Q.Solve for $x$: $\dfrac{|x-2|-1}{|x-2|-2} \le 0$.

We treat this as a sign analysis problem. Let $t = |x-2| \ge 0$. The inequality becomes

The rational expression is $\le 0$ when numerator and denominator have opposite signs (or numerator zero). Critical points: $t=1$ (zero numerator) and $t=2$ (undefined).

Sign chart for $t \ge 0$:

• For $0 \le t < 1$: numerator negative, denominator negative → positive (no).
• For $1 \le t < 2$: numerator $\ge 0$, denominator negative → $\le 0$ (yes).
• For $t > 2$: both positive → positive (no).

So $1 \le t < 2$, i.e. $1 \le |x-2| < 2$.

This gives two intervals:

and

Concept and Intuition

When you see an expression like $\frac{|x-2|-1}{|x-2|-2} \le 0$, the natural instinct might be to jump into cases for the absolute value. But there's a cleaner way: notice that $|x-2|$ appears twice in exactly the same form. This is a classic signal to substitute $t = |x-2|$, turning the inequality into a rational inequality in $t$, which is much simpler to handle.

The key idea: a rational expression $\frac{A}{B} \le 0$ means the numerator and denominator have opposite signs (or the numerator is zero). We never multiply both sides by the denominator unless we know its sign — instead, we use a sign chart.

After solving for $t$, we translate back: $|x-2| = t$ means $x$ is at distance $t$ from $2$, so $x = 2 \pm t$ (or $x$ in an interval if $t$ is a range).

Let's work through it.

Step-by-step solution

1. Substitute to simplify

Let $t = |x-2|$. Since $|x-2| \ge 0$ for all real $x$, we have $t \ge 0$. The inequality becomes:

2. Find critical points

The expression changes sign where numerator or denominator is zero:

• Numerator zero: $t-1 = 0 \implies t = 1$
• Denominator zero: $t-2 = 0 \implies t = 2$ (excluded, since division by zero is undefined)

These points split the non-negative real line into intervals: $[0,1)$, $(1,2)$, $(2,\infty)$.

3. Sign analysis

We test a value from each interval:

• For $t=0$ (in $[0,1)$): $\frac{0-1}{0-2} = \frac{-1}{-2} = \frac12 > 0$
• For $t=1.5$ (in $(1,2)$): $\frac{1.5-1}{1.5-2} = \frac{0.5}{-0.5} = -1 < 0$
• For $t=3$ (in $(2,\infty)$): $\frac{3-1}{3-2} = \frac{2}{1} = 2 > 0$

The inequality $\le 0$ is satisfied where the expression is negative or zero. That happens on $(1,2)$ (negative) and at $t=1$ (zero). So:

4. Convert back to $x$

Recall $t = |x-2|$. So we need:

This is a compound inequality. Let's solve each part.

Part A: $|x-2| \ge 1$

This means $x-2 \le -1$ or $x-2 \ge 1$, i.e.:

Part B: $|x-2| < 2$

This means $-2 < x-2 < 2$, i.e.:

5. Intersect the conditions

We need both conditions to hold simultaneously. So take the intersection of:

• $x \le 1$ or $x \ge 3$
• $0 < x < 4$

Let's do this carefully:

• From $x \le 1$ and $0 < x < 4$: we get $0 < x \le 1$
• From $x \ge 3$ and $0 < x < 4$: we get $3 \le x < 4$

So the solution in $x$ is:

6. Check endpoints

• $x=1$: $|1-2|=1$, expression $= \frac{0}{-1}=0$, included.
• $x=3$: $|3-2|=1$, expression $=0$, included.
• $x=0$: $|0-2|=2$, denominator zero, excluded.
• $x=4$: $|4-2|=2$, denominator zero, excluded.

Everything checks.