press the ‘esc’ key to exit at any time pass with pearson maths achievement standard 90147 use...

Press the ‘Esc’ key to exit at any time

PASS WITH PEARSON

Maths Achievement Standard 90147

Use straightforward algebraic methods and solve equations (1.1) Gwenda Hill

© Pearson Education New Zealand 2005 Start

How to use this CD and Legal Notices

Algebra Main Menu

Achievement Merit Excellence

Press to go to desired section.

ExitTitle Page

This CD consists of:1. A Main Menu (this page) returned to by pressing:

2. Three Sections within the main menu accessed by:

3. Exercise groups within each section accessed by:

4. Questions within each exercise group accessed by:

Back to Main Menu

Back to ‘Section’ Menu

Exercise ‘1’

Q1.

Achievement MenuPress the button to go to the menu page for each exercise group.

Exercise A9.

Exercise A11.

Exercise A13.

Exercise A7.

Exercise A5.

Exercise A3.

Exercise A1.

Exercise A15.

Exercise A10.

Exercise A12.

Exercise A14.

Exercise A8.

Exercise A6.

Exercise A4.

Exercise A2.

Exercise A16.

Like terms

More simplifying

Expand and simplify

Factorising quadratics

Algebraic exponent factors

Finding linear rule

Solve linear equations

Solve factorised quadratics

Simplifying

Expand and simplify

Factorise common factors

Algebraic exponent powers

Substitution

More linear rules

More linear equations

Achievement review

Back to Main Menu

Q5.

Q6.

Q7.

Q4.

Q3.

Q2.

Q1.

Exercise A1 Menu

Like Terms

Are the following pairs like or unlike terms? 4a, 3a

Are the following pairs like or unlike terms? 3x, 2x2

Are the following pairs like or unlike terms? 4a2, 3a3

Are the following pairs like or unlike terms? 5p, 3 x p

Select the like terms to 6x from: 5x, 7, -2x, x2, 4xy

Select the like terms to 2a2 from: 4a, 4a4, -2a2, 5a2, 45a2y

Select the like terms to 7 from: 6x, -7, 4, 2a5, 1/4

Answer each question then press the button to see the correct answer orscroll to each question/answer using the ‘next’ arrow (below).

Back to Achievement Menu Next

Exercise A1 Question 1

1. Are the following pair like or unlike terms? 4a, 3a

Only variable is a, Both have power 1 so Like

NextBack to Exercise A1 Menu


2. Are the following pair like or unlike terms? 3x, 2x2

Only variable is x, but One is x and other is x2

so Unlike

NextPrevious Back to Exercise A1 Menu


3. Are the following pair like or unlike terms? 4a2, 3a3

Only variable is a, One has power 2 and the other

3, so Unlike



4. Are the following pair like or unlike terms? 5p, 3 p

Only variable is p and 3 p = 3p

Both have power 1, so Like



5. Select the like terms to 6x from: 5x, 7, -2x, x2, 4xy

Only variable is x, 5x, -2x, x2, 4xy have x x2 has power 2 not 1, 4xy also has y so 5x and -2x



6. Select the like terms to 2a2 from:

4a, 4a4, -2a2, 5a2, 45a2y Only variable is a with power 2, -2a2, 5a2 and 45a2y have a2

45a2y also has y so -2a2 and 5a2



7. Select the like terms to 7 from: 6x, -7, 4, 2a5, 1/4

No variable – just a constant -7, 4 and 1/4 are the constants so like terms are: -7, 4 and 1/4

Previous Back to Exercise A1 Menu Next Menu

Exercise A2 Menu

Simplifying expressions

4a + 3a

4a2 + 3a3

4a2 – 2a2 + 5a2

5x + 6y + 3x – 2y

Q1.

Q8.

3x – 2x

5p – 3p + 7p

6x – 7 – 4x + 2

4x + 2x2 – 3x

Q3.

Q5.

Q7.

Q2.

Q4.

Q6.

Back to Achievement Menu

Simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below).

