What is 2.4: Simultaneous equations?

A simultaneous solution is an ordered pair satisfying every equation; graphically, solutions are intersection points. Use elimination when coefficients align conveniently, or substitute an expression from the linear equation into the nonlinear equation.

How many marks are 2.4 questions worth?

2.4 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 2.4 appeared in?

We have not yet indexed a verified past-paper appearance for 2.4 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

2.4

Solve simultaneous equations in two variables by elimination and by substitution, including one linear and one quadratic equation.

Draft — not yet indexed

Simultaneous equations

Worked answers and methods for 2.4 on Edexcel A-level Maths 9MA0.

Explanation

A simultaneous solution is an ordered pair satisfying every equation; graphically, solutions are intersection points. Use elimination when coefficients align conveniently, or substitute an expression from the linear equation into the nonlinear equation.
For example, substituting $y=5-x$ into $xy=6$ gives $x(5-x)=6$ , whose roots generate the two intersection pairs. After finding one coordinate, substitute each possible value back separately; losing the second quadratic root or mismatching coordinates are common errors.
If substitution produces a quadratic, its two real roots may give two intersection points.
Pair each value of the first coordinate with the value obtained from the same substitution, then verify both coordinates in both original equations.
Equations involving powers such as $y=2^x$ can similarly be reduced by substituting the exponential expression.

Worked example

Find every solution of $y=x+1$ and $x^2+y^2=25$ .

1.Substitute $y=x+1$ : $x^2+(x+1)^2=25$ , so $2x^2+2x-24=0$ .
2.Dividing by $2$ gives $(x+4)(x-3)=0$ , hence $x=3$ or $x=-4$ .
3.Using $y=x+1$ gives the pairs $(3,4)$ and $(-4,-3)$ .

Answer: $(x,y)=(3,4)$ or $(-4,-3)$

Common mistakes

Don't keep only one root of the quadratic produced by substitution and lose a valid intersection.
Don't pair an $x$ -value with the $y$ -value belonging to the other root.

Exam tip

State every ordered pair and check each pair in both original equations.

Worked practice

Tier 1 · Easy

Solve

x+y=9

and

x-y=1

simultaneously.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$x=5$ , $y=4$	2
Notes
Add the equations to eliminate $y$ : $2x=10$ , so $x=5$ . Substitution into $x+y=9$ gives $y=4$ .

(2 marks)

Tier 2 · Standard

Find the coordinates of the points of intersection of the curve

y=x^2-3x+1

and the line

y=2x-5

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$(2,-1)$ and $(3,1)$	4
Notes
Equating the two expressions for $y$ gives $x^2-3x+1=2x-5$ , so $x^2-5x+6=0$ . Hence $(x-2)(x-3)=0$ , giving $x=2$ or $x=3$ . Substitution into $y=2x-5$ gives $y=-1$ or $y=1$ , so the intersections are $(2,-1)$ and $(3,1)$ .

(4 marks)

Tier 3 · Hard

Solve the system

x+2y=7

and

x^2+y^2=13

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$(x,y)=(3,2)$ or $\left(-\dfrac15,\dfrac{18}{5}\right)$	5
Notes
From the line, $x=7-2y$ . Substitution gives $(7-2y)^2+y^2=13$ , so $5y^2-28y+36=0$ . The discriminant is $64$ , hence $y=(28\pm8)/10$ , giving $y=2$ or $y=18/5$ . Then $x=7-2y$ gives $x=3$ or $x=-1/5$ respectively.

(5 marks)

Tier 1 · Easy

Solve

3x+2y=19

and

2x-y=1

simultaneously.

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$x=3$ , $y=5$	3
Notes
Doubling the second equation gives $4x-2y=2$ . Adding this to $3x+2y=19$ gives $7x=21$ , so $x=3$ . Then $2(3)-y=1$ , giving $y=5$ .

(3 marks)

Tier 2 · Standard

Find every solution of

y=2x+1

and

xy=15

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$(x,y)=\left(\dfrac52,6\right)$ or $(-3,-5)$	4
Notes
Substitute $y=2x+1$ into $xy=15$ : $x(2x+1)=15$ , so $2x^2+x-15=0=(2x-5)(x+3)$ . Therefore $x=5/2$ or $x=-3$ . The corresponding values from $y=2x+1$ are $6$ and $-5$ .

