to view answer.In previous sections, we have covered solving for x in inequalities and equations with absolute values. Solving inequalities that include absolute values can be slightly more complicated. We’ll first look at a simple example:
Solve for x. That is, find the values of x that make the inequality true.
$$\left| x \right| < 3$$
1. Without performing any algebraic manipulations on this yet, let’s first just think about what it means. We want all of the values of x that, when we take the absolute value, the result is less than 3. So, first, we know that all of the numbers from 0 up to 3 (but not including 3) are certainly less than 3 when we take their absolute value – after all, they are all positive to begin with, so their absolute values are also positive.
So, does this mean that the inequality is true for all x less than 3?
What about -10? If we substitute -10 into the inequality, we get:
$$\eqalign{
& \left| { - 10} \right| < 3 \cr
& 10 < 3 \cr} $$
This is not true! So, our solution is not all x less than 3.
What we really need is for x to be less than three and greater than -3. Negative numbers greater than -3 (like -2.5 and -1.2) have an absolute value less than 3.
2. Thus, the solution set is:
$$ - 3 < x < 3$$
In other words, all values of x between -3 and 3 make the inequality true.
Solve for x – that is, find the values of x that make the inequality true.
$$\left| {3x - 5} \right| \leq 2$$
1. By the same argument as in Example 1, the expression inside the absolute value must be both less than (or equal to) 2 and greater than (or equal to) -2 in order for its absolute value to be less than (or equal to) 2. Thus, we can rewrite the inequality without absolute values by writing:
$$ - 2 \leq 3x - 5 \leq 2$$
2. Now, we have a string of inequalities, so we can add and subtract anything we want to all three parts of the inequality. We can also multiply or divide positive numbers to all three parts of the inequality. However, if we divide by a negative number, remember that this reverses the inequality.
3. We want to find the values of x that make the inequality true, and we can do this by isolating x. To start, we add 5 to all three parts of the inequality:
$$ - 2 + 5 \leq 3x - 5 + 5 \leq 2 + 5$$
$$3 \leq 3x \leq 7$$
4. Finally, we divide each part by 3. Since we are dividing by a positive number, the direction of the inequality remains the same:
$${\Large\frac{3}{3}} \leq {\Large\frac{{3x}}{3}} \leq {\Large\frac{7}{3}}$$
Which gives the final solution of:
$$1 \leq x \leq {\Large\frac{7}{3}}$$
This inequality means that both properties of x are true: x is greater than 1 and x is also less than $$\frac{7}{3}$$. For the same meaning, we could write:
$$x \geq 1\;\;\;and\;\;x \leq {\Large\frac{7}{3}}$$
Both forms of this solution mean that all values of x between (and including) 1 and $${\Large\frac{7}{3}}$$ make the original inequality true.
5. If we want to check our answer, we should do 3 things:
First: Check the end-points of the inequality: substitute 1 and then $${\Large\frac{7}{3}}$$ in for x. Each of these should give you exactly 2:
$$\left| {3\left( 1 \right) - 5} \right| = \left| { - 2} \right| = 2$$
$$\left| {3\left( {{\Large\frac{7}{3}}} \right) - 5} \right| = \left| {7 - 5} \right| = \left| 2 \right| = 2$$
Second: Check a value that you believe is in the solution set: choose a number between 1 and $$\frac{7}{3}$$. Since any number between those numbers is a solution, substituting any number between them should give you a true statement:
Let’s choose $$x = 2$$, since $$1 \leq 2 \leq {\Large\frac{7}{3}}$$
Substituting:
$$\left| {3\left( 2 \right) - 5} \right| = \left| {6 - 5} \right| = \left| 1 \right| = 1$$ and $$1 \leq 2$$ is true, as our solution implies.
Third: Check a value that you believe lies outside the solution set: choose a number either less than 1 or greater than $${\Large\frac{7}{3}}$$. Whichever number you choose should make the original inequality false.