Next


1. Simplify the following expression: 4a + 3a

Like terms = 7a

Back to Exercise A2 Menu NextPrevious


2. Simplify the following expression: 3x – 2x

Like terms = 1x = x

Back to Exercise 2 Menu NextPrevious


3. Simplify the following expression, where possible:4a2 + 3a3

No like terms as different powers

So not possible to simplify



4. Simplify the following expression: 5p – 3p + 7p

Like terms = 2p + 7p = 9p



5. Simplify the following expression: 4a2 – 2a2 + 5a2

Like terms = 2a2 + 5a2

= 7a2



6. Simplify the following expression: 6x – 7 – 4x + 2

Like terms together 6x – 4x and -7 + 2 = 2x + -5



7. Simplify the following expression: 5x + 6y + 3x – 2y

Like terms together 5x + 3x and 6y – 2y = 8x + 4y



8. Simplify the following expression: 4x + 2x2 – 3x

Like terms together

4x – 3x = x + 2x2

Back to Exercise 2 MenuPrevious Next Menu

Exercise A3 Menu

More simplifying

Q5. Q6.

Q7.

Q4.Q3.

Q2.Q1.

Q8.

4a 3a

4a2 3a3

4a2 -2a3 a2

5x 6y 3x 2y

3x -2x

5p 3p2 p

6x -7x 2

4x 2x2 3xy




1. Simplify the following expression: 4a 3a

4 3 = 12, a a = a2

so 12a2

Back to Exercise A3 Menu Next


2. Simplify the following expression: 3x -2x

3 -2 = -6, x x = x2

so -6x2



3. Simplify the following expression: 4a2 3a3

4 3 = 12, a2 a3 = a5

so 12a5



4. Simplify the following expression: 5p 3p2 p

5 3 1 = 15, p p2 p = p1+2+1

= p4

so 15p4



5. Simplify the following expression: 4a2 -2a3 a2

4 -2 1 = -8, a2 a3 a2 = a7

so -8a7



6. Simplify the following expression: 6x -7x 2

6 -7 2 = -84, x x = x2

so -84x2



7. Simplify the following expression: 5x 6y 3x 2y

5 6 3 2 = 180, x x = x2, y y

= y2

so 180x2 y2



8. Simplify the following expression: 4x 2x2 3xy

4 2 3 = 24, x x2 x = x4

so 24x4 y

Back to Exercise A3 MenuPrevious Next Menu

Exercise A4 MenuExpand and simplify

Q5.

Q4.Q3.

Q2.Q1.

Q10.Q9.

Q8.Q7.

Q6.

4 (a + 3)

a (a + 5)

4a (2a + 3)

5x (6y – 3x)

2x2 (3xy2 + 2x2y)

3 (x – 2)

p (p – 3)

6x ( -7x + 2)

-4x ( 2x – 3y)

3x2y (xy2 – 2x3y)


Expand and simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below).


4 a = 4a, 4 3 = 12

so 4a + 12

1. Expand and simplify: 4 (a + 3)



2. Expand and simplify: 3 (x – 2)

3 x = 3x, 3 2 = 6

so 3x – 6



3. Expand and simplify: a (a + 5)

a a = a2, a 5 = 5a

so a2 + 5a



p p = p2, p 3 = 3p

so p2 – 3p

4. Expand and simplify: p (p – 3)



4a 2a = 8a2, 4a 3 = 12a

so 8a2 + 12a

5. Expand and simplify: 4a (2a + 3)



6x -7x = -42x2, 6x 2 = 12x so -42x2 + 12x or 12x – 42x2

6. Expand and simplify: 6x (-7x + 2)



7. Expand and simplify: 5x (6y – 3x)

5x 6y = 30xy, 5x 3x = 15x2

so 30xy – 15x2



8. Expand and simplify: -4x (2x – 3y)

-4x 2x = -8x2, -4x 3y = -12xy so -8x2 – -12xy therefore -8x2 + 12xy or 12xy –

8x2



9. Expand and simplify: 2x2 (3xy2 + 2x2y)

2x2 3xy2 = 6x3y2, 2x2 2x2y = 4x4y

so 6x3y2 + 4x4y



10.Expand and simplify: 3x2y (xy2 – 2x3y)

3x2y xy2 = 3x3y3, 3x2y 2x3y = 6x5y2

so 3x3y3 + 6x5y2

Back to Exercise A4 MenuPrevious Next Menu

Exercise A5 Menu

Expand and simplify

Q5.

Q4.Q3.

Q2.Q1.

