Edexcel International A Level Maths · P1 · Pure Mathematics 1
Simultaneous equations and inequalities
28 free, high-quality past-paper-style questions on Simultaneous equations and inequalities for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 236 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P1-2-B-01P1 · ASEasy6 marks
Two equations connect the real numbers x and y: y − 3x = 1 2x2 + y2 = 18
(a)
Show that substituting the first equation into the second gives 11x2 + 6x − 17 = 0.
(2)
(b)
Hence solve the simultaneous equations, giving any non-integer values as exact fractions.
A parabola has equation y = x2 − 6x + 10. A straight line with equation y = x − 2 cuts the parabola at the points P and Q, where P is closer to the y-axis.
Figure: the parabola and the line meeting at P and Q
(a)
Find the coordinates of P and of Q.
(4)
(b)
Hence find the set of values of x for which x2 − 6x + 10 < x − 2.
(2)
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Answer
Mark
Notes
Part (a)
x2 − 6x + 10 = x − 2
M1
x2 − 7x + 12 = 0 ⇒ (x − 3)(x − 4) = 0 ⇒ x = 3, 4
A1
substitutes into y = x − 2
M1
P(3, 1), Q(4, 2)
A1
Part (b)
uses the x-coordinates of P and Q as critical values: the parabola is below the line between them
M1
3 < x < 4
A1
Exam tip: “Hence” means use part (a): the curve lies below the line exactly between the two intersection points.
Figure 1 shows the graph of y = f(x), where f(x) = a + bx − x2 and a and b are constants. The curve passes through the three labelled points. The horizontal line y = 5 crosses the curve at the points P and Q.
(a)
Using Figure 1, write down the set of values of x for which f(x) > 0.
(1)
(b)
Find the value of a and the value of b.
(2)
(c)
Use algebra to find the set of values of x for which f(x) ≥ 5. Give your answer in set notation.
(3)
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Answer
Mark
Notes
Part (a)
the curve is above the x-axis between (−2, 0) and (4, 0): −2 < x < 4
B1
Part (b)
(0, 8) gives a = 8
B1
substitutes (4, 0) or (−2, 0): 8 + 4b − 16 = 0, so b = 2
B1
Part (c)
8 + 2x − x2 ≥ 5 ⇒ x2 − 2x − 3 ≤ 0
M1
(x − 3)(x + 1) = 0: critical values −1 and 3
A1
inside region, end values included: {x : −1 ≤ x ≤ 3}
A1
Exam tip: P and Q are where f(x) = 5, so their x-coordinates are the critical values. The curve is on or above the line between them, so the answer is the inside region with the end values included.
Figure 1 shows two number lines. Number line A shows the solution set of the inequality 3(x − k) < x + 7, where k is a constant. Number line B shows the solution set of a second inequality in x.
(a)
Find the value of k.
(3)
(b)
Show that the inequality x/2 ≤ (2x + 1)/3 has the solution set shown on number line B.
(2)
(c)
Write down, in set notation, the set of values of x that satisfy both inequalities.
(1)
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Answer
Mark
Notes
Part (a)
expands: 3x − 3k < x + 7
M1
x < (7 + 3k)/2
A1
line A shows x < 5, so (7 + 3k)/2 = 5 ⇒ k = 1
A1
Part (b)
multiplies both sides by 6: 3x ≤ 4x + 2
M1
−2 ≤ x, i.e. x ≥ −2: a filled circle at −2 with the line to the right, as on B
A1
Part (c)
{x : −2 ≤ x < 5}
B1
Exam tip: On a number line an open circle means the end value is not included (< or >) and a filled circle means it is included (≤ or ≥). Carry the same choice into your set notation.
A farmer fences a rectangular paddock and divides it into two pens with one straight fence, as shown in Figure 1. The paddock measures x m by y m, and the dividing fence is parallel to the sides of length y m. Exactly 54 m of fencing is used in total, including the dividing fence, and the area of the paddock is 120 m2.
(a)
Use Figure 1 to explain why 2x + 3y = 54.
(1)
(b)
Show that y2 − 18y + 80 = 0.
(2)
(c)
Hence find the two possible pairs of dimensions of the paddock.
