In this lesson, I’ll continue with inequations involving modulus. We’ll build upon the knowledge gained in the previous lesson.
Example 1 Solve for x: |x – 4| > 5
Solution We’re already familiar with inequations of the form |x| > 5. The solution to this inequation is
x < –5 or x > 5
We can easily convert the original inequation to this familiar form.
To do that, we’ll substitute x – 4 = y, and our equation will look like |y| > 5. And now we can solve it easily. We’ll get
y < –5 or y > 5
This gives us
x – 4 < –5 or x – 4 > 5
Finally, we have
x < –1 or x > 9
Check and substitute these values in the original inequation if you wish to.
After you get enough practice, there wont be a need to make that substitution. You can directly start with the second step (i.e., x – 4 > 5 etc.).
Let’s take another example.
Example 2 Find the value of x if |x + 3| ≤ 4.
Solution Once again, we can substitute x + 3 as y (or z, if you like) and get |y| ≤ 4.
And from the previous lesson, we know that this has the solution
–4 ≤ y ≤ 4
which gives
–4 ≤ x + 3 ≤ 4
Finally, we’ll get
–4 – 3 ≤ x ≤ 4 – 3
which simplifies to
–7 ≤ x ≤ 1
Once again, we could have directly started from the second step, without the substitution.
Example 3 Find the value of x if |x – 6| ≥ 0.
Solution Recall that |x| is always non-negative. This follows from the definition, and also from the fact that |x| represents distance, something which is always non-negative.
In fact, the modulus of any expression is always non-negative. That means, |x – 6| is greater than or equal to zero for any value of x.
Therefore, the solution to the given inequality is
–∞ < x < ∞
In other words, x can be any real number. Why don’t you substitute some values of x and verify the same?
In case you’re a bit confused, go back to the previous lesson and go through the solution of the inequation |x| ≥ 0. Then, you can substitute x – 6 = y in this example and solve |y| ≥ 0.
Example 4 Find the value of x if |x + 1| < –2.
Solution We’ll again use the fact that |x| is always non-negative. This implies that |x + 1| is also always non-negative.
But the given inequation asks us to find the values of x such that |x + 1| < –2.
This is not possible, given the fact we just considered. Hence, the given inequation has no solution.
Example 5 Find the value of x for which |x – 5| < 2.
Solution We’ve already covered a similar problem before. But here, we’ll use the visual approach to solve the problem.
Recall that |x – 5| denotes the distance of the number x from 5. (We’ve covered this already here.)
So, the inequation is asking us to find the number x, such that its distance from 5 is less than 2. Now, there are two numbers such that their distance from 5 equals 2.
These numbers area 5 + 2 and 5 – 2, or 7 and 3. But we want this distance to remain less than 2.
Where do you think can x lie? Explore the same in the applet below.
As you can see, the favourable region is between 3 and 7 (that is, between 5 – 2 and 5 + 2).
Therefore, the solution to this inequality is
3 < x < 7
Try using the previous method (i.e., substituting x – 5 = y) and see if you get the same solution. I’ll take one last example before moving forward.
Example 6 Find the value of x for which |x + 7| ≥ 3.
Solution Remember that |x – a| is the distance between x and a on the number line. Here, we have ‘+’ sign instead of a ‘–’. So, we’ll simply rewrite the expression on the left as |x –(–7)|.
Now this equation asks, “Find the number x, whose distance from –7 is greater than or equal to 3.”
Just like the previous example, let’s first look at the numbers, whose distance from –7 equals 3. There are two values again: –7 + 3 and –7 – 3, or –4 and –10.
But we want this distance to be greater than or equal to 3. Where can x lie in this case? Here’s another applet for you to explore.
As you can see, the favourable region is outside –10 and –4 (that is, outside –7 – 3 and –7 + 3).
Therefore, the solution to this inequality is
x ≤ –10 or x ≥ –4
Once again, this visual method wins. But as I mentioned earlier, this might not always work, or at least things might not be as easy to visualize.
See you in the next lesson then, where we’ll discuss a few slightly more complicated inequations involving modulus.