Q6.

Q11.

Q10.Q9.

Q8.Q7.

Q12.

(a + 3) (a + 4)

(a – 3) (a + 5)

(a + 4) (2a + 3)

(5x + y) (x – y)

(a – b) (a + b)

(a + b)2

(x + 5) (x + 2)

(p – 6) (p – 3)

(x – 3) (-7x + 2)

(x + 4) (2x – 3)

(x – 3)2

(2x – 5) (2x + 5)

Next

Expand and simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below).



1. Expand and simplify: (a + 3) (a + 4)

First Outside Inside Last aa=a2 a4=4a 3a=3a 34=12 so a2 + 4a + 3a + 12 = a2 + 7a + 12



2. Expand and simplify: (x + 5) (x + 2)

First Outside Inside Last xx=x 2 x2=2x 5x=5x

52=10 so x2 + 2x + 5x + 10 = x2 + 7x + 10



3. Expand and simplify: (a – 3) (a + 5)

First Outside Inside Last aa=a2 a5=5a -3a=-3a -35=-15 so a2 + 5a – 3a – 15 = a2 + 2a – 15



4. Expand and simplify: (p – 6) (p – 3)

First Outside Inside Last pp=p2 p-3=-3p -6p=-6p -6–3=18 so p2 – 3p – 6p + 18 = p2 – 9p + 18



First Outside Inside Last a2a=2a2 a3=3a 42a=8a 43=12 so 2a2 + 3a + 8a + 12 = 2a2 + 11a + 12

5. Expand and simplify: (a + 4) (2a + 3)



6. Expand and simplify: (x – 3) (-7x + 2)


First Outside Inside Last x-7x=-7x2 x2=2x -3x –7x=21x -

32=-6 so -7x2 + 2x + 21x – 6 = -7x2 + 23x – 6


7. Expand and simplify: (5x + y) (x – y)

First Outside Inside Last 5xx =5x2 5x-y=-5xy yx=xy y-y=-

y2

so 5x2 – 5xy + xy + -y2

= 5x2 – 4xy – y2



8. Expand and simplify: (x + 4) (2x – 3)


First Outside Inside Last x2x=2x2 x-3=-3x 42x=8x 4-3=-12 so 2x2 – 3x + 8x – 12 = 2x2 + 5x – 12


9. Expand and simplify: (a – b) (a + b)


First Outside Inside Last aa=a2 ab=ab -ba=-ab -bb=-b2

so a2 + ab – ab – b2

= a2 – b2


10. Expand and simplify: (x – 3)2


= (x–3) (x–3) First Outside Inside Last xx=x2 x–3=-3x -3x=-3x -3-3=9 so x2 – 3x – 3x + 9 = x2 – 6x + 9


11. Expand and simplify: (a + b)2


= (a + b) (a + b)First Outside Inside Lastaa=a2 ab=ab ba=ab bb=b2 so a2 + ab + ab + b2 = a2 + 2ab + b2


12. Expand and simplify: (2x – 5)(2x + 5)

Back to Exercise A5 MenuPrevious

First Outside Inside Last 2x2x=4x2 2x5=10x -52x=-10x -55=-25 so 4x2 + 10x – 10x – 25 = 4x2 – 25

Next Menu

Exercise A6 MenuFactorising common factors

Q1. 3a – 15

a2 + 4a

2xy + 2xt

12x + 3p

15a2b3 + 25ab2

4x + 24

p – 6pq

3q + 15q2

9x2y + 30xy2

30x2y – 21x2y3Q10.

Q5.

Q4.Q3.

Q2.

Q9.

Q8.Q7.

Q6.

Next

Fully factorise the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below).