(3)
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Answer
Mark
Notes
Part (a)
two sides of length x, and three lengths y (two sides and the dividing fence), give 2x + 3y = 54 (given answer)
Exam tip: Make x the subject of the linear equation and substitute into xy = 120. Check both pairs in the original conditions: 2(15) + 3(8) = 54 with 15 × 8 = 120, and 2(12) + 3(10) = 54 with 12 × 10 = 120.
Figure 1 shows the lines l1 and l2 on a grid of unit squares. The line l1 is drawn solid and l2 is drawn dashed. The shaded region R is bounded by l1, l2 and the y-axis. Points on l1 and on the y-axis are included in R; points on l2 are not.
(a)
Find an equation of l1 and an equation of l2.
(2)
(b)
Write down three inequalities that together define R.
(3)
(c)
Determine whether the point (2, 5) lies in R. Give a reason for your answer.
(1)
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Answer
Mark
Notes
Part (a)
l1: gradient 1, y-intercept 1: y = x + 1
B1
l2: through (0, 7) and (7, 0): x + y = 7
B1
Part (b)
y ≥ x + 1
B1
x + y < 7
B1
x ≥ 0
B1
Part (c)
2 + 5 = 7, so (2, 5) lies on l2, which is excluded: (2, 5) is not in R
B1
Exam tip: Test a point that is clearly inside the region, such as (1, 4), in every inequality before you finish. It must satisfy all of them; if it fails one, that sign is the wrong way round.
The equation x2 + 2px + (3p + 4) = 0, where p is a constant, has two distinct real roots.
(a)
Show that p2 − 3p − 4 > 0.
(2)
(b)
Find the set of possible values of p, giving your answer in set notation.
(4)
(c)
Suppose instead that p is equal to one of the critical values found in (b), so that the equation has a repeated root. Find the repeated root for each of these two values of p.
(2)
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Answer
Mark
Notes
Part (a)
uses b2 − 4ac > 0: (2p)2 − 4(3p + 4) > 0
M1
4p2 − 12p − 16 > 0, divides by 4 (given answer)
A1
Part (b)
solves p2 − 3p − 4 = 0: (p − 4)(p + 1) = 0
M1
critical values −1 and 4
A1
chooses the outside region
M1
{p : p < −1} ∪ {p : p > 4}
A1
Part (c)
repeated root is −b/2a = −p, or substitutes and factorises
M1
p = 4: x = −4; p = −1: x = 1
A1
Common error: Writing −1 < p < 4 because that looks “neater”. For “> 0” on a ∪-shaped graph the answer is the two outside pieces.
A community garden plans a rectangular vegetable bed. Its width is x metres and its length is 4 metres more than its width. The edging available limits the perimeter to at most 40 m, and the bed must have an area of at least 32 m2.
Figure: the vegetable bed (not to scale)
(a)
Write down an inequality in x for the perimeter condition and solve it.
(2)
(b)
Show that the area condition gives x2 + 4x − 32 ≥ 0.
(1)
(c)
Solve x2 + 4x − 32 ≥ 0 and explain which part of the solution applies to the bed.
(3)
(d)
State the set of possible widths, and find the largest possible area of the bed.
(2)
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Answer
Mark
Notes
Part (a)
2x + 2(x + 4) ≤ 40
M1
x ≤ 8
A1
Part (b)
x(x + 4) ≥ 32, so x2 + 4x − 32 ≥ 0 (given answer)
B1
Part (c)
(x + 8)(x − 4) = 0, critical values −8 and 4
M1
x ≤ −8 or x ≥ 4
A1
a width must be positive, so x ≥ 4
A1
Part (d)
4 ≤ x ≤ 8 (ft their (a) and (c))
A1
largest area when x = 8: 8 × 12 = 96 m2
B1
Exam tip: In a context, check each piece of a quadratic inequality's solution against reality: a negative width is a correct solution of the inequality but not of the problem.
On one set of axes a student draws the curve y = 9 − x2 and the line y = x + 3. The line crosses the curve at the points A and B, where A has the smaller x-coordinate. The region R consists of the points that lie above the line and on or below the curve. Points on the line are not part of R.
