There are many opportunities for mistakes with absolute-value inequalities, so let's cover this topic slowly and look at some helpful pictures along the way. When we're done, I hope you will have a good picture in your head of what is going on, so you won't make some of the more common errors. Once you catch on to how these inequalities work, this stuff really isn't so bad.

–6 < 2x + 3 < 6 [this is the pattern for "less than"]

–6 – 3 < 2x + 3 – 3 < 6 – 3

–9 < 2x < 3

–9/2 < x < 3/2 Then the solution to | 2x + 3 | < 6 is the interval –9/2 < x < 3/2.

2x – 3 < –5 or 2x – 3 > 5 [this is the pattern for "greater than"]

2x < –2 or 2x > 8

x < –1 or x > 4 This PAIR of inequalities is the solution;

the solution to | 2x – 3 | > 5 consists of the two intervals x < –1 and x > 4.

Let's first return to the original definition of absolute value: "| x | is the distance of x from zero." For instance, since both –2 and 2 are two units from zero, we have | –2 | = | 2 | = 2:

With this definition and picture in mind, let's look at some absolute value inequalities.

Suppose you're asked to graph the solution to | x | < 3. The solution is going to be all the points that are less than three units away from zero. Look at the number line:

The number 1 will work, as will –1; the number 2 will work, as will –2. But 4 will not work, and neither will –4, because they are too far away. Even 3 and –3 won't work (though they're right on the edge). But 2.99 will work, as will –2.99. In other words, all the points between –3 and 3, but not actually including –3 or 3, will work in this inequality. Then the solution looks like this:

The open circles at the ends of the blue line indicate "up to, but not including, these points." Your book might use parentheses instead of circles. Translating this picture into algebraic symbols, you find that the solution is –3 < x < 3.

This pattern for "less than" absolute-value inequalities always holds: Given the inequality | x | < a, the solution is always of the form –a < x < a. Even when the exercises get more complicated, the pattern still holds. Copyright © Elizabeth Stapel 2014 All Rights Reserved Solve | 2x + 3 | < 6.Since this is a "less than" absolute-value inequality, the first step is to clear the absolute value according to the pattern. Then I'll solve the linear inequality.| 2x + 3 | < 6–6 < 2x + 3 < 6 [this is the pattern for "less than"]

–6 – 3 < 2x + 3 – 3 < 6 – 3

–9 < 2x < 3

–9/2 < x < 3/2 Then the solution to | 2x + 3 | < 6 is the interval –9/2 < x < 3/2.

The other case for absolute value inequalities is the "greater than" case. Let's first return to the number line, and consider the inequality | x | > 2.

The solution will be all points that are more than two units away from zero. For instance, –3 will work, as will 3; –4 will work, as will 4. But –1 will not work, and neither will 1, because they're too close. Even –2 will not work, and neither will 2 (although they're right on the edge). But 2.01 will work, as will –2.01. In other words, the solution will be two separate sections: one being all the points more than two units from zero off to the left, and the other being all the points more than two units from zero off to the right. The solution looks like this:

Translating this solution into symbols, we get "x < –2 or x > 2". That is, the solution is TWO inequalities, not one. DO NOT try to write this as one inequality. If you try to write this solution as

"–2 > x > 2", you will probably be counted wrong: if you take out the x in the middle, you'll see that you would be saying "–2 > 2", which certainly isn't true. Take the extra half a second, and write the solution correctly.

2x – 3 < –5 or 2x – 3 > 5 [this is the pattern for "greater than"]

2x < –2 or 2x > 8

x < –1 or x > 4 This PAIR of inequalities is the solution;

the solution to | 2x – 3 | > 5 consists of the two intervals x < –1 and x > 4.

The number would be greater then 36.

Step-by-step explanation:

Let x represent the unknown number. The opposite of this number would be represented by -x.

1/3 of this is found by multiplying by 1/3; this gives us

1/3(-x)

Since it is less than 12, this gives us

1/3(-x) < 12

To cancel the 1/3, we can multiply both sides by 3:

1/3(-x)(3) < 12(3)

-x < 36

We can cancel the negative by dividing both sides by -1. When we multiply or divide an inequality by a negative number we flip the inequality:

-x/-1 < 36/-1

x > 36

Our number must be greater than 36.

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