| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Line intersecting general conic |
| Difficulty | Moderate -0.8 This is a straightforward simultaneous equations problem requiring substitution of y=5-x into a quadratic, then solving the resulting quadratic equation. It's a standard Core 1/2 exercise with clear steps and no conceptual challenges, making it easier than the average A-level question which typically requires more problem-solving or integration of multiple techniques. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02f Solve quadratic equations: including in a function of unknown |
| VIIIV SIHI NI III IM ION OC | VIIV SIHI NI JIIIM ION OC | VI4V SIHI NI JIIIM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y = 5-x \Rightarrow x^2 + x + (5-x)^2 = 51\) or \(x = 5-y \Rightarrow (5-y)^2 + (5-y) + y^2 = 51\) | M1 | Attempts to rearrange linear equation to \(y=\ldots\) or \(x=\ldots\) and attempts to fully substitute into second equation |
| \(2x^2 - 9x - 26 = 0\) or \(2y^2 - 11y - 21 = 0\) | M1 | Collect terms to produce a 2 or 3 term quadratic \(= 0\). The "\(=0\)" may be implied by later work |
| Correct quadratic in \(x\) or \(y\) | A1 | As above |
| \((2x-13)(x+2) = 0 \Rightarrow x = \ldots\) or \((2y+3)(y-7) = 0 \Rightarrow y = \ldots\) | dM1 | Attempt to factorise and solve or complete the square and solve or use correct quadratic formula for a 3 term quadratic and obtain at least one value of \(x\) or \(y\). Dependent on both previous M marks |
| \(x = 6.5,\ x = -2\) or \(y = -1.5,\ y = 7\) | A1 cso | Correct answers for either both values of \(x\) or both values of \(y\) (possibly unsimplified) |
| Substitutes their \(x\) into \(y = 5-x\) or their \(y\) into \(x = 5-y\) | M1 | Substitute at least one value to find \(y\) or vice versa |
| \(x = 6.5\left(\text{or } \frac{13}{2}\right),\ x = -2\) and \(y = -1.5\left(\text{or } -\frac{3}{2}\right),\ y = 7\) | A1 cso | Fully correct solutions and simplified. Coordinates do not need to be paired |
## Question 2:
| Working | Mark | Guidance |
|---------|------|----------|
| $y = 5-x \Rightarrow x^2 + x + (5-x)^2 = 51$ **or** $x = 5-y \Rightarrow (5-y)^2 + (5-y) + y^2 = 51$ | M1 | Attempts to rearrange linear equation to $y=\ldots$ or $x=\ldots$ and attempts to fully substitute into second equation |
| $2x^2 - 9x - 26 = 0$ **or** $2y^2 - 11y - 21 = 0$ | M1 | Collect terms to produce a 2 or 3 term quadratic $= 0$. The "$=0$" may be implied by later work |
| Correct quadratic in $x$ or $y$ | A1 | As above |
| $(2x-13)(x+2) = 0 \Rightarrow x = \ldots$ **or** $(2y+3)(y-7) = 0 \Rightarrow y = \ldots$ | dM1 | Attempt to factorise and solve or complete the square and solve or use correct quadratic formula for a 3 term quadratic and obtain at least one value of $x$ or $y$. Dependent on both previous M marks |
| $x = 6.5,\ x = -2$ **or** $y = -1.5,\ y = 7$ | A1 cso | Correct answers for either both values of $x$ or both values of $y$ (possibly unsimplified) |
| Substitutes their $x$ into $y = 5-x$ or their $y$ into $x = 5-y$ | M1 | Substitute at least one value to find $y$ or vice versa |
| $x = 6.5\left(\text{or } \frac{13}{2}\right),\ x = -2$ and $y = -1.5\left(\text{or } -\frac{3}{2}\right),\ y = 7$ | A1 cso | Fully correct solutions and simplified. Coordinates do not need to be paired |
---2. Use algebra to solve the simultaneous equations
$$\begin{aligned}
x + y & = 5 \\
x ^ { 2 } + x + y ^ { 2 } & = 51
\end{aligned}$$
You must show all stages of your working.\\
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VIIIV SIHI NI III IM ION OC & VIIV SIHI NI JIIIM ION OC & VI4V SIHI NI JIIIM ION OO \\
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\hfill \mbox{\textit{Edexcel C12 2018 Q2 [7]}}