1. Factorise fully: 3a – 15


= 3 a – 3 5= 3 (a – 5)


2. Factorise fully: 4x + 2


= 4 x + 4 6= 4 (x + 6)


3. Factorise fully: a2 + 4a


= a a + a 4= a (a + 4)


4. Factorise fully: p – 6pq


= p 1 – p 6q= p (1 – 6q)


5. Factorise fully: 2xy + 2xt


= 2 x y + 2 x t= 2x (y + t)


6. Factorise fully: 3q + 15q2


= 3 q + 3 5 q q= 3q (1 + 5q) Note: Must have the 1


7. Factorise fully: 12x + 3p


= 3 4x + 3 p= 3 (4x + p)


8. Factorise fully: 9x2y + 30xy2


= 3 3 x x y + 3 10 x y y

Common factors are 3, x, y= 3xy(3x + 10y)


9. Factorise fully: 15a2b3 + 25ab2


= 3 5 a a b b b + 5 5 a b b

Common factors are 5, a, b and b= 5ab2 (3ab + 5)


10. Factorise fully: 30x2y – 21x2y3


= 3 10 x x y – 3 7 x x y

y y Common factors are 3, x2 and y= 3x2y (10 – 7y2)

Next Menu

Exercise A7 Menu

Factorising quadratics

Q1. a2 + 4a + 3

a2 + 4a – 5

a2 + 6a – 16

x2 + x – 20

a2 – 9

x2 + 4x + 4

x2 – 4x – 5

p2 – 6p + 8

k2 – 6k – 7

a2 + 6a + 9Q10.

Q5.

Q4.Q3.

Q2.

Q9.

Q8.Q7.

Q6.

Next

Fully factorise the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below).



1. Factorise fully: a2 + 4a + 3


No common factors

Pairs that multiply to give 3 are 1, 3 and -1, -3

1 + 3 = 4so (a + 1)(a + 3) or (a + 3)(a + 1)


2. Factorise fully: x2 + 4x + 4


No common factors

Pairs that multiply to give 4 are:1, 4; 2, 2; -1, -4 and -2, -2

2 + 2 = 4so (x + 2)(x + 2) = (x + 2)2


3. Factorise fully: a2 + 4a – 5


No common factors

Pairs that multiply to give -5 are: 1, -5 and 5, -1

5 + -1 = 4so (a + 5)(a – 1) or (a – 1)(a + 5)


4. Factorise fully: x2 – 4x – 5


No common factors

Pairs that multiply to give -5 are: 1, -5 and 5, -1

1 + -5 = -4so (x + 1)(x – 5) or (x – 5)(x + 1)


5. Factorise fully: a2 + 6a – 16


No common factors

Pairs that multiply to give -16 are:1, -16; 2, -8; 4, -4; 8, -2 and 16, -1

8 + -2 = 6so (a + 8)(a – 2) or (a – 2)(a + 8)


6. Factorise fully: p2 – 6p + 8


No common factors

Pairs that multiply to give 8 are:1, 8; 2, 4; -1, -8 and -2, -4

-2 + -4 = -6so (p – 2)(p – 4) or (p – 4)(p – 2)


7. Factorise fully: x2 + x – 20


No common factors

Pairs that multiply to give -20 are: 1, -20; 2, -10; 4, -5; 5, -4; 10, -2 and 20, -1x means 1x and 5 + -4 = 1so (x + 5)(x – 4) or (x – 4)(x + 5)


8. Factorise fully: k2 – 6k – 7


No common factorsPairs that multiply to give -7 are: 1, -7 and

7, -11 + -7 = -6so (k + 1)(k – 7) or (k – 7)(k + 1)


9. Factorise fully: a2 – 9


No common factors

Can be written as: a2 + 0p – 9 Pairs that multiply to give -9 are:

1, -9; 3, -3 and 9, -1 3 + -3 = 0so (a + 3)(a – 3) or (a – 3)(a + 3)


10. Factorise fully: a2 + 6a + 9


No common factors

Pairs that multiply to give 9 are: 1, 9; 3, 3; -1, -9 and 3, -3 3 + 3 = 6so (a + 3)(a + 3) = (a + 3)2

Next Menu

Exercise A8 MenuAlgebraic exponent powers

Q3.

Q1. (3x)2

Q4.

Q2. (4y)3

Next



(5a2)2 (2x2y)4


1. Simplify: (3x)2


= 3x 3x= 9x2


2. Simplify: (4y)3


= 4y 4y 4y = 43y3

= 64y3


3. Simplify: (5a2)2


= 5a2 5a2

= 52(a2)2

= 25a4


4. Simplify: (2x2y)4


= 24 (x2)4 (y)4

= 16 x8 y4

= 16x8y4

Next Menu

Exercise A9 MenuAlgebraic exponent factors

4

8

xy

x

12

15

x

xy

23

6

x

xy

3 2

2 3

12

15

x y

x y

2

2

36

60

a

ab

2

2

2

4

x

x y

2 436

60

p q

pq

2

3

(5 )

60

x

xy

Q4.Q3.Q2.Q1.

