Hello again.
In this lesson, we’ll discuss a few examples of equations related to modulus. This lesson is kind of a continuation of the previous one, so make sure you go through that one first.
Let’s begin.
Example 1 Solve for x: |x – 2| + |x + 4| = 8
Solution As we discussed in the previous lesson, to solve such equations, we’ll divide the number line into some regions. In each such region, the expressions within the modulus ‘walls’ retain their sign.
As a result, we’ll be able to remove the modulus, freeing the variables and making the equation easily solvable.
The first expression |x – 2| can be resolved, according to the definition, as
\( |x-2| = \begin{cases}
x – 2& \text{ if } x\ge 2 \\
2 – x& \text{ if } x < 2
\end{cases} \)
And, the second expression can similarly be resolved as,
\( |x+4| = \begin{cases}
x +4& \text{ if } x\ge -4 \\
– x – 4& \text{ if } x < -4
\end{cases} \)
Drag the slider in the applet below to observe how these two expressions change their signs.
Similar to the equation in the previous lesson, this equation also keeps changing its form depending on the region in which x lies.
So once again, we’ll divide the number line into separate regions, in which the equation ‘retains’ its form, and try to solve the equation in each region.
Case 1: x ≥ 2
When x ≥ 2, |x – 2| turns into x – 2, and stays like that. Also, x ≥ 2 means x > –4, and therefore |x + 4| turns into x + 4 (and stays like that).
The equation becomes something like
(x – 2) + (x + 4) = 8
which gives x = 3.
Try substituting this value in the original equation to verify. We’re now done with the region x ≥ 2. Let’s move to the next one.
Case 2: –4 ≤ x < 2
First, let’s think why we chose to restrict x to this region.
If x < 2, then |x – 2| becomes 2 – x (and stays 2 – x), but |x + 4| may equal x + 4 or – x – 4, depending on whether x ≥ –4 or x < –4.
So, if we restrict x to the region –4 ≤ x < 2, |x + 4| will also retain its form, i.e., x + 4.
The original equation will turn into
(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 equal the RHS. Try substituting some values of x to verify this.
This means that there’s no solution of this equation when –4 ≤ x < 2. We’re left with one last case now.
Case 3: x < –4
Here, |x – 2| = 2 – x (because x < –4 implies x < 2 as well) and |x + 4| will turn into –x – 4 (and stay like that).
Our equation now becomes
2 – x – 4 – x = 8
This gives x = –5. Once again, substitute this in the original equation to verify. And, since we’ve covered the whole number line, there are no more solutions to find.
The rest of the examples will use the same method. But do go through them anyways, as you’ll come across some interesting situations in each.
Example 2 Solve for u: |u + 5| – |u + 1| = 2
Solution Let’s start by making cases. We’ll move a little faster this time.
Case 1: u ≥ –1
In this case, |u + 1| = u + 1 and |u + 5| = u + 5. The equation becomes
(u + 5) – (u + 1) = 2
which gives 4 = 2.
This means that there won’t be any solution in this region, as the LHS would always remain equal to 4. Pick any random values of u ≥ –1 to verify this.
Case 2: –5 ≤ u < 1
In this case, |u + 1| = – u – 1 and |u + 5| = u + 5. The equation becomes
(u + 5) – (– u – 1) = 2
which gives u = –2.
Case 3: u < –5
In this case, |u + 1| = – u – 1 and |u + 5| = – u – 5. The equation becomes
(–u – 5) – (– u – 1) = 2
which gives –4 = 2. Another strange situation. No solution!
There’s a geometrical meaning attached to why we don’t get any solution in Cases 1 and 3. Recall that |u + 5| and |u + 1| are respectively the distances between u and –5 and u and –1.
And, |u + 5| – |u + 1| is the difference of these two distances. If P represents u on the number line, and A and B respectively represent –5 and –1, then the equation is simply asking us
“Find a point P on the number line such that PA – PB = 2”
So, what happens when P moves to the right of –1 or to the left of –5? Have a look in the applet below. You can drag the point P.
When P moves to the right of –1, then PA – PB is always equal to AB (which equals 4). That’s why we get 4 on the LHS in Case 1.
Similarly, when P lies to the left of –5, then PB – PA is always equal to AB, or 4. This means that PA – PB = –4. Go check the LHS of Case 3 to corelate.
In both these cases PA – PB cannot equal 2. Hence, no solution.
What about between A and B?
If P is right in the middle of A and B, then PA – PB equals 0. And, if we move P towards B, the value of PA – PB increases to 4. Somewhere between 0 and 4, PA – PB will also equal 2, the RHS of the equation.
