Hello. Now that we’ve discussed equations involving modulus, we’ll extend this knowledge to solving inequations of a similar form (i.e., linear inequations).
Example 1 Solve the equation |x| > 3.
Solution I’ll illustrate two different methods, both following from the definitions that we discussed previously.
Method 1
The definition of modulus tells me that
\( |x| = \begin{cases}
x& \text{ if } x\ge 0 \\
-x& \text{ if } x < 0
\end{cases} \)
So, when x ≥ 0, |x| can be replaced by x, and therefore we’ve to solve the inequality x > 3.
And, when x < 0, |x| equals –x, and this time we’ve to solve –x > 3.
Well, there’s nothing much left to solve: The first case gives x > 3, while the second one gives x < –3.
In other words, the inequality |x| > 3 is true for all values of x > 3 and x < –3.
In interval notation, the solution can be represented as
x ϵ (–∞, –3) U (3, ∞)
Let’s try substituting some of these values in the original inequation, just to be sure.
We’ll take x = 5. Then, |x| = |5| = 5 > 3, which means that x = 5 satisfies the inequality. If we take x = –6, then |x| = |–6| = 6 > 3, again implying that –6 is a valid solution.
Try some more values on your own, both inside and outside the solution set, to verify.
Method 2
We’ll now use the second definition. Recall that |x| represents the distance of x from 0 on the number line.
We can therefore rephrase this as “Find a number whose distance from the origin is greater than 3”.
First, let’s find the numbers whose distance from the origin equals 3. There are two such numbers, as we’ve seen earlier, 3 and –3.
Now, we want those numbers, whose distance from 0 is greater than 3. Can you figure it out?
If x = 3, then the distance of x from 0 is exactly 3. If we want this distance to increase (i.e., become greater than 3), then we’ve to move towards the right of 3. That is, in the region where x > 3.
Similarly, when x = –3, then it is at a distance of 3 from 0. And, if we want this distance to become greater than 3, then we’ve to move to the left of –3. That is, we’ve to stay in the region where x < –3.
Drag the point represented by x in the applet below to understand what I mean.
Combining these two cases, we’ll get the solution as
x < –3 or x > 3
Just like equations, this visual method is much easier to grasp. But visualizing like this becomes complicated when more terms are involved, which is when we’ll have to switch to the first method.
Let’s try another one.
Example 2 Find the values of y for which |y| < 5.
Solution Geometrically, this question asks, “Find all numbers on the number line such that their distance from the origin is less than 5.”
Let’s first find the numbers whose distance from the origin is exactly 5. These numbers, as we already know, are 5 and –5.
Now, we wish to find those numbers whose distance from the origin is less than 5. Where do you think all those numbers are?
Play around with the point representing y to explore.
As you would’ve seen, the required numbers are those that lie within the region bounded by –5 and 5. That is, all values of y such that –5 < y < 5.
Well, that’s it. The solution to the inequality |y| < 5 is
–5 < y < 5
In interval notation, the solution can be represented as
y ϵ (–5, 5)
Let’s also take a look at the analytical method.
To solve |y| < 5, we’ll make two cases, based on the definition of |y|.
Case 1: y ≥ 0
Here, |y| = y and therefore, the original inequation becomes y < 5, which represents the solution itself.
Just like in the case of equations, the solution y < 5 is valid only in the case when y ≥ 0.
To verify this, let’s substitute y = – 7 in the original inequation, where y < 5, but y ≱ 0. We’ll get the LHS as |–7|, which equals 7 and 7 ≮ 5 (i.e. doesn’t satisfy the inequality).
Therefore, for this case, we need to combine the solution and the case condition. We’ll get the final solution as
0 ≤ y < 5
Let’s look at the other case now.
Case 2: y < 0
Here, |y| = – y. The original inequation becomes – y < 5, which gives y > – 5.
Combining the solution with the case condition, we’ll get
–5 < y < 0
We’re done with all the cases. So, the final solution is
–5 < y < 0 OR 0 ≤ y < 5
which can be combined and expressed as
–5 < y < 5
We get the same solution as we got in the visual method. Once again, try to get a grip on making these ‘cases’ and combining them with the solutions you get.
