Simultaneous equations and inequalities - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 17 exam-style questions on Simultaneous equations and inequalities for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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17 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

Explain why \(x^2 - 4x + 7 > 0\) for all real values of \(x\).

Hence find the set of values of \(x\) for which
\[(x^2 - 4x + 7)(2x - 5) \leqslant 0\]
Give your answer as an inequality in \(x\).

Model answer

\(x^2 - 4x + 7 = (x - 2)^2 + 3\)
\((x - 2)^2 \geqslant 0\) for all real \(x\), so \(x^2 - 4x + 7 \geqslant 3 > 0\) for all real \(x\).
So \((x^2 - 4x + 7)(2x - 5)\) has the same sign as \(2x - 5\), and is zero only when \(2x - 5 = 0\).
\((x^2 - 4x + 7)(2x - 5) \leqslant 0 \iff 2x - 5 \leqslant 0 \iff x \leqslant \dfrac{5}{2}\)

Explanation

Completing the square shows that \(x^2 - 4x + 7\) is at least 3 for every real \(x\), so it never changes the sign of the product. The inequality therefore reduces to the linear inequality \(2x - 5 \leqslant 0\), giving \(x \leqslant \dfrac{5}{2}\). Common errors are saying only that the discriminant is negative, without the positive \(x^2\) coefficient, and using a strict inequality.

Mark scheme

1 mark
M1 Completes the square, \(x^2 - 4x + 7 = (x - 2)^2 + 3\), or finds the discriminant \((-4)^2 - 4(1)(7) = -12\); a correct final answer of \(x \leqslant \dfrac{5}{2}\) with no working scores 1 mark (the B1)
An answer of \(x \leqslant \dfrac{5}{2}\) with no working scores 1 mark: the first two marks need the explanation. Substituting a few values of \(x\) scores M0 A0.
1 mark
A1 Explains that \((x - 2)^2 \geqslant 0\), so \(x^2 - 4x + 7 \geqslant 3 > 0\) for all real \(x\); or that the discriminant is negative and the coefficient of \(x^2\) is positive, so the parabola lies entirely above the \(x\)-axis
Both parts of the discriminant reason are needed: 'discriminant < 0' alone shows only that there are no real roots, and scores A0. '\((x - 2)^2\) is always positive' alone is not accepted (it can be 0), but \((x - 2)^2 + 3 \geqslant 3 > 0\) is fine.
1 mark
B1 \(x \leqslant \dfrac{5}{2}\)
Since the first factor is always positive, the product is less than or equal to 0 exactly when \(2x - 5 \leqslant 0\). Accept \(\{x : x \leqslant \dfrac{5}{2}\}\), \(x \leqslant 2.5\), typed x <= 5/2 and interval notation \(\left(-\infty, \dfrac{5}{2}\right]\). \(x < \dfrac{5}{2}\) scores B0, as the product is 0 when \(x = \dfrac{5}{2}\). Expanding to a cubic is not needed, but a correct final answer reached that way still scores this mark.

Question 2

4 marks

The simultaneous equations
\[ax + 4y = 6\]
\[3x + 2y = b\]
where \(a\) and \(b\) are constants, have infinitely many solutions.

Find the values of the constants \(a\) and \(b\).

Given instead that \(a\) takes this value but \(b\) does not, explain, with reference to the graphs of the two equations, why the equations have no solutions.

Model answer

Doubling the second equation gives \(6x + 4y = 2b\).
The equations have infinitely many solutions only if they are the same equation, so \(ax + 4y = 6\) must be \(6x + 4y = 2b\):
\(a = 6\) and \(2b = 6\), so \(b = 3\).
If \(a = 6\) but \(b ≠ 3\), the equations are \(6x + 4y = 6\) and \(6x + 4y = 2b\), i.e. \(y = \dfrac{3}{2} - \dfrac{3}{2}x\) and \(y = \dfrac{b}{2} - \dfrac{3}{2}x\). Their graphs are parallel lines with gradient \(-\dfrac{3}{2}\) and different \(y\)-intercepts, so they never meet: no pair \((x, y)\) satisfies both equations.

Explanation

Two linear equations have infinitely many solutions only when they describe the same straight line, so one must be a multiple of the other, which gives \(a = 6\) and \(b = 3\). With \(a = 6\) and \(b ≠ 3\) the lines have the same gradient but different intercepts, so they are parallel and distinct, and there is no solution. Common errors are finding only \(a\), and saying 'parallel' without noting that the lines are different.

