Solve |2x - 3| ≤ 5

Question

Solve the inequality $|2x - 3| \leq 5$ and represent the solution on a number line.

Solution — Step by Step

For any expression $|A| \leq k$ (where $k \geq 0$ ), the equivalent compound inequality is:

-k \leq A \leq k

This is because $|A|$ represents the distance of $A$ from zero, so “the distance is at most $k$ ” means $A$ lies between $-k$ and $k$ , inclusive.

Here $A = 2x - 3$ and $k = 5$ :

-5 \leq 2x - 3 \leq 5

Add 3 to all three parts:

-5 + 3 \leq 2x \leq 5 + 3

-2 \leq 2x \leq 8

Divide all three parts by 2:

-1 \leq x \leq 4

\boxed{x \in [-1, 4]}

All real numbers from $-1$ to $4$ , inclusive.

On a number line: filled circles at $-1$ and $4$ , with the segment between them shaded.

Why This Works

The absolute value $|2x - 3|$ measures how far $2x - 3$ is from zero. Saying this distance is at most 5 means $2x - 3$ lies within 5 units of 0 on both sides — between $-5$ and $5$ . Once we set up this double-sided inequality, it’s just arithmetic to solve for $x$ .

Geometrically, $|2x - 3| \leq 5$ is equivalent to saying that $x$ lies within $5/2 = 2.5$ units of $3/2 = 1.5$ on the number line. Centre = 1.5, radius = 2.5 → range from $1.5 - 2.5 = -1$ to $1.5 + 2.5 = 4$ . ✓

The “centre and radius” interpretation: for $|x - c| \leq r$ , the solution is the interval $[c - r, c + r]$ . For $|2x - 3| \leq 5$ , divide by 2 first: $|x - 3/2| \leq 5/2$ . Centre = 3/2, radius = 5/2 → solution $[-1, 4]$ .

Alternative Method

Split into two cases based on the sign of $2x - 3$ :

Case 1: $2x - 3 \geq 0$ (i.e., $x \geq 3/2$ ). Then $|2x - 3| = 2x - 3$ .

2x - 3 \leq 5 \implies x \leq 4

Combined with $x \geq 3/2$ : $3/2 \leq x \leq 4$ .

Case 2: $2x - 3 < 0$ (i.e., $x < 3/2$ ). Then $|2x - 3| = -(2x - 3) = 3 - 2x$ .

3 - 2x \leq 5 \implies -2x \leq 2 \implies x \geq -1

Combined with $x < 3/2$ : $-1 \leq x < 3/2$ .

Union of both cases: $-1 \leq x \leq 4$ . ✓

Same answer, just more steps.

Common Mistake

For $|A| \leq k$ , students sometimes write two separate inequalities: $A \leq k$ AND $A \geq k$ . The second should be $A \geq -k$ . Also, when flipping direction for the negative case, the inequality sign must reverse: $-A \leq k \implies A \geq -k$ . Forgetting this flip is the classic sign error in absolute value problems.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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