Figure: the curve, the line and the region R
(a)
Find the coordinates of A and B.
(4)
(b)
Write down two inequalities that together define R.
(2)
(c)
Determine whether the point (1, 5) lies in R. Show your reasoning.
(1)
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Answer
Mark
Notes
Part (a)
9 − x2 = x + 3
M1
x2 + x − 6 = 0 ⇒ (x + 3)(x − 2) = 0 ⇒ x = −3, 2
A1
substitutes into either equation for y
M1
A(−3, 0), B(2, 5)
A1
Part (b)
y > x + 3 (strict)
B1
y ≤ 9 − x2 (non-strict)
B1
Part (c)
at x = 1: line gives 4 and curve gives 8, and 4 < 5 ≤ 8, so (1, 5) lies in R
B1
Exam tip: Match each sign to how the boundary is described: a boundary that is excluded (drawn dashed) needs < or >; one that is included (drawn solid) needs ≤ or ≥.
Figure 1 shows the curve C with equation y = x2 − 2x + 3 and the point P on the y-axis. A straight line through P has gradient m. For two values of m the line touches C: at the point A and at the point B, as shown.
(a)
Show that the x-coordinates of any points where a line through P with gradient m meets C satisfy x2 − (m + 2)x + 4 = 0.
(2)
(b)
Find the two values of m for which the line touches C.
(3)
(c)
Find the coordinates of A and of B.
(2)
(d)
Find the set of values of m for which a line through P with gradient m meets C at two distinct points.
(2)
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Answer
Mark
Notes
Part (a)
P is (0, −1), so the line is y = mx − 1, and x2 − 2x + 3 = mx − 1
A ball is thrown upwards from a balcony. Its height above the ground, h metres, t seconds after it is thrown is modelled by h = p + qt − 5t2, where p and q are constants. Figure 1 shows the graph of h against t; the curve passes through the two labelled points.
(a)
Use the labelled points to find the value of p and the value of q.
(3)
(b)
Find the set of values of t for which the ball is more than 12 m above the ground.
(3)
(c)
Hence state how long the ball spends more than 12 m above the ground.
(1)
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Answer
Mark
Notes
Part (a)
(1, 12): p + q − 5 = 12, so p + q = 17
M1
(3, 2): p + 3q − 45 = 2, so p + 3q = 47
M1
subtracts: 2q = 30, q = 15, p = 2
A1
Part (b)
2 + 15t − 5t2 > 12 ⇒ t2 − 3t + 2 < 0
M1
(t − 1)(t − 2) = 0: critical values 1 and 2
A1
1 < t < 2
A1
Part (c)
2 − 1 = 1 second (ft their (b))
B1
Exam tip: One critical value, t = 1, is a labelled point on the graph, so the quadratic in (b) must have (t − 1) as a factor. Use that as a check on your rearrangement.
Figure 1 shows part of the curve with equation xy = 6, for x > 0, and the straight line ℓ, which passes through the two labelled points. The line ℓ meets the curve at the points A and B, where A is nearer the y-axis.
(a)
Find an equation of ℓ.
(1)
(b)
Find the coordinates of A and of B.
(4)
(c)
Using your answer to (b) and Figure 1, find the set of values of x, where x > 0, for which 6/x > 8 − 2x.
(2)
(d)
The line 2x + y = k, where k is a positive constant, is a tangent to the curve xy = 6. Find the exact value of k.
(2)
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Answer
Mark
Notes
Part (a)
gradient (0 − 4)/(4 − 2) = −2, so y = −2(x − 4): y = 8 − 2x (or 2x + y = 8)
B1
Part (b)
substitutes: x(8 − 2x) = 6
M1
2x2 − 8x + 6 = 0 ⇒ x2 − 4x + 3 = 0
A1
(x − 1)(x − 3) = 0 ⇒ x = 1 or 3
M1
A(1, 6), B(3, 2)
A1
Part (c)
the curve is above ℓ to the left of A and to the right of B
Common error: Multiplying 6/x > 8 − 2x through by x keeps the sign only because x > 0 here. For negative x the inequality would reverse, which is why the question restricts x.
