Moderate -0.3 This is a straightforward rational equation requiring clearing denominators by multiplying through by (x+1)(2x+1), then solving the resulting quadratic. It's slightly easier than average as it's a standard technique with no conceptual surprises, though it requires careful algebraic manipulation.
Multiplying throughout by \((2x+1)(x+1)\) or combining fractions and multiplying up. Condone a single numerical error, sign error or slip provided no conceptual error. Do not condone omission of brackets unless implied by subsequent work
Multiplying out, collecting like terms and forming quadratic \(= 0\). Follow through from their equation provided algebra is not significantly eased and it is a quadratic. Condone a further sign or numerical error or minor slip when rearranging
\(3x^2 - x - 2 = 0\)
A1
or \(6x^2 - 2x - 4 = 0\) oe www (not fortuitously obtained - check for double errors)
\((3x+2)(x-1) = 0\)
M1
Solving their three term quadratic provided \(b^2 - 4ac \geq 0\). Use of correct quadratic formula or factorising or completing the square oe
\(x = -\frac{2}{3}\) or \(1\)
A1
cao for both obtained www. Accept \(-\frac{4}{6}\) oe or exact decimal equivalent. SC B1: \(x=1\) with or without working
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{4x}{x+1} - \frac{3}{2x+1} = 1 \Rightarrow 4x(2x+1) - 3(x+1) = (x+1)(2x+1)$ | M1 | Multiplying throughout by $(2x+1)(x+1)$ or combining fractions and multiplying up. Condone a single numerical error, sign error or slip provided no conceptual error. Do not condone omission of brackets unless implied by subsequent work |
| $8x^2 + 4x - 3x - 3 = 2x^2 + 3x + 1 \Rightarrow 6x^2 - 2x - 4 = 0$ | DM1 | Multiplying out, collecting like terms and forming quadratic $= 0$. Follow through from their equation provided algebra is not significantly eased and it is a quadratic. Condone a further sign or numerical error or minor slip when rearranging |
| $3x^2 - x - 2 = 0$ | A1 | or $6x^2 - 2x - 4 = 0$ oe www (not fortuitously obtained - check for double errors) |
| $(3x+2)(x-1) = 0$ | M1 | Solving their three term quadratic provided $b^2 - 4ac \geq 0$. Use of correct quadratic formula or factorising or completing the square oe |
| $x = -\frac{2}{3}$ or $1$ | A1 | cao for both obtained www. Accept $-\frac{4}{6}$ oe or exact decimal equivalent. SC B1: $x=1$ with or without working |
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