Question 3 - A-Level Maths

Edexcel C1 2014 January — Question 3 7 marks

Exam Board	Edexcel
Module	C1 (Core Mathematics 1)
Year	2014
Session	January
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Simultaneous equations
Type	Line intersecting general conic
Difficulty	Moderate -0.3 This is a standard C1 simultaneous equations question involving a line and conic. Students substitute x = 2y + 1 into the quadratic, expand to get a quadratic in y, then solve. While it requires careful algebra across multiple steps, it follows a completely routine method with no conceptual challenges, making it slightly easier than average.
Spec	1.02c Simultaneous equations: two variables by elimination and substitution

3. Solve the simultaneous equations $$\begin{gathered} x - 2 y - 1 = 0 \\ x ^ { 2 } + 4 y ^ { 2 } - 10 x + 9 = 0 \end{gathered}$$

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Answer	Marks	Guidance
$x = 2y + 1$	M1	Rearrange $x - 2y - 1 = 0$ into $x = ...$, or $y = ...$, or $2y = ...$and attempt to fully substitute into 2nd equation. It does not need to be correct but a clear attempt must be made. Condone missing brackets $(2y+1)^2 + 4y^2 - 10 \times 2y + 1 + 9 = 0$.
$8y^2 - 16y = 0$	M1, A1	Collect like terms to produce a quadratic equation in $x$ (or $y$) $= 0$. Either $A(y^2 - 2y) = 0$ or $B(x^2 - 6x + 5) = 0$. Correct quadratic equation in $x$ (or $y$) = 0.
$8y(y-2) = 0$ Alt $y(8y-16) = 0$	M1	Attempt to solve, with usual rules. Check the first and last terms only for factorisation. See appendix for completing the square and use of formula. Condone a solution from cancelling in a case like $A(y^2 - 2y) = 0$. They must proceed to find at least one solution $x = ..$ or $y = ...$.
$y = 0, y = 2$
$y = 0$ in $x = 2y + 1 \Rightarrow x = 1$	M1	Substitute at least one value of their $x$ to find $y$ or vice versa. This may be implied by their solution—you will need to check!
$y = 2$ in $x = 2y + 1 \Rightarrow x = 5$
$x=1, y=0$ and $x=5, y=2$	A1, A1	Both $x$'s correct or both $y$'s correct or a correct matching pair. Accept as a coordinate. Do not accept correct answers that are obtained from incorrect equations. Both "pairs" correct. Accept as coordinates $(1,0)$ $(5,2)$.
Special Cases where candidates write down answers with little or no working as can be awarded above. One correct solution – B2. Two correct solutions – B2, B2. To score all 7 marks candidates must prove that there are only two solutions. This could be justified by a sketch.

| $x = 2y + 1$ | M1 | Rearrange $x - 2y - 1 = 0$ into $x = ...$, or $y = ...$, or $2y = ...$and attempt to fully substitute into 2nd equation. It does not need to be correct but a clear attempt must be made. Condone missing brackets $(2y+1)^2 + 4y^2 - 10 \times 2y + 1 + 9 = 0$.

| $8y^2 - 16y = 0$ | M1, A1 | Collect like terms to produce a quadratic equation in $x$ (or $y$) $= 0$. Either $A(y^2 - 2y) = 0$ or $B(x^2 - 6x + 5) = 0$. Correct quadratic equation in $x$ (or $y$) = 0.

| $8y(y-2) = 0$ Alt $y(8y-16) = 0$ | M1 | Attempt to solve, with usual rules. Check the first and last terms only for factorisation. See appendix for completing the square and use of formula. Condone a solution from cancelling in a case like $A(y^2 - 2y) = 0$. They must proceed to find at least one solution $x = ..$ or $y = ...$.

| $y = 0, y = 2$ | | |

| $y = 0$ in $x = 2y + 1 \Rightarrow x = 1$ | M1 | Substitute at least one value of their $x$ to find $y$ or vice versa. This may be implied by their solution—you will need to check!

| $y = 2$ in $x = 2y + 1 \Rightarrow x = 5$ | | |

| $x=1, y=0$ and $x=5, y=2$ | A1, A1 | Both $x$'s correct or both $y$'s correct or a correct matching pair. Accept as a coordinate. Do not accept correct answers that are obtained from incorrect equations. Both "pairs" correct. Accept as coordinates $(1,0)$ $(5,2)$.

| | | Special Cases where candidates write down answers with little or no working as can be awarded above. One correct solution – B2. Two correct solutions – B2, B2. To score all 7 marks candidates must prove that there are only two solutions. This could be justified by a sketch.

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3. Solve the simultaneous equations

$$\begin{gathered}
x - 2 y - 1 = 0 \\
x ^ { 2 } + 4 y ^ { 2 } - 10 x + 9 = 0
\end{gathered}$$

\hfill \mbox{\textit{Edexcel C1 2014 Q3 [7]}}

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Q1 4 Q2 5 Q3 7 Q4 4 Q5 5 Q6 11 Q7 10 Q8 7 Q9 12 Q10 10

Answer	Marks	Guidance
\(x = 2y + 1\)	M1	Rearrange \(x - 2y - 1 = 0\) into \(x = ...\), or \(y = ...\), or \(2y = ...\)and attempt to fully substitute into 2nd equation. It does not need to be correct but a clear attempt must be made. Condone missing brackets \((2y+1)^2 + 4y^2 - 10 \times 2y + 1 + 9 = 0\).
\(8y^2 - 16y = 0\)	M1, A1	Collect like terms to produce a quadratic equation in \(x\) (or \(y\)) \(= 0\). Either \(A(y^2 - 2y) = 0\) or \(B(x^2 - 6x + 5) = 0\). Correct quadratic equation in \(x\) (or \(y\)) = 0.
\(8y(y-2) = 0\) Alt \(y(8y-16) = 0\)	M1	Attempt to solve, with usual rules. Check the first and last terms only for factorisation. See appendix for completing the square and use of formula. Condone a solution from cancelling in a case like \(A(y^2 - 2y) = 0\). They must proceed to find at least one solution \(x = ..\) or \(y = ...\).
\(y = 0, y = 2\)
\(y = 0\) in \(x = 2y + 1 \Rightarrow x = 1\)	M1	Substitute at least one value of their \(x\) to find \(y\) or vice versa. This may be implied by their solution—you will need to check!
\(y = 2\) in \(x = 2y + 1 \Rightarrow x = 5\)
\(x=1, y=0\) and \(x=5, y=2\)	A1, A1	Both \(x\)'s correct or both \(y\)'s correct or a correct matching pair. Accept as a coordinate. Do not accept correct answers that are obtained from incorrect equations. Both "pairs" correct. Accept as coordinates \((1,0)\) \((5,2)\).
Special Cases where candidates write down answers with little or no working as can be awarded above. One correct solution – B2. Two correct solutions – B2, B2. To score all 7 marks candidates must prove that there are only two solutions. This could be justified by a sketch.