Figure 1 shows the curve with equation y = 8 − 2x − x2 and the straight line ℓ, which passes through the labelled points (−4, 0) and (0, 4). The line ℓ meets the curve at (−4, 0) and at the point B. The shaded region R lies on or above the x-axis and below both ℓ and the curve. Points on ℓ and on the x-axis are included in R; points on the curve, which is drawn dashed, are not.
(a)
Find an equation of ℓ.
(1)
(b)
Find the coordinates of B.
(3)
(c)
Write down three inequalities that together define R.
(3)
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Answer
Mark
Notes
Part (a)
gradient 4/4 = 1, y-intercept 4: y = x + 4
B1
Part (b)
8 − 2x − x2 = x + 4
M1
x2 + 3x − 4 = 0 ⇒ (x + 4)(x − 1) = 0
M1
x = −4 is the labelled point, so x = 1: B(1, 5)
A1
Part (c)
y ≥ 0
B1
y ≤ x + 4
B1
y < 8 − 2x − x2 (strict, as the curve is excluded)
B1
Exam tip: A region can lie below two boundaries at once. Each of y ≤ x + 4 and y < 8 − 2x − x2 forms only part of the edge of R, but every point of R satisfies both.
A small firm makes and sells x thousand units of a product each month. Its monthly revenue, £R thousand, is modelled by R = 40x − x2, and its monthly cost, £C thousand, is modelled by a linear function of x. Figure 1 shows the graphs of R and C against x; the cost line passes through the two labelled points. The graphs cross at E and F.
(a)
Find an expression for C in terms of x.
(2)
(b)
Find the value of x at E and at F.
(3)
(c)
The firm makes a profit when R > C. Find the set of values of x for which the firm makes a profit.
(2)
(d)
The monthly cost then rises by the same amount, £d thousand, for every value of x, so that the new cost line is parallel to the old one. Find the value of d for which the new cost line just touches the revenue curve.
(2)
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Answer
Mark
Notes
Part (a)
gradient (425 − 225)/(30 − 10) = 10
M1
C − 225 = 10(x − 10): C = 125 + 10x
A1
Part (b)
40x − x2 = 125 + 10x
M1
x2 − 30x + 125 = 0 ⇒ (x − 5)(x − 25) = 0
M1
x = 5 at E and x = 25 at F
A1
Part (c)
the revenue curve is above the cost line between E and F
Exam tip: Find the cost line from the two labelled points rather than by reading the grid. The gradient and intercept are then exact, so the break-even values come out exactly too.
Figure 1 shows a triangle ABC in which angle ABC is a right angle, AB = x cm and BC = (x + 3) cm. The area of the triangle is less than 20 cm2 and the length of AC is at least 3√5 cm.
(a)
Show that the area condition gives x2 + 3x − 40 < 0, and hence find the set of values of x allowed by this condition.
(3)
(b)
Use the condition on AC to find the set of values of x allowed by this condition.
(3)
(c)
Find the set of possible values of x, giving your answer in set notation.
Common error: Keeping x ≤ −6 in the answer to (b). It is correct algebra for the quadratic inequality, but a length cannot be negative, so that piece is rejected before combining.
Two inequalities in x are studied in this question, first separately and then on a graph. Show your working in every part.
(a)
Solve 3(1 − x) < 2x + 13.
(2)
(b)
Solve x2 ≤ 2x + 15.
(4)
(c)
Find the set of values of x for which both inequalities in (a) and (b) are true. Give your answer in set notation.
(2)
(d)
The curve y = x2 and the line y = 2x + 15 meet at two points. Using your working in (b), write down their coordinates.
(2)
(e)
Hence find the set of values of x for which the curve y = x2 is above the line y = 2x + 15 and 3(1 − x) < 2x + 13.