We’ll prefer to work with the number line method, whenever suitable. But in many problems, you might need the analytical method.
Let’s summarize what we’ve done so far
|x| > 3 => x < –3 or x > 3
|y| < 5 => –5 < y < 5
We can extend this knowledge to solve equations like |w| ≤ 6 and |z| ≥ 4 too.
|w| ≤ 6 => –6 ≤ w ≤ 6
|z| ≥ 4 => z ≤ –4 or z ≥ 4
Observe the pattern carefully. As you get more practice, you can begin to write solutions for such simple equations directly, without using the number line or making cases.
Let’s look at a few edge cases of these inequalities.
Example 3 Solve for x: |x| ≥ 0
Solution Looking at the previous patterns, we could write something like
x ≤ 0 or x ≥ 0
This means that x can pretty much be any real number. In other words, x ϵ R.
But let’s take another look at this inequation, geometrically.
The inequation |x| ≥ 0 is asking us to find all numbers whose distance from the origin is greater than or equal to zero (or, non-negative).
But that’s true for any number, isn’t it? Pick any number on the number line – its distance from the origin will always greater than or equal to zero.
See for yourself.
Therefore, the solution to the inequation is ‘x can be any real number’ or x ϵ R.
Can you solve the inequation |x| > 0 on your own? (Can x be any real number here as well?)
Example 4 Solve the inequation |x| < 0.
Solution This inequation is asking us to find a number x such that its distance from the origin is less than 0 (or negative).
Is that possible? Can distances be negative? Nope.
Again, see for yourself.
Therefore, the inequation has no solution. Or, |x| < 0 => x ϵ ø.
Can you solve the inequation |x| ≤ 0 on your own? (Does this have no solution as well?)
A couple more examples and we’ll be done.
Example 5 Solve the equation |y| ≥ –5.
Solution Before you begin to write something weird like
|y| ≥ –5 => y ≥ –5 or y ≤ –(–5),
let’s look at this inequation geometrically.
The inequation is asking us to find all numbers such that their distance from the origin is greater than or equal to –5.
Now, the distance can never be equal to –5. But can it be greater than –5? Yes!
The distance |y|, by definition, is always non-negative (i.e. |y| ≥ 0). Hence, it is always greater than –5 (or –23, –140, or any negative number).
Therefore, the solution to this inequality is
y ϵ R
That, is y can be any real number. Try substituting some values of y to verify.
Example 6 Solve the equation |z| < –4.
Solution Once again, we’ll not jump to write something like
|z| < –4 => –(–4) < z < –4 😖
as we learnt in the first couple of examples.
By definition, |z| ≥ 0. This means that |z| can never be less than –4. Sadly, this inequation has no solution.
Let’s take look at the analytical method too. We’ll make two cases, z ≥ 0 and z < 0.
Case 1: z ≥ 0
Here, the inequation becomes z < –4, which is the solution itself. But we’ll reject this solution as it violates the condition of Case 1. (Try substituting any value of z less than –4 to verify anyway).
Case 2: z < 0
Here, the inequation becomes –z < –4, which gives z > 4. Once again, we’ll reject this solution as it violates the condition of Case 2. (You can again substitute any value of z greater than 4 to verify.)
In both cases, we rejected the solutions. Therefore, the inequality |z| < –4 has no solution.
That brings us to the end of this lesson. Please go through all the examples once again and make sure you understand both the methods, and especially the edge cases.
Lesson Summary
The following tables summarize what we’ve learnt in this lesson.
1. a > 0
| Inequation | Solution |
| |x| > a | x < –a or x > a |
| |x| < a | –a < x < a |
| |x| ≥ a | x ≤ –a or x ≥ a |
| |x| < a | –a ≤ x ≤ a |
2. a = 0
| Inequation | Solution |
| |x| > a | x ≠ 0 |
| |x| < a | No solution |
| |x| ≥ a | –∞ < x < ∞ |
| |x| ≤ a | x = 0 |
3. a < 0
| Inequation | Solution |
| |x| > a | –∞ < x < ∞ |
| |x| < a | No solution |
| |x| ≥ a | –∞ < x < ∞ |
| |x| ≤ a | No solution |
In the next lesson, we’ll continue with slightly modified versions of these equations.