Hence, we’ll get a solution (i.e., between A and B). Let’s not be interested in what this solution is, but the fact that a solution exists.
Allow all this to sink in. 😑
Example 3 Solve for y: |y – 5| + |y – 1| = 4
Solution Let’s jump right into making the three cases.
Case 1: y ≥ 5
The equation becomes
(y – 5) + (y – 1) = 4
which gives y = 5.
Case 2: 1 ≤ y < 5
The equation becomes
(– y + 5) + (y – 1) = 4
which gives 4 = 4.
Hmm. What could that mean? Take a pause and think about it before you proceed.
The 4 = 4 means that when 1 ≤ y < 5, the LHS is always equal to 4, and hence always equal to the RHS. In other words, every value of y, such that 1 ≤ y < 5, is a solution of the original equation.
Let’s try a few.
If y = 1.3, then LHS = |1.3 – 5| + |1.3 – 1| = |–3.7| + |–0.3| = 3.7 + 0.3 = 4 = RHS
y = 4.2, then LHS = |4.2 – 5| + |4.2 – 1| = |–0.8| + |3.2| = 0.8 + 3.2 = 4 = RHS
y = 2, then LHS = |2 – 5| + |2 – 1| = |–3| + |1| = 3 + 1 = 4 = RHS
Awesome!
Case 3: y < 1
The equation becomes
(– y + 5) + (1 – y) = 4
which gives y = 1. Take a good look at this solution.
Is everything alright? Well, no.
The equation (– y + 5) + (1 – y) = 4 was only valid when we restricted y in the region y < 1. But that solution that we got is y = 1, which doesn’t lie in this region.
That means, this solution is invalid. And we’ll reject it.
But, if you substitute y = 1 in the original equation, you’ll find that LHS = RHS for this value of y.
What’s going on?
Well, go back to Case 2. We found that every value of y such that 1 ≤ y < 5 is a solution to the original equation. This solution includes y = 1. So, no need to worry.
But remember – any solution, that lies outside the region that we choose, must be rejected.
Let’s also take a look at what’s going on geometrically in Case 2.
Similar to the previous example, if P represents y on the number line, and A and B respectively represent 1 and 5, then the equation is asking us
“Find a point P on the number line such that PA + PB = 4”
So, what happens when P lies between A and B? Have a look in the applet below. You can drag the point P.
When P lies between A and B, then PA + PB is always equal to AB (which equals 4). That’s why we get 4 on the LHS in Case 2.
What about when P lies to the right of 5 (i.e., y > 5)?
PA + PB starts to increase from 4, and keeps on increasing as we move further to the right. Hence, PA + PB will never be equal to 4 again in this region. The same happens when P lies to the left to 1.
Therefore, we will not get any solution for y > 5 and y < 1.
Combining the solutions of Case 1 and Case 2, we’ll get the final solution as 1 ≤ y ≤ 5.
Allow this to sink in too. 😑😑
Example 4 Solve for z: |z – 3| + |z – 7| = 2
Solution I know you’re thinking ‘But I know all this!’, but this example has something new to show as well.
Case 1: z ≥ 7
The equation becomes
(z – 3) + (z – 7) = 2
which gives z = 6. Rejected!
Why? Because, we restricted z in the region z ≥ 7, and only those solution that lie in this region will be valid.
Let’s verify this by plugging z = 6 in the LHS. We’ll get
LHS = |6 – 3| + |6 – 7| = |3| + |–1| = 3 + 1 = 4 ≠ RHS
Told ya!
Case 2: 3 ≤ z < 7
The equation becomes
(z – 3) + (7 – z) = 2
which gives 4 = 2.
Bah! We’ve seen this absurdity before. Let’s move on.
Case 3: z < 3
The equation becomes
(3 – z) + (7 – z) = 2
which gives z = 4. Rejected again! You figure out why.
If we look at all the cases, this equation has no solution at all. 😥
What’s going on geometrically? Take a look at a similar applet. You can drag the point P.
When P lies between A and B, PA + PB equals AB, or 4. When we move P outside AB, PA + PB becomes greater than 4. And, it keeps increasing if we move further to the left of A or to the right of B.
In any case, PA + PB or |z – 3| + |z – 7| will never equal 2 (the RHS), and hence, the equation will have no solution. Case closed.
That’s all about equations for now. Why don’t you try solving the following equation on your own?
|x – 1| + |x – 2| + |x – 3| = 9
This will be a bit difficult to visualize on the number line. Try solving it by making cases.
Next, we’ll cover inequalities, the method of whose solutions will be very similar to equations. Later, we’ll cover everything again using graphs.
See you in the next lesson.