(2)
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Answer
Mark
Notes
Part (a)
3 − 3x < 2x + 13 ⇒ −10 < 5x
M1
x > −2
A1
Part (b)
x2 − 2x − 15 ≤ 0
M1
(x − 5)(x + 3) = 0 ⇒ critical values −3 and 5
A1
chooses the inside region
M1
−3 ≤ x ≤ 5
A1
Part (c)
−2 < x ≤ 5 (ft)
A1
{x : −2 < x ≤ 5} or {x : x > −2} ∩ {x : −3 ≤ x ≤ 5}
B1
Part (d)
(−3, 9)
B1
(5, 25)
B1
Part (e)
curve above the line for x < −3 or x > 5, combined with x > −2
M1
x > 5
A1
Model sketch
Figure: model sketch: the curve and the line meeting at two points
Exam tip: Draw both solution sets on one number line before combining them; the overlap, and which end values are included, can then be read straight off.
The curve C has equation y = (x − 1)(5 − x). The line L has equation 2y + x = 5.
Figure: the curve C, the line L and the region R
(a)
Write down the coordinates of the points where C crosses the x-axis.
(1)
(b)
Find the coordinates of the points where L meets C.
(5)
(c)
The finite region R contains the points that lie above L and on or below C. Points on L are not included in R. Write down inequalities that define R.
(3)
(d)
The line with equation y = k − x/2, where k is a constant, is parallel to L. Find the set of values of k for which this line meets C at two distinct points.
Figure 1 shows the curve C with equation y = x2 + bx + c, where b and c are constants, and the straight line L. C and L each pass through two of the labelled points. L crosses C at the points P and Q, where P is to the left of the y-axis.
(a)
Find the value of b and the value of c.
(2)
(b)
Find an equation of L.
(1)
(c)
Find the exact coordinates of P and of Q.
(4)
(d)
Write down the set of values of x for which C lies below L.
(1)
(e)
A line parallel to L is a tangent to C. Find an equation of this line and the coordinates of the point where it touches C.
(3)
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Answer
Mark
Notes
Part (a)
roots −1 and 3: y = (x + 1)(x − 3)
M1
= x2 − 2x − 3: b = −2, c = −3
A1
Part (b)
gradient (5 − 1)/2 = 2: y = 2x + 1
B1
Part (c)
x2 − 2x − 3 = 2x + 1 ⇒ x2 − 4x − 4 = 0
M1
x = (4 ± √32)/2 = 2 ± 2√2
A1
substitutes into y = 2x + 1
M1
P(2 − 2√2, 5 − 4√2), Q(2 + 2√2, 5 + 4√2)
A1
Part (d)
2 − 2√2 < x < 2 + 2√2
B1
Part (e)
y = 2x + k: x2 − 4x − 3 − k = 0 with 16 + 4(3 + k) = 0
M1
k = −7: y = 2x − 7
A1
repeated root x = 2: (2, −3)
A1
Exam tip: A line with the same gradient as L can be written y = 2x + k. Only one constant is unknown, so the single condition b2 − 4ac = 0 fixes it.
Figure 1 shows, on a grid of unit squares, the curve with equation y = x(x − a), where a is a positive constant, and a straight line l, drawn dashed, which passes through the two labelled points (0, −4) and (2, 2). The line crosses the curve at P and Q, where P has the smaller x-coordinate. The shaded region R is bounded by the x-axis, the curve and l. Points on the x-axis and on the curve are included in R; points on l are not.
(a)
Write down the value of a.
(1)
(b)
Find an equation of l.
(1)
(c)
Find the coordinates of P and of Q.
(4)
(d)
Write down three inequalities that together define R.
(3)
(e)
Find all the points with integer coordinates that lie in R.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
the curve crosses the x-axis at (2, 0): a = 2
B1
Part (b)
gradient (2 − (−4))/2 = 3: y = 3x − 4
B1
Part (c)
x2 − 2x = 3x − 4
M1
x2 − 5x + 4 = 0 ⇒ (x − 1)(x − 4) = 0
M1
x = 1 or x = 4
A1
P(1, −1), Q(4, 8)
A1
Part (d)
y ≥ 0
B1
y ≥ x2 − 2x
B1
y < 3x − 4
B1
Part (e)
R lies between x = 4/3 and x = 4, and x = 4 gives 8 ≤ y < 8, so only x = 2 and x = 3 are possible
M1
x = 2: 0 ≤ y < 2 gives (2, 0), (2, 1)
A1
x = 3: 3 ≤ y < 5 gives (3, 3), (3, 4)
A1
Common error: Including (2, 2) or (3, 5). These points lie on the dashed line, which is not part of R, whereas points on the solid curve and on the x-axis, such as (3, 3) and (2, 0), are included.
