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GCSE · Algebra and Coordinate Geometry - Simultaneous Equations

A5 Simultaneous Equations: Linear and Quadratic

AQA 8365 · Calculator allowed · about 120 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Solve the simultaneous equations 3x + 2y = 16 and x - y = 2.

(Total for Question 1 is 3 marks)

Solve the simultaneous equations 4x - 3y = 1 and 2x + y = 13.

(Total for Question 2 is 3 marks)

A line has equation y = x - 1 and a curve has equation y = x² - 7. Which one of these pairs gives both solutions of the simultaneous equations?

A) (3, 2) and (-2, -3)
B) (2, 3) and (-3, -2)
C) (3, 2) only
D) (7, 6) and (-2, -3)

(Total for Question 3 is 1 mark)

Solve the simultaneous equations y = x + 1 and y = x² - 4x + 5.

(Total for Question 4 is 5 marks)

Solve the simultaneous equations x + y = 3 and y = x² - 6x + 3.

(Total for Question 5 is 5 marks)

A rectangular allotment plot has a perimeter of 26 metres and an area of 40 square metres. The length of the plot is greater than its width. Form a pair of simultaneous equations and solve them to find the length and width of the plot.

(Total for Question 6 is 5 marks)

Solve the simultaneous equations xy = 20 and x + y = 9.

(Total for Question 7 is 5 marks)

A curve has equation x² + y² = 17. A line has equation y = x + 3. Find the coordinates of the two points where the line intersects the curve.

(Total for Question 8 is 6 marks)

Two numbers have a sum of 15. The sum of their squares is 113. Find the two numbers.

(Total for Question 9 is 5 marks)

Show that the line y = 3x + 2 does not intersect the curve y = x² + 2x + 9.

(Total for Question 10 is 4 marks)

A curve has equation x² + y² = 45. A line has equation x - y = 3. The line intersects the curve at two points, A and B.

(a)Find the coordinates of A and B.(6)

(b)Find the length of chord AB, giving your answer as a simplified surd.(3)

(Total for Question 11 is 9 marks)

A line has equation y = x + 4. A curve has equation y = x² - 4x - 2. The line intersects the curve at two points, A and B.

(a)Find the coordinates of A and B.(5)

(b)Find the coordinates of the midpoint of AB.(2)

(c)Find the length of AB, giving your answer as a simplified surd.(3)

(Total for Question 12 is 10 marks)

The line y = 3x + k is a tangent to the curve y = x² + 5x + 7. Find the value of k.

(Total for Question 13 is 5 marks)

A circle has equation x² + y² = 20. A line has equation y = 2x + c. Given that the line is a tangent to the circle, find the possible values of c.

(Total for Question 14 is 6 marks)

A curve has equation y = x² + px + 9, where p is a constant. The curve is a tangent to the line y = 4x at exactly one point. Given that p < 0, find the value of p and the coordinates of the point of contact.

(Total for Question 15 is 6 marks)

Two numbers, x and y, satisfy x + y = 5 and x² + y² = 15. Find the values of x and y, giving your answers as simplified surds.

(Total for Question 16 is 6 marks)

A square has side length x cm. A rectangle has length (x + 3) cm and width y cm. The square and the rectangle have equal areas. The perimeter of the rectangle is 26 cm. Given that x > 0, find the value of x and the value of y.

(Total for Question 17 is 7 marks)

A line has equation y = kx + 4, where k is a constant. A curve has equation y = x² + 2x + 8.

(a)Show that the x-coordinates of the points of intersection of the line and the curve satisfy x² + (2 - k)x + 4 = 0.(2)

(b)Hence find the range of values of k for which the line intersects the curve at two distinct points.(4)

(Total for Question 18 is 6 marks)

Solve the simultaneous equations 2x² - y² = 7 and x + y = 1.

(Total for Question 19 is 6 marks)

A curve has equation y = x² - 6x + c, where c is a constant. A line has equation y = 2x - 5. The line intersects the curve at two points whose x-coordinates differ by 8. Find the value of c.

