Let’s come back to the original problem: |x – 1| + |x – 4| = 7.

The idea here is to get rid of those *walls* of the modulus, so that we can add
the two trapped expressions.

The first expression will be x – 1 when x ≥ 1 and 1 – x when x < 1.

And, the second expression will equal to x – 4 when x ≥ 4 and 4 – x when x < 4.

Drag the slider in the applet below to understand what I mean.

Basically, this equation keeps changing its form depending on the region in which x lies. So, what we’ll do is solve this equation in separate regions in which it ‘retains’ its form. Stay with me.

Suppose x ≥ 4. Then in this region, |x – 4| = x – 4. Also, x ≥ 4 implies x > 1, which means that |x – 1| = x – 1.

Also, |x – 4| ‘remains’ x – 4 and |x – 1| remains x – 1 (and neither of them becomes 4 – x or 1 – x).

That is, the equation retains its form. Now we can solve the equation.

If x ≥ 4, |x – 1| + |x – 4| = 7 implies x – 1
+ x – 4 = 7. This gives 2x = 12 or **x = 6**.

Let’s try substituting this value in the original equation to see if things are okay.

We’ll get the LHS as |6 – 1| + |6 – 4| = |5| + |2| = 5 + 2 = 7, which is equal to the RHS. Great!

We’re now done with the region x ≥ 4. Let’s move to the next one.

If x < 4, then |x – 4| becomes 4 – x, but |x – 1| may equal x – 1 or 1 – x, depending on whether x ≥ 1 or x < 1.

So, what we’ll do is restrict ourselves to the region 1 ≤ x < 4. The idea behind this remains the same – the equation will ‘retain’ its form in this region. That is, |x – 4| will always equal 4 – x and |x – 1| will always equal x – 1.

Let’s solve the equation then.

|x – 1| + |x – 4| = 7 implies x – 1 + 4 – x = 7, which gives 3 = 7.

Well, that’s strange. Can you make sense of it?

This means that when x lies between 1 and 4, the LHS will always be 3.

Let’s try substituting some values of x.

If x = 2, then LHS = |2 – 1| + |2 – 4| = |1| + |–2| = 1 + 2 = 3.

If x = 1.8, then LHS = |1.8 – 1| + |1.8 – 4| = |0.8| + |–2.2| = 0.8 + 2.2 = 3.

If x = 3.9, then LHS = |3.9 – 1| + |3.9 – 4| = |2.9| + |–0.1| = 2.9 + 0.1 = 3.

I hope you believe me now. Why don’t you try some more values anyways?

We’re left with one last case: x < 1.

Here, |x – 1| = 1 – x and since x < 1 also means x < 4, |x – 4| will equal 4 – x.

Our equation now becomes 1 – x + 4 – x = 7. This gives 5 – 2x = 7, which gives x = –1.

Well, let’s substitute just to be sure.

LHS = |–1 – 1| + |–1 – 4| = |–2| + |–5| = 2 + 5 = 7 = RHS. Yay!

And, we’re done. We’ve covered all possible cases: x ≥ 4, 1 ≤ x < 4 and x < 1 together cover the whole number line.

Let’s summarize!

To solve an equation of the form |x – 1| + |x – 4| = 7, we divided the number line into some ‘regions’, such that in a particular region, the expressions inside the modulus retain their signs. Or, |x – 4| remained x – 4 (or 4 – x) throughout that region.

So, when x ≥ 4, the equation became x – 1 + x – 4 = 7, which gave x = 6.

When 1 ≤ x < 4, we got x – 1 + 4 – x = 7, which gave a strange equation, 3 = 7, implying that there’s no solution in this region.

Finally, when x < 1, we got 1 – x + 4 – x = 7, which gave x = –1.

I’ll stop here. I know you’re confused. More examples on the way in the next lesson!