Let’s choose $$x = 0$$, since $$0 < 1$$.
Substituting:
$$\left| {3\left( 0 \right) - 5} \right| = \left| {0 - 5} \right| = \left| 5 \right| = 5$$ And $$5 \leq 2$$ is false, as our solution implies.
Thus, since all three of these checks worked, our solution is correct.
Solve for x – that is, find the values of x that make the inequality true.
$$\left| x \right| > 7$$
1. In this inequality, we have an absolute value of x that is greater than 7. So, let’s think about the values that would make this inequality true.
First, since the absolute value of a number is always positive, we can easily see that if $$x > 7$$, then inequality is true. For instance: $$
\left| {20} \right| = 20 > 7$$ is true.
2. What other numbers would make the inequality true? In particular, which negative numbers make the inequality true?
Any negative number with an absolute value greater than 7 will make the inequality true. So, any $$x < - 7$$ will make the inequality true. For instance: $$\left| { - 15}\right| = 15 > 7$$ is true.
3. Thus, our solution set is: All values of x such that: $$x > 7\;\;\;or\;\;\;x < - 7$$
Note that the or is important here. Either x is less than -7 ORit is greater than 7. It cannot be both. However, in either situation (x is less than -7 ORit is greater than 7), that value of x will make the original inequality true. That is why they must both be included in the solution set.
Solve for x – that is, find the values of x that make the inequality true.
$$\left| { - 8x + 1} \right| \geq 7$$
1. When we take the absolute value of the expression inside the absolute value bars, we want the result to be greater than 7. Therefore, just like in Example 3, we want either:
$$ - 8x + 1 \geq 7$$
or:
$$- 8x + 1 \leq - 7$$
So, we either want the expression to be greater than positive 7, or less than -7 so that its absolute value is greater than 7.
2. So, now, we just solve each inequality to find the values of x that will make these inequalities, and the original absolute value inequality true:
First, we have:
$$ - 8x + 1 \geq 7$$
Subtract 1 from each side:
$$ - 8x \geq 6$$
Divide each side by -8. REMEMBER that when we divide (or multiply) by a negative number on each side, we must switch the direction of the inequality:
$$x \leq - {\Large\frac{6}{8}}$$ and the fraction simplifies, so we get: $$x \leq -{\Large\frac{3}{4}}$$
Next, we have:
$$ - 8x + 1 \leq - 7$$
Subtract 1 from each side:
$$ - 8x \leq - 8$$
Divide each side by -8, and once again, remember to switch the inequality:
$$x \geq {\Large\frac{{ - 8}}{{ - 8}}}$$
So, we get: $$x \geq 1$$
3. Thus, our solution set, for the original inequality are all the values of x such that:
$$x \leq -{\Large \frac{3}{4}}\;\;\;or\;\;x \geq 1$$
So, substituting any number less than or equal to $$ - {\Large\frac{3}{4}}$$ or greater than or equal to 1 will make the original inequality true.
4. Once again, you can check by substituting $$ - {\Large\frac{3}{4}}$$ and 1 into the original inequality. If your answer is correct, those will each give a value of 7.
Then, substitute some number that is in the solution set, like 2. This will give an answer that makes the inequality true.
Then, substitute some number that is outside the solution set, like 0. This will make the inequality false.
Some practice problems to check your skills:
1. $$\left| x \right| \geq 14$$
Think about it for a moment and then access this link to view answer.
2. $$\left| {x + 9} \right| < 3$$
Think about it for a moment and then access this link to view answer.
3. $$\left| {9x - 7} \right| > - 3$$
Think about it for a moment and then access this link to view answer.
4. $$\left| {10x - 5} \right| > 25$$
Think about it for a moment and then access this link to view answer.
5. $$\left| { - 2x + \frac{1}{2}} \right| \geq 3$$
Think about it for a moment and then access this link to view answer.
6. $$\left| {5 - 3x} \right| < 10$$
Think about it for a moment and then access this link to view answer.