Additional Mathematics 4037/23 — October/November 2020

Walkthrough

The modulus $\left|3x - 2\right|$ equals $3x - 2$ when $3x - 2$ is non-negative, and $-(3x - 2)$ when it is negative. So the single equation becomes two linear equations, one for each case.

In the first case we solve $3x - 2 = 4 + x$, giving $2x = 6$ and $x = 3$. In the second case the left-hand side is $2 - 3x$, so we solve $2 - 3x = 4 + x$, giving $-2 = 4x$ and $x = -0.5$.

Each candidate solution should be checked in the original equation: both check out here ($7 = 7$ for $x = 3$ and $3.5 = 3.5$ for $x = -0.5$), so both are accepted.

Key Takeaways

• An equation of the form $\left|ax + b\right| = c(x)$ is solved by considering two cases: $ax + b = c(x)$ and $ax + b = -c(x)$.
• Checking solutions against the original equation guards against extraneous roots (important when the right-hand side could be negative).

Common Mistakes

• Solving only one case and giving just $x = 3$ — the mark scheme awards a separate M1/A1 for the second case, so one root only scores half the marks.
• Sign slips when negating: writing $-3x - 2 = 4 + x$ instead of $2 - 3x = 4 + x$ loses the second root entirely.
• Forgetting to verify that the right-hand side is non-negative; if a case produced a value making $4 + x < 0$, it would have to be rejected.

Things to Be Careful About

• Both solutions are required — the mark scheme gives B1 for $x = 3$ and then M1 A1 for setting up and solving $2 - 3x = 4 + x$ to get $x = -0.5$ ("oe" means any equivalent correct form is accepted).
• Answers may be given as decimals or fractions ($-0.5$ or $-\frac{1}{2}$); both are accepted.
• Show the setup of the second equation explicitly — the M1 is for forming $2 - 3x = 4 + x$, so jumping straight to $x = -0.5$ risks losing the method mark.