Q5. Q8.Q7.Q6.

Next




1. Simplify: 4

8

xy

x

4 4

8

xy x

x4

y

x

22

y


Common factors on top and bottom are:

4 and x4xy = 4x y 8x = 4x

2


2. Simplify: 12

15

x

xy

12 3

15

x x

xy

4

3x

55

4

y y


Common factors on top and bottom are: 3 and x

12x = 3x 4 15xy = 3x 5y


3. Simplify:

23

6

x

xy

23 3

6

x x

xy3

x

x

22y

x

y


Common factors on top and bottom are: 3 and x

3x2 = 3x x 6xy = 3x 2y


4. Simplify:

3 2

2 3

12

15

x y

x y

2 23 2

2 3

312

15

x yx y

x y

2 2

4

3

x

x y

55

4

y

x

y


Common factors on top and bottom are: 3, x2 and y2

12x3y2 = 3x2 y2 4x 15x2y3 = 3x2 y2 5y


5. Simplify:

2

2

36

60

a

ab

2

2

36 12

60

a a

ab

3

12

a

a

2 2

3

5 5b

a

b


Common factors on top and bottom are: 12 and a(Note: It does not matter if you only notice one or a smaller factor – this just means you will take more steps as you find other common factors)

36a2 = 12a 3a 60ab2 = 12a 5b2


6. Simplify:

2 2

2

2 2

4

x x

x y

2

1

2x

22

1

y y

2

2

2

4

x

x y


Common factors on top and bottom are: 2 and x2 2x2 = 2x2 1, 4x2y = 2x2 2y


7. Simplify:

2 4 1236

60

pqp q

pq

33

12

pq

pq

3

5

3

5

pq

2 436

60

p q

pq


Common factors on top and bottom are: 12, p and q 36p2q4 = 12pq 3pq3, 60pq = 12pq 5


8. Simplify:

2

3

25 5

60

x x

xy

5

5

x

x

3 3

5

112 2

x

y y

2

3

(5 )

60

x

xy


Top is 25x2

Common factors on top and bottom are: 5 and x 25x2 = 5x 5x, 60xy3 = 5x 12y3

Next Menu

Exercise A10 Menu

Substitution

If A = 0.5(a + b)h, find A when a = 3, b = 7 and h = 6

If k = 0.6b2, find k when b = 4

S = . What is S when a = 3, c = 5 and d = 2?

The volume of a sphere is found by where r is radius.Find the volume of a sphere with radius 5.

A new symbol has the meaning that x y = 3x2 – y. Find the value of -2 4.

To convert Fahrenheit temperature, f, to Celsius, C,use the formula . Convert 360o Fahrenheit to oCelsius.

15a

c d34

3V r

59 ( 32)C f

Q5.

Q6.

Q4.

Q3.

Q2.

Q1.

Next

Substitute within the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below).



1. If A = 0.5(a + b)h, find A when a = 3, b = 7 and h = 6


A = 0.5 (3 + 7) 6

= 30


2. If k = 0.6b2, find k when b = 4


k = 0.6 (4)2

= 9.6


3. S = . What is S when a = 3, c = 5 and d = 2?

15a

c d

15 15 3

5 2

a

c d


= (15 3) (5 – 2)= 15


4. The volume of a sphere is found by where r is radius. Find the volume of a sphere with radius 5.

343V r

343 (5)


Note: There was only 1 sig fig in the radius, so it might be acceptable to write 500 to 1 sig fig. However, you must show working steps and it is a good idea to show the unrounded answer.

V =

= 4 3 53

= 523.5987756= 524 to 3 sig fig


5. A new symbol has the meaning that x y = 3x2 – y. Find the value of -2 4


-2 takes the place of x and 4 takes the place of y-2 4 = 3(-2)2 – 4 Note: Use brackets to square a negative

= 8


6. To convert Fahrenheit temperature, f, to Celsius, C, use the formula . Convert 360o Fahrenheit to o Celsius.

59 ( 32)C f

59 (360 32)C

Previous Back to Exercise A10 Menu

= 182.222222 so 182.2 Celsius to 1 d.p.

