Solve Practice Questions of Inequality
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.
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). But2.99 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 2000-2011 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.
The pattern for "greater than" absolute-value inequalities always holds: the solution is always in two parts. Given the inequality | x | > a, the solution always starts by splitting the inequality into two pieces:x < –a or x > a. For instance:
- Solve | 2x – 3 | > 5.
The first thing to do is clear the absolute value bars by splitting the inequality into two. Then I'll solve the two linear inequalities.
| 2x – 3 | > 5
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.
There is another situation you might encounter: You'll be given a pair of inequalities, and you'll be asked to find the corresponding absolute-value inequality. This process can feel a bit weird, so I'll give a couple examples of how it works.
- Find the absolute-value inequality statement that corresponds to the inequality
–2 < x < 4.
I first look at the endpoints. Negative two and four are six units apart. Half of six is three. So I want to adjust this inequality so that it relates to –3 and 3, instead of to –2 and 4. To accomplish this, I will adjust the ends by subtracting 1 from all three "sides":
–2 < x < 4
–2 – 1 < x – 1 < 4 – 1
–3 < x – 1 < 3
Since the last line above is in the "less than" format, the absolute-value inequality will be of the form "absolute value of something is less than 3". I can convert this nicely to
| x – 1 | < 3
- Find the absolute-value inequality statement that corresponds to the inequalities
x < 19 or x > 24.
I first look at the endpoints. Nineteen and 24 are five units apart. Half of five is 2.5. So I want to adjust the inequality so it relates to –2.5 and 2.5, instead of relating to19 and 24. Since 19 –
(–2.5) =21.5 and 24 – 2.5 = 21.5, I need to subtract 21.5 all around:
x < 19 or x > 24
x – 21.5 < 19 – 21.5 or x – 21.5 > 24 – 21.5
x – 21.5 < –2.5 or x – 21.5 > 2.5
Since the last line above is the "greater than" format, the absolute-value inequality will be of the form "absolute value of something is greater than or equal to 2.5". I can convert this nicely to:
| x – 21.5 | > 2.5
Warning: There is one "trick" type of question for this kind of problem, where they'll try to trip you up on homework or tests. They'll ask you to solve something like "| x + 2 | < –1". But can an absolute value ever be negative, let alone be less than a negative? No! So there is no solution to this inequality; it doesn't even make sense. Don't waste a lot of time trying to "solve" this; just write down "no solution".
Similarly, if you're given something like "| x – 2 | > –3", the solution would, naturally, be "all x" or "all real numbers", since the absolute value, being itself positive or zero, must then always be greater than any negative number.
hope it helps.