| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Simultaneous with substitution elimination |
| Difficulty | Standard +0.3 This is a structured simultaneous equations problem with clear guidance ('show that' followed by 'hence solve'). Part (a) requires straightforward substitution and algebraic manipulation. Part (b) involves solving a quadratic in x² (standard technique) then back-substituting. While it has multiple steps and requires careful algebra, the path is clearly signposted and uses routine P1 techniques, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2xy - 3x^2 = 50\); \(y - x^3 + 6x = 0 \Rightarrow 2x(x^3 - 6x) - 3x^2 = 50\) | M1 | Attempts substitution of \(y\) from one equation into the other |
| \(\Rightarrow 2x^4 - 12x^2 - 3x^2 - 50 = 0 \Rightarrow 2x^4 - 15x^2 - 50 = 0\) | A1* | CSO; correct bracketing with at least one intermediate line shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((2x^2 + 5)(x^2 - 10) = 0 \Rightarrow x^2 = ...\) | M1 | Non-calculator method to solve quadratic in \(x^2\); allow factorisation, formula, or completing the square |
| \(x^2 = 10\) | A1 | Correct solution; \(x^2 = -\frac{5}{2}\) scores 0 unless recovered |
| \(\Rightarrow y = (\sqrt{10})^3 - 6\sqrt{10} = ...\) | M1 | Uses at least one \(x\) value of form \(p\sqrt{a}\), \(a>0\); non-calculator method for \(y\) |
| One solution pair: \(x = \sqrt{10}\), \(y = 4\sqrt{10}\) | A1 | At least one correct pair; need not be fully simplified but must be single terms |
| Both solutions: \(x = \sqrt{10}, y = 4\sqrt{10}\) and \(x = -\sqrt{10}, y = -4\sqrt{10}\) | A1 | CSO; both pairs correct, clearly paired; no extra incorrect answers |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2xy - 3x^2 = 50$; $y - x^3 + 6x = 0 \Rightarrow 2x(x^3 - 6x) - 3x^2 = 50$ | M1 | Attempts substitution of $y$ from one equation into the other |
| $\Rightarrow 2x^4 - 12x^2 - 3x^2 - 50 = 0 \Rightarrow 2x^4 - 15x^2 - 50 = 0$ | A1* | CSO; correct bracketing with at least one intermediate line shown |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2x^2 + 5)(x^2 - 10) = 0 \Rightarrow x^2 = ...$ | M1 | Non-calculator method to solve quadratic in $x^2$; allow factorisation, formula, or completing the square |
| $x^2 = 10$ | A1 | Correct solution; $x^2 = -\frac{5}{2}$ scores 0 unless recovered |
| $\Rightarrow y = (\sqrt{10})^3 - 6\sqrt{10} = ...$ | M1 | Uses at least one $x$ value of form $p\sqrt{a}$, $a>0$; non-calculator method for $y$ |
| One solution pair: $x = \sqrt{10}$, $y = 4\sqrt{10}$ | A1 | At least one correct pair; need not be fully simplified but must be single terms |
| Both solutions: $x = \sqrt{10}, y = 4\sqrt{10}$ and $x = -\sqrt{10}, y = -4\sqrt{10}$ | A1 | CSO; both pairs correct, clearly paired; no extra incorrect answers |
---6. In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
\begin{enumerate}[label=(\alph*)]
\item Given that
$$2 x y - 3 x ^ { 2 } = 50$$
and
$$y - x ^ { 3 } + 6 x = 0$$
show that
$$2 x ^ { 4 } - 15 x ^ { 2 } - 50 = 0$$
\item Hence solve the simultaneous equations
$$\begin{aligned}
2 x y - 3 x ^ { 2 } & = 50 \\
y - x ^ { 3 } + 6 x & = 0
\end{aligned}$$
Give your answers in fully simplified surd form.\\
\includegraphics[max width=\textwidth, alt={}, center]{3cf69966-e825-4ff0-a6e8-c5dfdc92c53f-14_2257_52_312_1982}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2022 Q6 [7]}}