(4 marks)

Tier 3 · Hard

Solve the simultaneous equations

x+y=7

and

x^2-xy+y^2=13

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$(x,y)=(3,4)$ or $(4,3)$	5
Notes
Use $y=7-x$ in the second equation. This gives $x^2-x(7-x)+(7-x)^2=13$ , so $3x^2-21x+36=0$ . Dividing by $3$ gives $(x-3)(x-4)=0$ . Thus $x=3$ or $x=4$ , and $y=7-x$ gives the ordered pairs $(3,4)$ and $(4,3)$ .

(5 marks)

Tier 2 · Standard

A theatre sells

160

tickets. An adult ticket costs £

14

and a student ticket costs £

9

. The total ticket income is £

1860

. Find the number of each type of ticket sold.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$84$ adult tickets and $76$ student tickets	4
Notes
Let $a$ and $s$ be the adult and student ticket counts. Then $a+s=160$ and $14a+9s=1860$ . Substituting $s=160-a$ into the income equation gives $14a+9(160-a)=1860$ , so $5a=420$ and $a=84$ . Hence $s=160-84=76$ .

(4 marks)

Tier 3 · Hard

The simultaneous equations

(k+1)x+2y=7

and

3x+(k-1)y=5

do not have a unique solution for certain real values of

k

. Find these values and determine whether each system has no solution or infinitely many solutions.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$k=\sqrt7$ or $k=-\sqrt7$ ; for each value the system has no solution.	6
Notes
Eliminating $y$ gives a coefficient of $(k+1)(k-1)-6=k^2-7$ on $x$ . A unique solution fails when $k^2-7=0$ , so $k=\pm\sqrt7$ . For infinitely many solutions, the coefficient and constant ratios would also have to agree. Comparing the $x$ -coefficients and constants would require $3/(k+1)=5/7$ , giving $k=16/5$ , which is neither $\sqrt7$ nor $-\sqrt7$ . Therefore the equations are inconsistent for both exceptional values, so each system has no solution.

(6 marks)

Tier 3 · Hard

The line

y=2x+1

meets the curve

y=x^2-4x-2

at points

A

and

B

. Without finding the coordinates of

A

and

B

separately, find the exact coordinates of the midpoint of

AB

and the exact length

AB

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Midpoint $(3,7)$ $AB=4\sqrt{15}$	6
Notes
Equating the two expressions for $y$ gives $x^2-4x-2=2x+1$ , so the intersection inputs are the roots of $x^2-6x-3=0$ . Their sum is $6$ , so the midpoint has $x$ -coordinate $3$ and hence $y$ -coordinate $2(3)+1=7$ . The discriminant is $48$ , so the two $x$ -coordinates differ by $\sqrt{48}=4\sqrt3$ . Since $y=2x+1$ , the corresponding $y$ -coordinates differ by $8\sqrt3$ . Therefore $AB=\sqrt{(4\sqrt3)^2+(8\sqrt3)^2}=4\sqrt{15}$ .

(6 marks)

Q10

Tier 3 · Hard

10.

The line

y=2x+k

and the circle

x^2+y^2=20

are given, where

k

is a real constant. Find the values of

k

for which the line is tangent to the circle and the corresponding point of contact in each case. State also the values of

k

for which there are two distinct intersections and no intersections.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$k=10$ gives contact point $(-4,2)$ ; $k=-10$ gives contact point $(4,-2)$ . There are two distinct intersections for $\|k\|<10$ and no intersections for $\|k\|>10$ .	6
Notes
Substitution of $y=2x+k$ into the circle gives $5x^2+4kx+k^2-20=0$ . Its discriminant is $(4k)^2-20(k^2-20)=400-4k^2$ . Tangency requires this to be zero, so $k=\pm10$ . At a repeated root, $x=-4k/10=-2k/5$ . Thus $k=10$ gives $x=-4$ , $y=2$ , while $k=-10$ gives $x=4$ , $y=-2$ . The discriminant is positive when $\|k\|<10$ , giving two distinct intersections, and negative when $\|k\|>10$ , giving no intersections.

(6 marks)

Verified exam appearances

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