Question 4 - A-Level Maths

Edexcel C1 2011 June — Question 4 7 marks

Exam Board	Edexcel
Module	C1 (Core Mathematics 1)
Year	2011
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Simultaneous equations
Type	Line intersecting general conic
Difficulty	Moderate -0.3 This is a straightforward simultaneous equations problem requiring substitution of a linear equation into a quadratic, then solving the resulting quadratic. While it involves a difference of squares (hyperbola), the method is standard C1 fare with no conceptual challenges—slightly easier than average due to the clean algebraic manipulation and simple arithmetic.
Spec	1.02c Simultaneous equations: two variables by elimination and substitution 1.02f Solve quadratic equations: including in a function of unknown

4. Solve the simultaneous equations $$\begin{aligned} x + y & = 2 \\ 4 y ^ { 2 } - x ^ { 2 } & = 11 \end{aligned}$$

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Either path leads to acceptable working.

Path 1:

Answer	Marks	Guidance
$y^2 = 4 - 4x + x^2$	M1
$4(4 - 4x + x^2) - x^2 = 11$ or $4(2-x)^2 - x^2 = 11$	M1
$3x^2 - 16x + 5 = 0$	A1	Correct 3 terms
$(3x - 1)(x - 5) = 0$, $x = \frac{1}{3}$ or $x = 5$	M1
$y = \frac{5}{3}$ or $y = -3$	A1

Path 2:

Answer	Marks	Guidance
$x^2 = 4 - 4y + y^2$	M1
$4y^2 - (4 - 4y + y^2) = 11$ or $4y^2 - (2 - y)^2 = 11$	M1
$3y^2 + 4y - 15 = 0$	A1	Correct 3 terms
$(3y - 5)(y + 3) = 0$, $y = \frac{5}{3}$ or $y = -3$	M1
$x = \frac{1}{3}$ or $x = 5$	A1
Final answer (either path): $y = \frac{5}{3}, y = -3$ or $x = \frac{1}{3}, x = 5$	M1 A1
Notes		1st M: Squaring to give 3 or 4 terms (need a middle term). 2nd M: Substitute to give quadratic in one variable (may have just two terms). 3rd M: Attempt to solve a 3 term quadratic. 4th M: Attempt to find at least one y value (or x value). (The second variable) This will be by substitution or by starting again. If y solutions are given as x values, or vice-versa, penalise accuracy, so that it is possible to score M1 M1 A1 M1 A0 M1 A0. "Non-algebraic" solutions: No working, and only one correct solution pair found (e.g. $x = 5, y = -3$): M0 M0 A0 M1 A0 M1 A0. No working, and both correct solution pairs found, but not demonstrated: M0 M0 A0 M1 A1 M1 A1. Both correct solution pairs found, and demonstrated: Full marks are possible (send to review)

Either path leads to acceptable working.

**Path 1:**
$y^2 = 4 - 4x + x^2$ | M1 |
$4(4 - 4x + x^2) - x^2 = 11$ or $4(2-x)^2 - x^2 = 11$ | M1 |
$3x^2 - 16x + 5 = 0$ | A1 | Correct 3 terms
$(3x - 1)(x - 5) = 0$, $x = \frac{1}{3}$ or $x = 5$ | M1 |
$y = \frac{5}{3}$ or $y = -3$ | A1 |

**Path 2:**
$x^2 = 4 - 4y + y^2$ | M1 |
$4y^2 - (4 - 4y + y^2) = 11$ or $4y^2 - (2 - y)^2 = 11$ | M1 |
$3y^2 + 4y - 15 = 0$ | A1 | Correct 3 terms
$(3y - 5)(y + 3) = 0$, $y = \frac{5}{3}$ or $y = -3$ | M1 |
$x = \frac{1}{3}$ or $x = 5$ | A1 |

Final answer (either path): $y = \frac{5}{3}, y = -3$ or $x = \frac{1}{3}, x = 5$ | M1 A1 |

Notes | | 1st M: Squaring to give 3 or 4 terms (need a middle term). 2nd M: Substitute to give quadratic in one variable (may have just two terms). 3rd M: Attempt to solve a 3 term quadratic. 4th M: Attempt to find at least one y value (or x value). (The second variable) This will be by substitution or by starting again. If y solutions are given as x values, or vice-versa, penalise accuracy, so that it is possible to score M1 M1 A1 M1 A0 M1 A0. "Non-algebraic" solutions: No working, and only one correct solution pair found (e.g. $x = 5, y = -3$): M0 M0 A0 M1 A0 M1 A0. No working, and both correct solution pairs found, but not demonstrated: M0 M0 A0 M1 A1 M1 A1. Both correct solution pairs found, and demonstrated: Full marks are possible (send to review)

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4. Solve the simultaneous equations

$$\begin{aligned}
x + y & = 2 \\
4 y ^ { 2 } - x ^ { 2 } & = 11
\end{aligned}$$

\hfill \mbox{\textit{Edexcel C1 2011 Q4 [7]}}

This paper (10 questions)

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Q1 3 Q2 7 Q3 5 Q4 7 Q5 7 Q6 7 Q7 6 Q8 10 Q9 9 Q10 14

Answer	Marks	Guidance
\(y^2 = 4 - 4x + x^2\)	M1
\(4(4 - 4x + x^2) - x^2 = 11\) or \(4(2-x)^2 - x^2 = 11\)	M1
\(3x^2 - 16x + 5 = 0\)	A1	Correct 3 terms
\((3x - 1)(x - 5) = 0\), \(x = \frac{1}{3}\) or \(x = 5\)	M1
\(y = \frac{5}{3}\) or \(y = -3\)	A1

Answer	Marks	Guidance
\(x^2 = 4 - 4y + y^2\)	M1
\(4y^2 - (4 - 4y + y^2) = 11\) or \(4y^2 - (2 - y)^2 = 11\)	M1
\(3y^2 + 4y - 15 = 0\)	A1	Correct 3 terms
\((3y - 5)(y + 3) = 0\), \(y = \frac{5}{3}\) or \(y = -3\)	M1
\(x = \frac{1}{3}\) or \(x = 5\)	A1
Final answer (either path): \(y = \frac{5}{3}, y = -3\) or \(x = \frac{1}{3}, x = 5\)	M1 A1
Notes		1st M: Squaring to give 3 or 4 terms (need a middle term). 2nd M: Substitute to give quadratic in one variable (may have just two terms). 3rd M: Attempt to solve a 3 term quadratic. 4th M: Attempt to find at least one y value (or x value). (The second variable) This will be by substitution or by starting again. If y solutions are given as x values, or vice-versa, penalise accuracy, so that it is possible to score M1 M1 A1 M1 A0 M1 A0. "Non-algebraic" solutions: No working, and only one correct solution pair found (e.g. \(x = 5, y = -3\)): M0 M0 A0 M1 A0 M1 A0. No working, and both correct solution pairs found, but not demonstrated: M0 M0 A0 M1 A1 M1 A1. Both correct solution pairs found, and demonstrated: Full marks are possible (send to review)