P2.4Simultaneous equations

Edexcel A level Maths (9MA0) · Pure mathematics › Algebra and functions

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Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy5 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Solve the simultaneous equations
\(y = 3x - 4\)
\(y = x^{2} - 2x\)[5]

Show the answer and mark scheme

Answer: \(x = 1, \ y = -1\) and \(x = 4, \ y = 8\)

M1 for substituting the linear equation into the quadratic: \(x^{2} - 2x = 3x - 4\)
A1 for a correct three-term quadratic: \(x^{2} - 5x + 4 = 0\)
dM1 for solving their quadratic
A1 for \(x = 1\) and \(x = 4\)
A1 for \(y = -1\) and \(y = 8\), correctly paired

Worked solution: \(x^{2} - 2x = 3x - 4\)
\(x^{2} - 5x + 4 = 0\)
\((x - 1)(x - 4) = 0\), so \(x = 1\) or \(x = 4\).
Using \(y = 3x - 4\): when \(x = 1\), \(y = -1\); when \(x = 4\), \(y = 8\).

Question 2Medium6 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Solve the simultaneous equations
\(x + y = 1\)
\(x^2 + 4xy + 4y^2 = 4\)[6]

Show the answer and mark scheme

Answer: \(x = 0, \ y = 1\) and \(x = 4, \ y = -3\)

M1 for substituting \(y = -x + 1\) into the second equation
M1 for expanding the brackets correctly (at least two of the three terms in \(x\) correct)
A1 for a correct quadratic, e.g. \(x^{2} - 4x = 0\)
dM1 for solving their quadratic
A1 for \(x = 0\) and \(x = 4\)
A1 for \(y = 1\) and \(y = -3\), correctly paired

Worked solution: Substitute \(y = -x + 1\):
\(x^2 + 4x(-x + 1) + 4(-x + 1)^2 = 4\)
\(x^{2} - 4x = 0\)
\(x(x - 4) = 0\), so \(x = 0\) or \(x = 4\).
Then \(y = 1\) and \(y = -3\) respectively.

Question 3Hard8 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
The circle \(C\) has equation \((x + 3)^2 + (y - 1)^2 = 7\).
The line \(l\) has equation \(y = 2x + k\), where \(k\) is a constant.

(a) Show that the \(x\)-coordinates of any points of intersection of \(l\) and \(C\) satisfy
\(5x^2 + (4k + 2)x + (k^2 - 2k + 3) = 0\)[3]

(b) Given that \(l\) is a tangent to \(C\), find the exact possible values of \(k\).[5]

Show the answer and mark scheme

(a) Answer: Substitute \(y = 2x + k\) into the equation of \(C\), expand and collect terms.

M1 for substituting \(y = 2x + k\) into the equation of \(C\)
M1 for expanding, e.g. \((2x + k - 1)^2 = 4x^2 + 4(k - 1)x + (k - 1)^2\)
A1* for collecting terms to obtain the given equation with no errors

Worked solution: Writing \(C\) as \((x + 3)^2 + (y - 1)^2 = 7\) and substituting \(y = 2x + k\):
\((x + 3)^2 + (2x + k - 1)^2 = 7\).
Expanding: \(x^{2} + 6x + 9 + 4x^2 + 4(k - 1)x + (k - 1)^2 = 7\).
Collecting terms gives \(5x^2 + (4k + 2)x + (k^2 - 2k + 3) = 0\).

(b) Answer: \(k = 7 \pm \sqrt{35}\)

M1 for using \(b^2 - 4ac = 0\) with their coefficients from part (a)
A1 for a correct equation in \(k\), e.g. \((4k + 2)^2 - 20(k^2 - 2k + 3) = 0\)
A1 for simplifying to \(k^{2} - 14k + 14 = 0\)
dM1 for solving their quadratic in \(k\)
A1 for \(k = 7 \pm \sqrt{35}\)

Worked solution: Tangent means the quadratic in part (a) has equal roots, so \(b^2 - 4ac = 0\).
\((4k + 2)^2 - 20(k^2 - 2k + 3) = 0\).
Expanding and dividing by \(-4\): \(k^{2} - 14k + 14 = 0\).
\(k = \frac{14 \pm \sqrt{140}}{2} = 7 \pm \sqrt{35}\).

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