Solve (x + 1)/3 + (x - 2)/4 = 2. Give your answer as a fraction in its simplest form.
(Total for Question 10 is 4 marks)
11
Chloe bakes two cakes using a 4 kg bag of flour. The first cake uses (x + 2)/3 kg of flour and the second cake uses (2x - 1)/5 kg of flour. All of the flour in the bag is used between the two cakes.
(a)Form an equation in x for the total mass of flour used, in kg.(1)
(b)Solve your equation to find the value of x.(3)
(c)Hence find the mass of flour used in the first cake, giving your answer in kg correct to 2 decimal places.(2)
(Total for Question 11 is 6 marks)
12
Solve 5/(x + 2) = 3/(x - 1)
(Total for Question 12 is 4 marks)
13
Solve 3/x + 2/(x + 1) = 1. Give your solutions correct to 2 decimal places.
(Total for Question 13 is 5 marks)
14
Show that x/(x - 2) - (x - 3)/(x + 1) = (6x - 6) / [(x - 2)(x + 1)], where x ≠ 2 and x ≠ -1.
(Total for Question 14 is 4 marks)
15
Write 2/x + 3/(x + 2) - 1/(x - 1) as a single fraction in its simplest form.
(Total for Question 15 is 4 marks)
16
Solve (2x - 1)/(x - 3) - (x + 1)/(x + 2) = 1. Give your solution correct to 2 decimal places.
(Total for Question 16 is 5 marks)
17
Simplify fully: [1/x + 1/y] divided by [1/x - 1/y]. Give your answer as a single fraction in terms of x and y.
(Total for Question 17 is 4 marks)
18
Show that 4/(x2 - 1) + 3/(x - 1) = (3x + 7) / [(x - 1)(x + 1)], where x ≠ 1 and x ≠ -1.
(Total for Question 18 is 4 marks)
Mark scheme · 7.6 Algebraic Fractions
Question 1
M1 cancels a common numerical or algebraic factor, e.g. 6/9 or x2/x or y3/y2 seen
A1 2xy/3 oe cao
Answer: 2xy/3
Question 2
M1 factorises the numerator as 4(x + 3)
A1 4 cao (x ≠ -3)
Answer: 4
Question 3
M1 attempts to factorise x2 + 2x - 15 as (x + 5)(x - 3)
A1 correct factorisation (x + 5)(x - 3) seen
A1 x + 5 oe cao (x ≠ 3)
Answer: x + 5
Question 4
M1 factorises x2 - 4 as (x - 2)(x + 2) and 3x + 3 as 3(x + 1)
M1 cancels the common factors (x + 1) and (x + 2)
A1 3x - 6 oe (3(x - 2)) cao
Answer: 3x - 6
Question 5
M1 multiplies by the reciprocal of the second fraction: (x2-9)/(2x) * 4x2/(x+3)
M1 factorises x2 - 9 as (x - 3)(x + 3) and cancels the common factor (x + 3)
A1 2x2 - 6x oe (2x(x - 3)) cao
Answer: 2x2 - 6x
Question 6
M1 uses a common denominator of 12
M1 correct numerator 4x + 3(x - 1)
A1 (7x - 3)/12 oe cao
Answer: (7x - 3)/12
Question 7
M1 uses the common denominator (x + 1)(x - 2)
M1 correct numerator 3(x - 2) + 2(x + 1)
A1 numerator simplified to 5x - 4 oe
A1 (5x - 4)/(x2 - x - 2) oe (denominator expanded or left factorised)
Answer: (5x - 4)/(x2 - x - 2)
Question 8
M1 factorises the numerator as (x - 3)(x + 3)
M1 factorises the denominator as (x + 3)(x - 2)
A1 (x - 3)/(x - 2) oe cao (x ≠ -3, x ≠ 2)
Answer: (x - 3)/(x - 2)
Question 9
M1 combines the fractions over the common denominator (x + 3): (2x + 1) - (x - 2) all over (x + 3)
A1 1 cao (x ≠ -3)
Answer: 1
Question 10
M1 multiplies every term by 12 (the LCM of 3 and 4)