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4.6H Forming and Solving Equations: Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Solve 3(2x - 1) = 5x + 4.

(Total for Question 1 is 2 marks)

The perimeter of a rectangle is 2x + 14. The length is x + 4 and the width is x - 2. Form an equation and solve for x.

(Total for Question 2 is 3 marks)

Solve the pair of simultaneous equations
3x - y = 7
2x + y = 4

(Total for Question 3 is 3 marks)

Expand and rearrange to form a quadratic, then solve: (x - 2)(x + 5) = 3x + 4.

(Total for Question 4 is 4 marks)

A taxi company charges a fixed booking fee of 2 pounds plus 1.6 pounds per mile. Sophie paid 10.6 pounds for a journey. Form an equation and find how many miles Sophie travelled.

(Total for Question 5 is 4 marks)

Show that the equation 2x² - 3x - 5 = 0 has roots x = (3 ± √49)/4 and hence give the exact roots. Give all working.

(Total for Question 6 is 5 marks)

The sum of three consecutive odd integers is 51. Let the integers be n, n + 2 and n + 4. Form an equation and solve for n. State the three integers.

(Total for Question 7 is 4 marks)

A cylinder has radius r and height h. The volume V is given by V = π r² h. Rearrange this formula to make h the subject, then find h when V = 400pi and r = 4.

(Total for Question 8 is 5 marks)

Show that for all x not equal to 2, (x² - 4)/(x - 2) = x + 2. Hence, or otherwise, evaluate (x² - 4)/(x - 2) when x = 7 and comment on the value at x = 2.

(Total for Question 9 is 6 marks)

Find the value(s) of k for which the linear equation kx + 6 = 2k - x has no solution, exactly one solution, or infinitely many solutions. Explain your reasoning.

(Total for Question 10 is 4 marks)

Mark scheme · 4.6H Forming and Solving Equations: Higher Tier Practice

Question 1

M1 expand and collect terms correctly, e.g. 6x - 3 = 5x + 4 or equivalent method
A1 x = 7 cao
Answer: 7

Question 2

M1 form equation using perimeter 2(length + width) = 2x + 14, eg 2((x + 4) + (x - 2)) = 2x + 14
M1 collect terms and simplify correctly to a linear equation: 4x + 4 = 2x + 14
A1 x = 5 cao
Answer: 5

Question 3

M1 add or eliminate y to get 5x = 11 or equivalent elimination
M1 divide to find x = 11/5
A1 x = 11/5, y = -2/5 cao
Answer: x = 11/5, y = -2/5

Question 4

M1 expand LHS to x² + 3x - 10 or equivalent expansion
M1 rearrange to standard form x² + 3x - 10 - 3x - 4 = 0 giving x² - 14 = 0
M1 take square roots or factor correctly leading to x² = 14
A1 x = √14 or x = -√14 (both values required) cao
Answer: x = √14 or x = -√14

Question 5

M1 form equation 2 + 1.6m = 10.6 or equivalent
M1 subtract 2 to get 1.6m = 8.6
M1 divide to get m = 8.6/1.6
A1 m = 5.375 miles or 5.375 cao (allow 5.38 awrt) cao
Answer: 5.375

Question 6

M1 substitute into quadratic formula or complete square to get x = (3 ± √9 + 40)/4 or equivalent
M1 show discriminant step leading to √49
M1 write x = (3 ± √49)/4
M1 simplify √49 = 7 and substitute
A1 final exact roots x = (3 + 7)/4 = 5/2 and x = (3 - 7)/4 = -1 arise cao
Answer: x = 5/2 or x = -1

Question 7

M1 form equation n + (n + 2) + (n + 4) = 51
M1 collect terms to 3n + 6 = 51 and solve to n = 15
M1 state the three integers correctly
A1 integers are 15, 17 and 19 cao
Answer: 15, 17, 19

Question 8

M1 rearrange to h = V / (π r²) or equivalent
M1 substitute V = 400pi and r = 4 to get h = 400pi / (π * 16)
M1 cancel π and simplify 400 / 16 correctly
M1 obtain h = 25
A1 h = 25 units cao
Answer: 25

Question 9

M1 factorise numerator to (x - 2)(x + 2) or equivalent
M1 cancel (x - 2) to obtain x + 2, with domain note x ≠ 2
M1 state simplified form (x + 2) and that it holds for x ≠ 2
M1 substitute x = 7 to get 9
M1 state that expression is undefined at x = 2 because original denominator is 0
A1 final details: (x² - 4)/(x - 2) = x + 2 for x ≠ 2, value at 7 is 9, at 2 expression is undefined cao
Answer: Simplifies to x + 2 for x ≠ 2; value at x = 7 is 9; expression is undefined at x = 2.

Question 10

M1 bring x terms together: kx + x = 2k - 6 giving x(k + 1) = 2k - 6
M1 consider k + 1 = 0 case (k = -1) and substitute to check
M1 state that if k = -1 then equation becomes 0x + 6 = -2 - 6 so 6 = -8 which is false hence no solution
A1 for k ≠ -1 there is exactly one solution x = (2k - 6)/(k + 1); no k gives infinitely many solutions. cao
Answer: k = -1 gives no solution; for k ≠ -1 there is exactly one solution x = (2k - 6)/(k + 1); no value of k gives infinitely many solutions.

Forming and Solving Equations: Higher Tier Practice - Worksheets, Questions and Revision