Solving |x + 1| = |2x − 4|

With a modulus on both sides, $|A| = |B|$ splits into two cases: $A = B$ and $A = -B$. Both sides are non-negative, so unlike a modulus equal to a variable, both candidates are usually valid, though checking is still good practice.

Solve $|x + 1| = |2x - 4|$. [5]

The working

Step 1, set up the two cases: $x + 1 = 2x - 4 \qquad \text{or} \qquad x + 1 = -(2x - 4) \quad \text{(M1)}$

Step 2, solve the first case: $x + 1 = 2x - 4 \;\Rightarrow\; 1 + 4 = 2x - x \;\Rightarrow\; x = 5 \quad \text{(A1)}$

Step 3, solve the second case (distribute the minus to both terms): $x + 1 = -2x + 4 \;\Rightarrow\; 3x = 3 \;\Rightarrow\; x = 1 \quad \text{(A1)}$

Step 4, check both in the original:

• $x = 5$: $|6| = |6|$ ✓
• $x = 1$: $|2| = |-2|$ ✓

Both valid: $x = 1$ or $x = 5$. (M1, A1)

Where the marks are won and lost

• The second case is $x + 1 = -(2x - 4)$, distribute the minus to both terms: $-2x + 4$.
• Both sides being moduli means both candidates are typically valid, but a quick check confirms it and catches any slip.
• Squaring both sides ($(x+1)^2 = (2x-4)^2$) is an alternative route that reaches the same two solutions.

Common mistakes

• Distributing the minus to only one term: $-(2x - 4) = -2x - 4$ (wrong; it’s $-2x + 4$).
• Setting up only one case.
• Assuming a solution must be rejected (here both are valid).

Full method: Modulus Equations & Inequalities notes. Topic home: Equations, Inequalities and Graphs pillar.