| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Line intersecting quadratic curve |
| Difficulty | Moderate -0.8 Part (a) is a standard C1 simultaneous equations question requiring substitution of a linear equation into a quadratic, then solving the resulting quadratic—routine technique with no conceptual challenges. Part (b) is a direct application of the solutions from (a) to solve a quadratic inequality, requiring only knowledge of which region satisfies the inequality. Both parts are textbook exercises with straightforward methods, making this easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02f Solve quadratic equations: including in a function of unknown1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| \(5 - 2x = 2x^2 - 3x - 16\) | M1 A1 | |
| \((2x - 7)(x + 3) = 0\) | M1 A1 | |
| \(x = -3, x = \frac{7}{2}\) and \(y = 11, y = -2\) | M1 A1ft | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Using critical values \(x = -3, x = \frac{7}{2}\) | M1 | |
| \(x < -3, x > \frac{7}{2}\) | M1 A1ft | (3 marks) |
## Part (a)
$5 - 2x = 2x^2 - 3x - 16$ | M1 A1 |
$(2x - 7)(x + 3) = 0$ | M1 A1 |
$x = -3, x = \frac{7}{2}$ and $y = 11, y = -2$ | M1 A1ft | (6 marks)
## Part (b)
Using critical values $x = -3, x = \frac{7}{2}$ | M1 |
$x < -3, x > \frac{7}{2}$ | M1 A1ft | (3 marks)
**Total: 9 marks**
---\begin{enumerate}[label=(\alph*)]
\item Solve the simultaneous equations
$$y + 2x = 5,$$
$$2x^2 - 3x - y = 16.$$ [6]
\item Hence, or otherwise, find the set of values of $x$ for which
$$2x^2 - 3x - 16 > 5 - 2x$$ [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q6 [9]}}