Question 5 - A-Level Maths

Edexcel C1 2010 January — Question 5 7 marks

Exam Board	Edexcel
Module	C1 (Core Mathematics 1)
Year	2010
Session	January
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Simultaneous equations
Type	Line intersecting quadratic curve
Difficulty	Moderate -0.3 This is a standard C1 simultaneous equations question requiring substitution of the linear equation into the quadratic, forming a quadratic in one variable, then solving. It's slightly easier than average because the linear equation is already solved for y, and the resulting quadratic is straightforward to solve, but it still requires multiple steps and careful algebraic manipulation.
Spec	1.02c Simultaneous equations: two variables by elimination and substitution

5. Solve the simultaneous equations $$\begin{array} { r } y - 3 x + 2 = 0 \\ y ^ { 2 } - x - 6 x ^ { 2 } = 0 \end{array}$$

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Question 5:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$y = 3x - 2$, $(3x-2)^2 - x - 6x^2 (= 0)$	M1	Obtaining an equation in $x$ only (or $y$ only). Condone missing "= 0". Condone sign slips e.g. $(3x+2)^2 - x - 6x^2 = 0$, but not other algebraic mistakes
$9x^2 - 12x + 4 - x - 6x^2 = 0$, $3x^2 - 13x + 4 = 0$	M1 A1cso	Multiplying out $(3x-2)^2$, must lead to 3-term quadratic $ax^2+bx+c$ where $a\neq 0, b\neq 0, c\neq 0$, and collecting terms
$(3x-1)(x-4) = 0$, $x = \frac{1}{3}$ (or exact equivalent), $x = 4$	M1 A1	Solving 3-term quadratic; both $x$ values
$y = -1$, $y = 10$	M1 A1	Using $x$ value to find $y$ value; both $y$ values. Solutions need not be "paired"

Alternative method:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$x = \frac{y+2}{3}$, $y^2 - \frac{y+2}{3} - 6\left(\frac{y+2}{3}\right)^2 = 0$	M1
$y^2 - \frac{y+2}{3} - 6\left(\frac{y^2+4y+4}{9}\right) = 0$, $y^2 - 9y - 10 = 0$	M1 A1
$(y+1)(y-10) = 0$, $y = -1$, $y = 10$	M1 A1
$x = \frac{1}{3}$, $x = 4$	M1 A1

## Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 3x - 2$, $(3x-2)^2 - x - 6x^2 (= 0)$ | M1 | Obtaining an equation in $x$ only (or $y$ only). Condone missing "= 0". Condone sign slips e.g. $(3x+2)^2 - x - 6x^2 = 0$, but not other algebraic mistakes |
| $9x^2 - 12x + 4 - x - 6x^2 = 0$, $3x^2 - 13x + 4 = 0$ | M1 A1cso | Multiplying out $(3x-2)^2$, must lead to 3-term quadratic $ax^2+bx+c$ where $a\neq 0, b\neq 0, c\neq 0$, and collecting terms |
| $(3x-1)(x-4) = 0$, $x = \frac{1}{3}$ (or exact equivalent), $x = 4$ | M1 A1 | Solving 3-term quadratic; both $x$ values |
| $y = -1$, $y = 10$ | M1 A1 | Using $x$ value to find $y$ value; both $y$ values. Solutions need not be "paired" |

**Alternative method:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = \frac{y+2}{3}$, $y^2 - \frac{y+2}{3} - 6\left(\frac{y+2}{3}\right)^2 = 0$ | M1 | |
| $y^2 - \frac{y+2}{3} - 6\left(\frac{y^2+4y+4}{9}\right) = 0$, $y^2 - 9y - 10 = 0$ | M1 A1 | |
| $(y+1)(y-10) = 0$, $y = -1$, $y = 10$ | M1 A1 | |
| $x = \frac{1}{3}$, $x = 4$ | M1 A1 | |

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5. Solve the simultaneous equations

$$\begin{array} { r } 
y - 3 x + 2 = 0 \\
y ^ { 2 } - x - 6 x ^ { 2 } = 0
\end{array}$$

\hfill \mbox{\textit{Edexcel C1 2010 Q5 [7]}}

This paper (10 questions)

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Q1 3 Q2 6 Q3 5 Q4 7 Q5 7 Q6 8 Q7 9 Q8 7 Q9 13 Q10 10

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\(x = \frac{y+2}{3}\), \(y^2 - \frac{y+2}{3} - 6\left(\frac{y+2}{3}\right)^2 = 0\)	M1
\(y^2 - \frac{y+2}{3} - 6\left(\frac{y^2+4y+4}{9}\right) = 0\), \(y^2 - 9y - 10 = 0\)	M1 A1
\((y+1)(y-10) = 0\), \(y = -1\), \(y = 10\)	M1 A1
\(x = \frac{1}{3}\), \(x = 4\)	M1 A1