(Total for Question 20 is 6 marks)

Mark scheme · A5 Simultaneous Equations: Linear and Quadratic

Question 1

M1 rearrange x - y = 2 to x = y + 2 (or equivalent elimination step)
M1 substitute and solve to find one value, e.g. 5y = 10 leading to y = 2
A1 x = 4 and y = 2 cao
Answer: x = 4, y = 2

Question 2

M1 rearrange 2x + y = 13 to y = 13 - 2x (or equivalent elimination step)
M1 substitute and solve to find one value, e.g. 10x = 40 leading to x = 4
A1 x = 4 and y = 5 cao
Answer: x = 4, y = 5

Question 3

B1 A selected cao
Answer: A) (3, 2) and (-2, -3)

Question 4

M1 substitute x + 1 for y (or equivalent elimination of y)
A1 rearrange to a correct 3-term quadratic equal to zero, x² - 5x + 4 = 0 oe
M1 factorise or otherwise solve the quadratic, (x - 1)(x - 4) = 0
A1 x = 1 and x = 4 cao
A1 y = 2 and y = 5 ft from their x values
Answer: (1, 2) and (4, 5)

Question 5

M1 rearrange x + y = 3 to y = 3 - x
M1 substitute into the quadratic and simplify to a 3-term equation equal to zero, x² - 5x = 0 oe
A1 correct equation x² - 5x = 0
M1 factorise, x(x - 5) = 0, to find x = 0 and x = 5
A1 y = 3 and y = -2 ft from their x values
Answer: (0, 3) and (5, -2)

Question 6

B1 form the perimeter equation l + w = 13 oe
B1 form the area equation lw = 40 oe
M1 eliminate one variable to form a 3-term quadratic equal to zero, e.g. l² - 13l + 40 = 0 oe
A1 factorise or otherwise solve, (l - 5)(l - 8) = 0, giving l = 5 or l = 8
A1 length = 8 m and width = 5 m oe, with correct units
Answer: Length = 8 m, width = 5 m

Question 7

M1 rearrange x + y = 9 to y = 9 - x and substitute into xy = 20
A1 correct 3-term quadratic equal to zero, x² - 9x + 20 = 0 oe
M1 factorise or otherwise solve, (x - 4)(x - 5) = 0
A1 x = 4 and x = 5 cao
A1 both pairs of solutions stated: (4, 5) and (5, 4) ft
Answer: (4, 5) and (5, 4)

Question 8

M1 substitute y = x + 3 into x² + y² = 17
A1 correct expansion, x² + x² + 6x + 9 = 17 oe
M1 rearrange to a correct 3-term quadratic equal to zero, x² + 3x - 4 = 0 oe
A1 correct quadratic x² + 3x - 4 = 0
M1 factorise or otherwise solve, (x + 4)(x - 1) = 0
A1 both coordinate pairs correct: (-4, -1) and (1, 4) cao
Answer: (-4, -1) and (1, 4)

Question 9

M1 square the sum, (x + y)² = x² + 2xy + y² = 225
M1 substitute x² + y² = 113 to find 2xy = 225 - 113
A1 xy = 56 oe
M1 form and solve the quadratic t² - 15t + 56 = 0, e.g. (t - 7)(t - 8) = 0
A1 the two numbers are 7 and 8 oe
Answer: 7 and 8

Question 10

M1 set 3x + 2 equal to x² + 2x + 9 and rearrange to x² - x + 7 = 0 oe
M1 calculate the discriminant b² - 4ac = 1 - 28
A1 discriminant = -27
A1 cso: since -27 < 0 there are no real solutions, so the line does not intersect the curve
Answer: Discriminant = -27 < 0, so no real intersection points exist.

Question 11

(a) M1 rearrange x - y = 3 to x = y + 3
(a) A1 correct expansion, (y + 3)² + y² = 45 oe, e.g. 2y² + 6y + 9 = 45
(a) M1 rearrange to a correct 3-term quadratic equal to zero, y² + 3y - 18 = 0 oe
(a) A1 correct quadratic y² + 3y - 18 = 0
(a) M1 factorise or otherwise solve, (y + 6)(y - 3) = 0
(a) A1 both coordinate pairs correct: (-3, -6) and (6, 3) cao
(a) Answer: A(-3, -6), B(6, 3)
(b) M1 use the distance formula with the coordinates from part (a) ft, √9² + 9²
(b) A1 √162 oe unsimplified
(b) A1 9sqrt(2) cao
(b) Answer: 9sqrt(2)

Question 12

(a) M1 set x + 4 equal to x² - 4x - 2
(a) A1 correct 3-term quadratic equal to zero, x² - 5x - 6 = 0 oe
(a) M1 factorise or otherwise solve, (x - 6)(x + 1) = 0
(a) A1 x = -1 and x = 6 cao
(a) A1 y = 3 and y = 10 ft from their x values
(a) Answer: A(-1, 3) and B(6, 10)
(b) M1 use the midpoint formula with the coordinates from part (a) ft
(b) A1 midpoint (2.5, 6.5) cao
(b) Answer: (2.5, 6.5)
(c) M1 use the distance formula with the coordinates from part (a) ft, √7² + 7²
(c) A1 √98 oe unsimplified
(c) A1 7sqrt(2) cao
(c) Answer: 7sqrt(2)