Figure 1 shows the cross-section of a tunnel. The arch is modelled by a parabola through the labelled points A(−5, 0), (5, 0) and (0, 5), where x and y are measured in metres and the x-axis is the level ground. A straight cable runs from A to a point B on the arch.
(a)
Show that the arch has equation y = 5 − x2/5.
(2)
(b)
A vehicle with a flat roof 3.2 m above the ground drives along the centre of the tunnel; the dashed line in Figure 1 shows the level of its roof. Find the set of values of x for which the arch is higher than 3.2 m, and hence state the condition on the width, w metres, of the roof if it is to pass without touching the arch.
(3)
(c)
The cable lies along the line y = m(x + 5), where m is a positive constant. Show that the x-coordinates of the points where this line meets the arch satisfy x2 + 5mx + 25m − 25 = 0.
(2)
(d)
Use the fact that x = −5 is one solution to find the x-coordinate of B in terms of m. Given that B has x-coordinate 2, find the value of m.
(2)
(e)
A straight support beam lies along the line y = c − x/2, where c is a constant, and touches the arch at exactly one point. Find the value of c.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
roots ±5: y = λ(25 − x2)
M1
(0, 5) gives λ = 1/5, so y = 5 − x2/5 (given answer)
A1
Part (b)
5 − x2/5 > 3.2 ⇒ x2 < 9
M1
−3 < x < 3
A1
w < 6
B1
Part (c)
5 − x2/5 = m(x + 5)
M1
multiplies by 5 and rearranges: x2 + 5mx + 25m − 25 = 0 (given answer)
A1
Part (d)
(x + 5)(x + 5m − 5) = 0, so x = 5 − 5m
M1
5 − 5m = 2 ⇒ m = 3/5
A1
Part (e)
5 − x2/5 = c − x/2 ⇒ 2x2 − 5x + 10(c − 5) = 0
M1
b2 − 4ac = 25 − 80(c − 5) = 0
M1
c = 85/16
A1
Exam tip: When one intersection is already known, its x-coordinate gives a factor of the quadratic, here (x + 5). Factorising is quicker than the formula and keeps the other root exact.
The line l has equation y = kx − 2, where k is a constant, and the curve C has equation y = x2 + x + 2. The x-coordinates of any points where l meets C satisfy a quadratic equation whose discriminant is D. Figure 1 shows the graph of D against k; it crosses the vertical axis at the labelled point and the k-axis at P and Q.
(a)
Show that D = k2 − 2k − 15.
(2)
(b)
Find the values of k at P and Q. Hence find, in set notation, the set of values of k for which l meets C at two distinct points.
(4)
(c)
For the positive value of k for which l is a tangent to C, find the coordinates of the point of contact.
(3)
(d)
Solve the inequality (k + 1)/2 − (2k − 3)/5 > 1.
(2)
(e)
Find the set of values of k that satisfy the inequality in (d) and for which l does not meet C.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
x2 + x + 2 = kx − 2 ⇒ x2 + (1 − k)x + 4 = 0
M1
D = (1 − k)2 − 16 = k2 − 2k − 15 (given answer)
A1
Part (b)
(k − 5)(k + 3) = 0
M1
P: k = −3, Q: k = 5
A1
D > 0 where the graph is above the k-axis, outside PQ
M1
{k : k < −3} ∪ {k : k > 5}
A1
Part (c)
k = 5: x2 − 4x + 4 = 0
M1
(x − 2)2 = 0, x = 2
A1
y = 5 × 2 − 2: (2, 8)
A1
Part (d)
multiplies every term by 10: 5(k + 1) − 2(2k − 3) > 10
M1
5k + 5 − 4k + 6 > 10 ⇒ k > −1
A1
Part (e)
l misses C when D < 0: −3 < k < 5
M1
combined with k > −1: −1 < k < 5
A1
Exam tip: The graph of D turns the line–curve question into a quadratic inequality in k: above the k-axis means two points, on it means a tangent, and below it means no points.
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