Hello again. This lesson will discuss some slightly complicated inequations involving modulus.
The methods used here will be quite similar to those used in this lesson. So, make sure you go through that one first.
Let’s begin.
Example 1 Solve for x: |x – 2| + |x + 4| ≥ 8
Solution As already discussed, we’ll divide the number line into separate regions, in which the inequation ‘retains’ its form, and try to solve the inequation in each region.
Case 1: x ≥ 2
In this case, |x – 2| = x – 2 and |x + 4| = x + 4. The inequation becomes
x – 2 + x + 4 ≥ 8
which gives
x ≥ 3
You can substitute any value of x greater than or equal to 3 in the original inequation to verify. That’s all for this case. Let’s move to the next one.
Case 2: –4 ≤ x < 2
In this case, |x – 2| = 2 – x and |x + 4| = x + 4. The original inequation becomes
2 – x + x + 4 ≥ 8
which simplifies to
6 ≥ 8
This means that when x lies between –4 and 2, the LHS will always be 6, which can never be greater than or equal the RHS, which will always equal 8. Try substituting some values of x to verify this.
This means that there’s no solution of this inequation when –4 ≤ x < 2. We’re left with one last case now.
Case 3: x < –4
Here, |x – 2| = 2 – x and |x + 4| = –x – 4. Our inequation now becomes
2 – x – 4 – x ≥ 8
This gives x ≤ –5. Once again, substitute this in the original inequation to verify. And, since we’ve covered the whole number line, there are no more solutions to find.
Let’s summarize solutions now.
When we considered the region x ≥ 2, we got the solution as x ≥ 3.
When we considered the region –4 ≤ x < 2, we got no solution.
Finally, when we considered region x < –4, we got the solution as x ≤ –5.
Combining all these, the final solution can be written as
x ≤ –5 or x ≥ 3
In set notation, this can be written as
x ∈ (-∞, -5] U [3, ∞)
Let’s move on to the next example.
Example 2 Solve for u: |u + 1| + |u – 3| < 2
Solution Let’s start again by making cases.
Case 1: u ≥ 3
In this case, |u – 3| = u – 3 and |u + 1| = u + 1. The inequation becomes
(u + 1) + (u – 3) < 2
which gives u < 2.
Now, this doesn’t mean that this inequation is true for any value of u less than 2. This solution is only valid when u ≥ 3.
That is, u should simultaneously satisfy u ≥ 3 and u < 2. This is something that isn’t possible – a number cannot be simultaneously greater than 3 and smaller than 2.
Hence, there’s no solution in this region.
Case 2: –1 ≤ u < 3
In this case, |u – 3| = –u + 3 and |u + 1| = u + 1. The inequation becomes
(u + 1) + (–u + 3) < 2
which gives 4 < 2, something which isn’t true.
That means, when we restrict ourselves to the region –1 ≤ u < 3, the LHS simplifies to 4, which can never be less than the RHS.
Once again, there’s no solution in this case.
Case 3: u < –1
In this case, |u – 3| = –u + 3 and |u + 1| = –u – 1. The inequation becomes
(–u – 1) + (–u + 3) < 2
which gives u > –1.
Given that we’re restricted to the region u < –1, it seems like we didn’t get any solution in this case as well.
We’ve run out of regions. In each case, we didn’t get any solution. Therefore, this inequation has no solution. Sad.
Example 3 Solve for y: |y – 5| – |y – 1| ≤ 6
Solution Let’s jump right into making the three cases.
Case 1: y ≥ 5
The inequation becomes
(y – 5) – (y – 1) ≤ 6
which gives –4 ≤ 6, something which is always true.
That means, if we substitute any value of y greater than or equal to 5, then the LHS will always be less than or equal to the RHS.
Therefore, the solution in this case is
y ≥ 5
Case 2: 1 ≤ y < 5
The inequation becomes
(–y + 5) – (y – 1) ≤ 6
which gives y ≥ 0. Similar to previous cases, we have to restrict ourselves in the region 1 ≤ y < 5. Combining this restriction with the solution, we’ll get the final solution as
1 ≤ y < 5
Case 3: y < 1
The inequation becomes
(–y + 5) – (1 – y) ≤ 6
which gives 4 ≤ 6, something which always true. And similar to Case 1, the solution in this case is
y < 1
Let’s summarize all solutions.
Case 1: y ≥ 5
Case 2: 1 ≤ y < 5
Case 3: y < 1
Combining these cases, we’ll get the solution as
–∞ < y < ∞
That means, the inequation is true for any real value of y. Nice inequation!
There are geometrical reasons for inequations being true for any value x and for them having no solutions. We’ve discussed these already in a previous lesson. So, I won’t be covering them here again.
But, do go back and try to apply geometrical reasoning to the second and third examples.
That’s all about inequations for now. To kill some time, try solving the following one on your own.
|x – 1| + |x – 2| + |x – 3| ≤ 9
Next, I’ll talk about the graph of the modulus function. And later, we’ll revisit some of these inequations and solve them using these graphs.
See you in the next lesson. Goodbye!