5. Equations Involving Modulus (Part 3)


Hello again. This lesson will discuss a few (somewhat) crazy equations involving modulus. Make sure that you understand both the methods covered in the previous lesson before proceeding.

Let’s begin by solving |x – 1| + |x – 2| = 5.

Hmm, somewhat crazy. I was right.

Before we start with this equation, I’ll take a little detour and solve a simpler equation: |x – 1| = 5.

I know we’ve covered these in the previous lesson, but I want to cover it again using a slightly different method – using the definition of modulus.

Let’s relook at the definition of modulus: \( |x| = \begin{cases} x \text{ if } x\ge 0 \\ -x \text{ if } x<0 \end{cases}\)

But here, we have |x – 1| instead of |x|. So, we’ll simply replace x by x – 1 in that definition.

That means, \( |x-1| = \begin{cases} x-1 \text{ if } x-1\ge 0 \\ -(x-1) \text{ if } x-1<0 \end{cases}\)

So far, so good.

Proceeding further, we have \( |x-1| = \begin{cases} x-1 \text{ if } x\ge 1 \\ 1-x \text{ if } x<1 \end{cases}\)

So, for example, if x = 3 (which satisfies x ≥ 1), then |x – 1| = x – 1 = 3 – 1 = 2.

We could have also plugged x = 3 directly in |x – 1| to get |3 – 1| = |2| = 2.

And, if x = –2 (which satisfies x < 1), then |x – 1| = 1 – x = 1 – (–2) = 3.

Once again, we could have plugged x = –2 in |x – 1| to get |–2 – 1| = |–3| = 3. We’ll get the same value.

In general, if we have the value of x, we can directly plug its value in the given expression to find the expression’s value.

But when we do not have a fixed value of x, or to perform operations on different expressions involving modulus, we generally use the definition. That is, |x – 1| = x – 1 when x ≥ 1 etc.

Let me take another example.

\( |x+3| = \begin{cases} x+3 \text{ if } x+3\ge 0 \\ -(x+3) \text{ if } x+3 <0 \end{cases}\) which simplifies a bit to \( |x+3| = \begin{cases} x+3 \text{ if } x\ge -3 \\ -x-3 \text{ if } x<-3 \end{cases}\)

Another example.

\( |x-4| = \begin{cases} x-4 \text{ if } x\ge 4 \\ 4-x \text{ if } x<4 \end{cases}\)

And another.

\( |2x-3| = \begin{cases} 2x-3 \text{ if } x\ge 3/2 \\ 3-2x \text{ if } x<3/2 \end{cases}\)

And another.

\( |x^2-5| = \begin{cases} x^2-5 \text{ if } x^2\ge 5 \\ 5-x^2 \text{ if } x^2<5 \end{cases}\)

I hope you get the idea.

In general, we can write \( |Exp| = \begin{cases} Exp \text{ if } Exp\ge 0 \\ -Exp \text{ if } Exp<0 \end{cases}\) where ‘Exp’ is any mathematical expression.

Now, let’s solve |x – 1| = 5.

\( |x-1| = \begin{cases} x-1 \text{ if } x\ge 1 \\ 1-x \text{ if } x<1 \end{cases}\)

This means that solving |x – 1| = 5 is equivalent to solving x – 1 = 5, when x ≥ 1 and solving 1 – x = 5 when x < 1.

Let’s do it then!

When x ≥ 1, we get x – 1 = 5 or x = 6. And when x < 1, we have 1 – x = 5 or x = –4.

Just to quickly verify, we can plug these solutions in the original equation.

Substituting x = 6, we get the LHS as |6 – 1| = |5| = 5, which is equal to the RHS.

Similarly on substituting x = -4, we get the LHS as |–4 – 1| = |–5| = 5, which is again equal to the RHS. Both solutions are correct that means.

I’ll end this lesson here and start with the original problem |x – 1| + |x – 2| = 5 in the next one.