part individually.

½ (x + 2) < 4

Multiply both sides 2 (the same as dividing both sides by a half)

(½ (x + 2))2 < (4)2,

(x + 2) < 8,

Subtract both sides by 2,

x + 2 - 2 < 8 - 2

x < 6

Now for the other part,

-3(x - 3) ≤ 12,

Divide both sides by -3, (remember, switch the inequality, as we are dividing by a negative)

-3(x - 3)
-3

12
-3

x - 3 ≥ -4,

x - 3 + 3 ≥ -4 + 3

x ≥ -1

So, we have our two parts to the “or” inequality.

x < 6 or x ≥ -1,

Now we graph this. Let’s graph the two parts on one line graph, Note something interesting; as this is an “or” inequality, we aren't looking for the values that satisfy both parts of the inequality. We are looking to graph the values that can satisfy one part or the other part or both.

Try and think of a number that is not either less than 6 or greater than or equal to -1. There is not one! This means that every number will satisfy this inequality!