Additional Mathematics 4037/21 — October/November 2010

Walkthrough

The modulus $\left|2x + 10\right|$ measures the distance of $2x + 10$ from zero, so it equals $7$ when the expression inside is either $+7$ or $-7$. This gives two simple linear equations.

First case: $2x + 10 = 7$, so $2x = -3$ and $x = -1.5$.

Second case: $2x + 10 = -7$, so $2x = -17$ and $x = -8.5$.

Both values check out, since substituting either into $2x + 10$ gives a number whose absolute value is $7$.

Key Takeaways

• An equation $|f(x)| = k$ (with $k > 0$) always splits into the two cases $f(x) = k$ and $f(x) = -k$.
• Both solutions must be given; giving only one loses an accuracy mark.

Common Mistakes

• Solving only $2x + 10 = 7$ and forgetting the negative case — the mark scheme explicitly awards marks for solving $2x + 10 = -7$ (or squaring to get $(2x + 10)^2 = 49$).
• Sign slips when moving the $10$: writing $2x = 7 + 10$ instead of $2x = 7 - 10$.
• Dropping the negative root as if it were extraneous — here both roots are valid because the expression inside the modulus is linear with no domain restriction.

Things to Be Careful About

• The answers are exact decimals ($-1.5$ and $-8.5$); no rounding issues arise.
• An equivalent accepted route is squaring: $(2x + 10)^2 = 49$, then $2x + 10 = \pm 7$ — but you must still obtain both roots.
• State both answers clearly; the scheme awards B1 for $-1.5$ and A1 for $-8.5$ after the M1 method mark for setting up the second case.