Mark scheme

1 mark
M1 Uses the fact that one equation must be a multiple of the other, e.g. doubles the second equation, \(6x + 4y = 2b\), and compares it with \(ax + 4y = 6\); or eliminates \(y\) to obtain \((a - 6)x = 6 - 2b\) and requires both sides to be zero; or a correct final answer of \(a = 6\) and \(b = 3\) with no working
An answer of \(a = 6\) and \(b = 3\) with no working scores 3 marks; the last mark needs the explanation.
1 mark
A1 \(a = 6\)
1 mark
A1 \(b = 3\)
Depends on the M mark.
1 mark
B1 Explains that the graphs are parallel straight lines (both with gradient \(-\dfrac{3}{2}\)) that are not the same line (e.g. they have different \(y\)-intercepts, \(\dfrac{3}{2}\) and \(\dfrac{b}{2}\)), so they never meet and no point satisfies both equations
Both 'parallel' (or equal gradients) and 'not the same line' are needed. Accept a reason based on \(6x + 4y = 6\) and \(6x + 4y = 2b\) with \(2b ≠ 6\). 'The lines do not cross' with no reason scores B0.

Question 3

7 marks

The curve \(C\) has equation \(y = 8 + 2x - x^2\) and the line \(l\) has equation \(y = 2x - 1\).

Using algebra, find the \(x\)-coordinates of the points where \(l\) meets \(C\).

Hence find the set of values of \(x\) for which \(C\) lies above \(l\) and below the \(x\)-axis. Find also the set of values of \(x\) for which \(l\) lies above the \(x\)-axis and below \(C\).

Give each set in set notation.

Model answer

\(8 + 2x - x^2 = 2x - 1\)
\(x^2 = 9\), so \(l\) meets \(C\) where \(x = -3\) and \(x = 3\).
\(C\) is n-shaped, so \(C\) lies above \(l\) for \(-3 < x < 3\).
\(C\) meets the \(x\)-axis where \(x^2 - 2x - 8 = 0\), \((x + 2)(x - 4) = 0\), so \(C\) is below the \(x\)-axis for \(x < -2\) or \(x > 4\).
\(C\) above \(l\) and below the \(x\)-axis: \(\{x : -3 < x < -2\}\)
\(l\) meets the \(x\)-axis where \(x = \dfrac{1}{2}\), and \(l\) has positive gradient, so \(l\) is above the \(x\)-axis for \(x > \dfrac{1}{2}\).
\(l\) above the \(x\)-axis and below \(C\): \(\left\{x : \dfrac{1}{2} < x < 3\right\}\)

Explanation

Equating the expressions for \(y\) gives \(x^2 = 9\), so the line and the curve meet where \(x = \pm 3\), and the n-shaped curve is above the line between these values. Each set is the overlap of two conditions, so you also need where \(C\) and \(l\) cross the \(x\)-axis (\(x = -2\), 4 and \(\dfrac{1}{2}\)); a sketch makes the overlaps clear. Common errors are losing the root \(x = -3\) and combining the conditions with \(\cup\) instead of finding where both hold.

Mark scheme

1 mark
M1 Sets \(8 + 2x - x^2 = 2x - 1\) and solves algebraically, e.g. \(x^2 = 9\); correct answers with no working score 3 marks (see comments)
An answer of \(x = -3\), \(x = 3\), \(\{x : -3 < x < -2\}\) and \(\{x : \dfrac{1}{2} < x < 3\}\) with no working scores 3 marks as a special case: the two B marks and 1 mark for the two correct sets. The intersections must be found using algebra for the M1 A1, and the combining of the conditions must be seen for the other marks.
1 mark
A1 \(x = -3\) and \(x = 3\)
Both needed.
1 mark
B1 \(C\) meets the \(x\)-axis where \(x = -2\) and \(x = 4\)
From \(x^2 - 2x - 8 = (x + 2)(x - 4) = 0\). May be implied by correct use in the sets.
1 mark
B1 \(l\) meets the \(x\)-axis where \(x = \dfrac{1}{2}\)
May be implied by correct use in the second set.
1 mark
M1 (dM1) Combines the conditions for at least one set, e.g. \(C\) is above \(l\) for \(-3 < x < 3\) and \(C\) is below the \(x\)-axis for \(x < -2\) or \(x > 4\), and finds where both hold
Depends on the first M mark. The overlap may be found on a sketch or a number line.
1 mark
A1 \(\{x : -3 < x < -2\}\)
Accept \(\{x : x > -3\} \cap \{x : x < -2\}\) and typed forms such as {x : -3 < x < -2}. Strict inequalities and set notation are required; interval notation such as \((-3, -2)\) scores A0.
1 mark
A1 \(\left\{x : \dfrac{1}{2} < x < 3\right\}\)
Depends on the second M mark: \(l\) is above the \(x\)-axis for \(x > \dfrac{1}{2}\) and below \(C\) for \(-3 < x < 3\). Accept \(\{x : 0.5 < x < 3\}\) and \(\{x : x > \tfrac{1}{2}\} \cap \{x : x < 3\}\). Strict inequalities and set notation are required; interval notation such as \((\tfrac{1}{2}, 3)\) scores A0.

Question 4

9 marks

When two resistors with resistances \(a\) ohms and \(b\) ohms are connected in series, their combined resistance is \((a + b)\) ohms. When they are connected in parallel, their combined resistance, \(R\) ohms, is given by
\[R = \dfrac{ab}{a + b}\]
An electrician has two resistors whose combined resistance is 20 ohms in series and 4.2 ohms in parallel.

