3.14H 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 5x - 7 = 3x + 9.

(Total for Question 1 is 2 marks)

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

(Total for Question 2 is 3 marks)

Solve the equation (2x + 3)/4 = x - 1.

(Total for Question 3 is 4 marks)

Show that the equation 3x² - 5x - 2 = 0 has solutions x = 2 and x = -1/3. Then factorise 3x² - 5x - 2 fully.

(a)Show that x = 2 and x = -1/3 are solutions.(2)

(b)Factorise 3x² - 5x - 2.(3)

(Total for Question 4 is 5 marks)

Solve the simultaneous equations
2u + 3v = 7
5u - v = 6

(a)Eliminate v and find u.(3)

(b)Find v using your value of u.(3)

(Total for Question 5 is 6 marks)

Solve the equation 4/(x - 2) + 1 = 3/(x - 2). State any values of x that are not allowed.

(a)Solve for x.(3)

(b)State any values of x not allowed and why.(3)

(Total for Question 6 is 6 marks)

Solve the equation 2(3 - t) = t² - 5t + 6. Show all steps and give exact answers.

(a)Rearrange the equation to write it in the form f(t) = 0.(3)

(b)Factorise and solve the equation fully, giving exact values of t.(3)

(c)Check whether both roots satisfy the original equation and state your conclusion.(1)

(Total for Question 7 is 7 marks)

Mark scheme · 3.14H Solving Equations: Higher Tier Practice

Question 1

M1 collect like terms, e.g. 5x - 3x = 9 + 7 or equivalent method
A1 x = 8 cao
Answer: 8

Question 2

M1 expand and collect terms, 6y - 3 = 4y + 5 or equivalent
M1 isolate y, 6y - 4y = 5 + 3
A1 y = 4 cao
Answer: 4

Question 3

M1 multiply both sides by 4 or equivalent method, 2x + 3 = 4x - 4
M1 collect like terms, 3 + 4 = 4x - 2x
A1 x = 7/2 cao
B1 allow either fraction or decimal form 7/2 or 3.5
Answer: 7/2 or 3.5

Question 4

(a) M1 substitute x = 2 and x = -1/3 into 3x² - 5x - 2 and show each gives 0
(a) A1 both substitutions give 0 cao
(a) Answer: Substituting x=2 gives 3(4)-10-2=12-12=0; substituting x=-1/3 gives 3(1/9)-5(-1/3)-2 = 1/3+5/3-2 = 6/3-2 = 2-2 = 0
(b) M1 use roots to write factors or perform factorisation to (x - 2)(3x + 1) or equivalent
(b) A1 3x² - 5x - 2 = (x - 2)(3x + 1) cao
(b) B1 expand to check or state factors correctly
(b) Answer: (x - 2)(3x + 1)

Question 5

(a) M1 multiply second equation by 3 and add to first to eliminate v or equivalent
(a) M1 compute resulting equation, 2u + 3v + 15u - 3v = 7 + 18
(a) A1 u = 25/17 cao
(a) Answer: u = 25/17
(b) M1 substitute u into one equation, e.g. 5u - v = 6
(b) M1 calculate v = 5u - 6 with u = 25/17 leading to v = 125/17 - 102/17
(b) A1 v = 23/17 cao
(b) Answer: v = 23/17

Question 6

(a) M1 subtract 3/(x-2) from both sides or combine fractions over common denominator
(a) M1 get 4/(x-2) - 3/(x-2) + 1 = 0 leading to 1/(x-2) + 1 = 0
(a) A1 x = 1 cao
(a) Answer: x = 1
(b) B1 identify x = 2 is not allowed because denominator x - 2 = 0
(b) B1 state that x = 2 would make left and right expressions undefined
(b) A1 x = 2 not allowed cao
(b) Answer: x = 2 is not allowed because it makes the denominator zero.

Question 7

(a) M1 expand both sides or expand left-hand side: 6 - 2t = t² - 5t + 6
(a) M1 collect all terms on one side, e.g. 0 = t² - 3t or equivalent
(a) A1 clear statement t² - 3t = 0 cao
(a) Answer: t² - 3t = 0
(b) M1 factorise the quadratic to t(t - 3) or equivalent
(b) M1 use zero product principle to obtain candidate roots t = 0 or t = 3
(b) A1 exact solutions t = 0 and t = 3 cao
(b) Answer: t = 0 and t = 3
(c) B1 both roots t = 0 and t = 3 satisfy the original equation (both give equality) oe
(c) Answer: Both t = 0 and t = 3 satisfy the original equation.

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