Question 13

M1 set 3x + k equal to x² + 5x + 7 and rearrange to x² + 2x + (7 - k) = 0 oe
A1 correct equation x² + 2x + (7 - k) = 0
M1 set the discriminant equal to zero (tangent condition), 2² - 4(7 - k) = 0
A1 correct simplified equation, 4k - 24 = 0 oe
A1 k = 6 cao
Answer: k = 6

Question 14

M1 substitute y = 2x + c into x² + y² = 20
A1 correct expansion, x² + 4x² + 4cx + c² = 20 oe
A1 correct 3-term quadratic equal to zero, 5x² + 4cx + (c² - 20) = 0 oe
M1 set the discriminant equal to zero (tangent condition), (4c)² - 4(5)(c² - 20) = 0
A1 correct simplified equation, c² = 100 oe
A1 c = 10 or c = -10, both values, cao
Answer: c = 10 or c = -10

Question 15

M1 set x² + px + 9 equal to 4x and rearrange to x² + (p - 4)x + 9 = 0 oe
M1 set the discriminant equal to zero (tangent condition), (p - 4)² - 36 = 0
A1 correct equation (p - 4)² = 36 oe, leading to p - 4 = 6 or p - 4 = -6
A1 p = 10 or p = -2 identified as the two possible values
A1 p = -2 selected, since p < 0
A1 point of contact (3, 12) ft from their value of p
Answer: p = -2; point of contact (3, 12)

Question 16

M1 square the sum, (x + y)² = x² + 2xy + y² = 25
M1 substitute x² + y² = 15 to find 2xy = 25 - 15
A1 xy = 5 oe
M1 form the quadratic t² - 5t + 5 = 0 and apply the quadratic formula
A1 t = (5 ± √5) / 2 oe
A1 x = (5 + √5)/2 and y = (5 - √5)/2, or the values the other way round, cao
Answer: x = (5 + √5)/2, y = (5 - √5)/2 (or vice versa)

Question 17

B1 form the equal-areas equation x² = (x + 3)y oe
B1 form the perimeter equation (x + 3) + y = 13 oe
M1 rearrange the perimeter equation to y = 10 - x and eliminate y, x² = (10 - x)(x + 3) oe
A1 correct 3-term quadratic equal to zero, 2x² - 7x - 30 = 0 oe
M1 factorise or otherwise solve, (2x + 5)(x - 6) = 0
A1 x = 6, rejecting x = -2.5 since x > 0
A1 y = 4 ft from their value of x
Answer: x = 6, y = 4

Question 18

(a) M1 set kx + 4 equal to x² + 2x + 8
(a) A1 cso: correctly rearrange to x² + (2 - k)x + 4 = 0
(a) Answer: x² + (2 - k)x + 4 = 0 (shown)
(b) M1 write the discriminant condition (2 - k)² - 4(1)(4) > 0
(b) M1 expand and simplify to k² - 4k - 12 > 0 oe
(b) A1 factorise or otherwise find the critical values k = 6 and k = -2
(b) A1 k > 6 or k < -2 oe
(b) Answer: k > 6 or k < -2

Question 19

M1 rearrange x + y = 1 to y = 1 - x
M1 substitute into 2x² - y² = 7 and expand, 2x² - (1 - 2x + x²) = 7 oe
A1 correct 3-term quadratic equal to zero, x² + 2x - 8 = 0 oe
M1 factorise or otherwise solve, (x + 4)(x - 2) = 0
A1 x = -4 and x = 2 cao
A1 both pairs of solutions stated: (-4, 5) and (2, -1) ft
Answer: (-4, 5) and (2, -1)

Question 20

M1 set 2x - 5 equal to x² - 6x + c and rearrange to x² - 8x + (c + 5) = 0 oe
A1 correct equation x² - 8x + (c + 5) = 0, with sum of roots = 8 and product of roots = c + 5
M1 use (x1 - x2)² = (x1 + x2)² - 4x1x2 with the given difference of 8
A1 correct equation 64 - 4(c + 5) = 64 oe
M1 solve the equation for c
A1 c = -5 cao
Answer: c = -5

Simultaneous Equations: Linear and Quadratic - Worksheets, Questions and Revision