Using algebra, find the resistance of each of these two resistors.

A second pair of resistors has resistances \(x\) ohms and \((20 - x)\) ohms, where \(0 < x < 20\). Their combined resistance in parallel must be greater than 3.2 ohms. Find the set of possible values of \(x\), giving your answer as an inequality in \(x\).

Show that, for any two resistors whose combined resistance in series is 20 ohms, the combined resistance in parallel is at most 5 ohms.

Model answer

\(a + b = 20\) and \(\dfrac{ab}{a + b} = 4.2\), so \(ab = 84\).
\(a(20 - a) = 84\)
\(a^2 - 20a + 84 = 0\)
\((a - 6)(a - 14) = 0\), so \(a = 6\) or \(a = 14\), and then \(b = 14\) or \(b = 6\).
The resistors are 6 ohms and 14 ohms.
Second pair: \(\dfrac{x(20 - x)}{20} > 3.2\)
\(20x - x^2 > 64\)
\(x^2 - 20x + 64 < 0\)
\((x - 4)(x - 16) < 0\), so \(4 < x < 16\)
For any pair with series resistance 20 ohms, the resistances are \(x\) and \(20 - x\), so \(R = \dfrac{x(20 - x)}{20} = \dfrac{20x - x^2}{20}\).
\(20x - x^2 = 100 - (x - 10)^2 \leqslant 100\), since \((x - 10)^2 \geqslant 0\)
So \(R \leqslant \dfrac{100}{20} = 5\): the combined resistance in parallel is at most 5 ohms (with equality when both resistors are 10 ohms).

Explanation

The two given resistances give \(a + b = 20\) and \(ab = 84\), and substituting \(b = 20 - a\) gives a quadratic whose roots, 6 and 14, are the two resistances. For the second pair the parallel resistance is \(\dfrac{x(20 - x)}{20}\), and both the inequality and the greatest value come from this quadratic: completing the square shows its maximum is 5, when \(x = 10\). Common errors are using \(ab = 4.2\) instead of \(\dfrac{ab}{20} = 4.2\), choosing the outside region and showing only one example in the last part.

Mark scheme

1 mark
M1 Uses \(\dfrac{ab}{20} = 4.2\) to obtain \(ab = 84\) and substitutes \(b = 20 - a\) (or equivalent) to form a quadratic in one variable; an answer with no working scores at most 2 marks (see comments)
An answer of 6 ohms and 14 ohms with no working scores 0 marks for the first part, as the question says 'using algebra'. A correct set \(4 < x < 16\) with no working scores 2 marks as a special case (the two A marks of the second part); the last part needs a complete argument.
1 mark
A1 \(a^2 - 20a + 84 = 0\) oe
Accept \(a(20 - a) = 84\) rearranged to a three-term quadratic.
1 mark
M1 (dM1) Solves their quadratic algebraically, e.g. \((a - 6)(a - 14) = 0\)
Depends on the first M mark.
1 mark
A1 The resistances are 6 ohms and 14 ohms
Both needed. Units are not required.
1 mark
M1 Forms \(\dfrac{x(20 - x)}{20} > 3.2\) (or = 3.2) and rearranges to a three-term quadratic, e.g. \(x^2 - 20x + 64 < 0\)
Not dependent on the earlier marks.
1 mark
A1 Critical values \(x = 4\) and \(x = 16\)
1 mark
A1 \(4 < x < 16\)
Depends on the previous M mark. Accept \(\{x : 4 < x < 16\}\), typed 4 < x < 16 and interval notation \((4, 16)\). Non-strict inequalities score A0 (at \(x = 4\) the resistance is exactly 3.2 ohms), as does the outside region.
1 mark
M1 Writes the parallel resistance as \(R = \dfrac{x(20 - x)}{20}\) and finds its greatest value: completes the square, \(20x - x^2 = 100 - (x - 10)^2\), or uses the vertex of the parabola at \(x = 10\), or forms \(x^2 - 20x + 20R = 0\) and uses \(b^2 - 4ac \geqslant 0\)
Alternative: differentiates \(\dfrac{20x - x^2}{20}\) and solves \(\dfrac{dR}{dx} = 0\) to get \(x = 10\). Testing particular values of \(x\) scores M0.
1 mark
A1* Shows that \(R = \dfrac{100 - (x - 10)^2}{20} \leqslant \dfrac{100}{20} = 5\) (or \(400 - 80R \geqslant 0\), so \(R \leqslant 5\)), with a conclusion
cso. A complete argument needs the greatest value 5, an explanation of why \(R\) cannot exceed it (e.g. \((x - 10)^2 \geqslant 0\), or real resistances need \(b^2 - 4ac \geqslant 0\)) and a conclusion. Showing only that \(R = 5\) when \(x = 10\) scores A0.

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