| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Solve rational equation |
| Difficulty | Moderate -0.3 This is a straightforward algebraic manipulation question requiring factorization of a quadratic, combining fractions over a common denominator, and solving a resulting linear equation. While it involves multiple steps, each technique is routine for C3 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{x+4}{(2x+1)(x+1)} - \frac{2}{2x+1} = \frac{(x+4)-2(x+1)}{(2x+1)(x+1)}\) | M1 | |
| \(= \frac{2-x}{(2x+1)(x+1)}\) | A1 | |
| \(\frac{2-x}{(2x+1)(x+1)} = \frac{1}{2}\) | M1 | |
| \(2(2-x) = 2x^2 + 3x + 1\) | M1 | |
| \(2x^2 + 5x - 3 = 0\) | M1 | |
| \((2x-1)(x+3) = 0\) | M1 | |
| \(x = -3, \frac{1}{2}\) | A1 | (6 marks) |
$\frac{x+4}{(2x+1)(x+1)} - \frac{2}{2x+1} = \frac{(x+4)-2(x+1)}{(2x+1)(x+1)}$ | M1 |
$= \frac{2-x}{(2x+1)(x+1)}$ | A1 |
$\frac{2-x}{(2x+1)(x+1)} = \frac{1}{2}$ | M1 |
$2(2-x) = 2x^2 + 3x + 1$ | M1 |
$2x^2 + 5x - 3 = 0$ | M1 |
$(2x-1)(x+3) = 0$ | M1 |
$x = -3, \frac{1}{2}$ | A1 | (6 marks)
---\begin{enumerate}
\item (a) Express
\end{enumerate}
$$\frac { x + 4 } { 2 x ^ { 2 } + 3 x + 1 } - \frac { 2 } { 2 x + 1 }$$
as a single fraction in its simplest form.\\
(b) Hence, find the values of $x$ such that
$$\frac { x + 4 } { 2 x ^ { 2 } + 3 x + 1 } - \frac { 2 } { 2 x + 1 } = \frac { 1 } { 2 } .$$
\hfill \mbox{\textit{Edexcel C3 Q1 [6]}}