A Modulus Equation That Needs Checking

When a modulus equals an expression containing the variable, the two-case split can produce a value that does not actually work, because the right-hand side must be non-negative. So this question adds a step: check each solution, and reject any that fails.

Solve $|x - 3| = 2x$. [4]

The working

Step 1, split into two cases ($x - 3$ is either $+2x$ or $-2x$): $x - 3 = 2x \qquad \text{or} \qquad x - 3 = -2x \quad \text{(M1)}$

Step 2, solve each: $x - 3 = 2x \;\Rightarrow\; -3 = x \;\Rightarrow\; x = -3$ $x - 3 = -2x \;\Rightarrow\; 3x = 3 \;\Rightarrow\; x = 1 \quad \text{(A1)}$

Step 3, check each in the original (the right-hand side $2x$ must be $\ge 0$):

• $x = -3$: LHS $= |-3 - 3| = 6$, RHS $= 2(-3) = -6$. $6 \ne -6$, reject (a modulus cannot equal a negative). (M1)
• $x = 1$: LHS $= |1 - 3| = 2$, RHS $= 2(1) = 2$. ✓ valid.

Conclusion: $x = 1$ only. (A1)

Where the marks are won and lost

• Both cases must be written and solved, that produces the two candidates.
• The check is the decisive step. $x = -3$ satisfies the algebra of one case but makes the RHS negative, so it cannot solve $|x - 3| = 2x$. Rejecting it, with a reason, earns the final mark.
• A quick alternative check: since $|x - 3| \ge 0$, we need $2x \ge 0$, i.e. $x \ge 0$, which immediately rules out $x = -3$.

Common mistakes

• Giving both $x = -3$ and $x = 1$ without checking.
• Assuming the negative case never matters (it does, when the RHS has a variable).
• Squaring both sides but then mishandling the extraneous root the same way.

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