Standard +0.3 This requires solving a modulus equation by considering cases (or using the geometric interpretation that x is equidistant from -3/2 and 1/2, giving x = -1/2), then substituting into another modulus expression. It's slightly above routine as it combines two modulus concepts, but the algebra is straightforward and it's a standard P2 exercise with no novel insight required.
Solve non-modular equation \((2x+3)^2 = (2x-1)^2\) or linear equation with signs of \(2x\) different
M1
Obtain \(x = -\frac{1}{2}\)
A1
Substitute negative value into expression and show correct evaluation of modulus at least once
M1
Obtain \(5 - 3 = 2\) with no errors seen
A1
Total
4
**Question 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Solve non-modular equation $(2x+3)^2 = (2x-1)^2$ or linear equation with signs of $2x$ different | M1 | |
| Obtain $x = -\frac{1}{2}$ | A1 | |
| Substitute negative value into expression and show correct evaluation of modulus at least once | M1 | |
| Obtain $5 - 3 = 2$ with no errors seen | A1 | |
| **Total** | **4** | |
2 Given that $x$ satisfies the equation $| 2 x + 3 | = | 2 x - 1 |$, find the value of
$$| 4 x - 3 | - | 6 x |$$
\hfill \mbox{\textit{CAIE P2 2019 Q2 [4]}}