GED Mathematical Reasoning: Inequalities II

Example 1

Solve: 3a - 15 < 8a < a + 7

 

The statement showing in example one is a “compound” inequality. A compound inequality combines two inequality statements into one. In other words, the statement can be divided into these two inequalities:

3a - 15 < 8a

8a < a + 7

To solve the compound inequality, we will solve the two inequalities separately and then combine the solution sets. Let’s get started by solving the inequality: 3a - 15 < 8a

No simplifying can be done on either side of the inequality sign, so we’ll move along to step 2 and add and subtract to get the variable terms on one side of the inequality and the constant terms on the other side.

Adding 15 to both sides leaves just “3a” on the left side.

3a - 15 +15 < 8a +15

3a< 8a +15

On the right side, we cannot add “8a” and 15 since they are not like terms, so we will leave it as: 8 “a” plus 15.

Next, we will subtract 8a from both sides. That will allow us to combine the variable terms, which are “like terms,” on the left side.

On the left side, 3a minus 8a is equal to -5a. On the right side, the 8a and -8a undo each other, leaving just 15.

3a - 8a < 8a - 8a +15

- 5a < 15

We now have a one-step inequality: -5a less than 15

Following step 3 of the process, we will now divide both sides by -5 to solve for “a.”

On the left side, -5 divided by -5 equals one, leaving just “a” on the left side.

On the right side, we need to divide 15 by -5, which is equal to -3.

- 5a \div -5 < 15 \div -5

- 5a \div -5 < 15 \div -5

And since we divided both sides by a negative number, we must flip the inequality sign.

 a > -3

So the solution to the inequality is: “a” greater than -3. Now let’s solve the other inequality:  8a < a + 7

8a < a + 7

No simplifying can be done on either side of the inequality sign, so we’ll move along to step 2 and add and subtract to get the variable terms on the left side of the inequality and the constant term on the right side. Let’s begin by subtracting “a” from both sides.

8a - a < a - a + 7

7a < 7

We now have a one-step inequality: 7a less than 7.

Following step 3 of the process, we will now divide both sides by 7 to solve for “a.”

 7a \div 7 < 7 \div 7

 a < 1

The solution to this inequality is: “a” less than 1.

Now, let’s put all this together.

The solution to the first inequality is:   a > -3

The solution to the second inequality is:   a < 1

Therefore the solution set to the compound inequality is:  -3 < a < 1

Visually, the solution set looks like this:

algeb71

The area between -3 and one is shaded.  And open circles are placed on the numbers -3 and 1 since they are NOT included in the solution set. You might think of the notation this way: Since “a” needs to be a number between -3 and one, we sandwich it between those numbers using the appropriate inequality signs.

Now let’s do a quick check. We’ll choose “a” equals 0, which is between -3 and one, to check our answer.

Check: n = 0
3a - 15 < 8a < a + 7
3(0) - 15 < 8(0) < (0) + 7
0 - 15 < 0 < 0 + 7
- 15 < 0 < 7 is TRUE

Is the statement true? Yes! The resulting statement is:  -15 less than zero less than 7. Make sense of it this way: Is zero greater than -15 and less than 7? In other words, is zero between the numbers -15 and 7? Yes! Therefore, we can be confident that our solution set is correct.

As a final note about inequalities, it is usually easier to make sense of the solution to a single inequality when the variable appears on the left side.

If ever you solve a single inequality and end up with a statement like:  4 < x , you may flip the statement around so that the variable is on the left side by flipping the order of the 4 and “x” but remember to flip the inequality sign, too, to preserve the truth of the statement.

 4 < x \rightarrow x > 4

So: 4 less than ‘x’ can be rewritten as:  “x” greater than 4.

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