Note: Other rounding acceptable.

Next Menu

Exercise A11 Menu

Finding linear rule

8, 12, 16, 20, …

10, 17, 24, 31, …

24, 27.5. 31, 34.5, …

35, 32, 29, 26, …

Q5.

Q7.

Q3.

Q1.

Next

22, 24, 26, 28, …

15, 15.5, 16, 16.5, …

104, 102, 100, 98, …

12.5, 12.1, 11.7, 11.3, ...

Q6.

Q4.

Q2.

Q8.

Find the rule for finding the nth term for the following sequences then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below).



1. Find the rule for finding the nth term for the sequence:8, 12, 16, 20, …


Difference between terms is 4Rule is from the 4 x table 4, 8, 12, 16, …Need 4 more to give the sequenceTherefore rule is 4n + 4




Difference between terms is 2Rule is from the 2 x table 2, 4, 6, 8,

…Need 20 more to give the sequenceTherefore rule is 2n + 20




Difference between terms is 7Rule is from the 7 x table 7, 14, 21, 28,

…Need 3 more to give the sequenceTherefore rule is 7n + 3


4. Find the rule for finding the nth term for the sequence:15, 15.5, 16, 16.5, …


Difference between terms is 0.5Multiplying 1, 2, 3, 4, … by 0.5 gives:

0.5, 1, 1.5, 2, …Need 14.5 more to give the sequenceTherefore rule is 0.5n + 14.5


5. Find the rule for finding the nth term for the sequence:24, 27.5. 31, 34.5, …


Difference between terms is 3.5Multiplying 1, 2, 3, 4, … by 3.5 gives:

3.5, 7, 10.5, 14, …Need 20.5 more to give the sequenceTherefore rule is 3.5n + 20.5




Difference between terms is -2Multiplying 1, 2, 3, 4, … by -2 gives:

-2, -4, -6, -8, …Need 106 more to give the sequenceTherefore rule is -2n + 106 or 106 – 2n




Difference between terms is -3Multiplying 1, 2, 3, 4, … by -3 gives:

-3, -6, -9, -12, …Need 38 more to give the sequenceTherefore rule is -3n + 38 or 38 – 3n


8. Find the rule for finding the nth term for the sequence:12.5, 12.1, 11.7, 11.3, ...


Difference between terms is –0.4Multiplying 1, 2, 3, 4, … by –0.4 gives:

–0.4, –0.8, –1.2, –1.6, …Need 12.9 more to give the sequenceTherefore rule is –0.4n + 12.9 or 12.9 – 0.4n

Next Menu

Exercise A12 Menu Linear rules diagram or table

Q1. A matches design. Find a rule and work out how many matches would be needed.

Q2. A rectangular shape. Find a rule to find the perimeter of the 10th rectangle that would be drawn in this sequence.

Q3. The telephone account. Complete the table to show the monthly costs. Write a rule to show how much is paid for n calls.

Q4. A rectangular courtyard is being tiled. Find a rule to find the number of tiles after n edges have been added.

Q5. The savouries table. Find an expression for the number of savouries allowed for n guests.

Q6. The amount of sewing pattern material. Write a general rule to give the amount of material for width (w)

Q7. The number of doors needed. If a block of flats is to have 17 storeys, how many doors are needed?

Next

Find the rule for finding the nth term for the following sequences then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution.


Answer


1. The following design is being created with matches. Find a rule and work out how many matches would be needed to make a design with 20 of the basic shapes.

basic shape7 matches

2 basic shapes12 matches

3 basic shapes17 matches

Sequence is 7, 12, 17, …Difference is 55 x table would give: 5, 10, 15, …Need 2 more each time so 5n + 220 basic shapes, n = 20,Need 5 20 + 2 = 102 matches


Answer


2. A rectangular shape is being increased by adding 1 metre to each side at the corner as shown. It starts as size 5 m by 4 m so has perimeter of 18 m.Find a rule to find the perimeter of the 10th rectangle that would be drawn in this sequence.

4m by 5mPerimeter = 18m



Sequence is 18, 26, 34,Difference is 88 x table is 8, 16, 24, …Need 10 more each time.Rule is 8n + 1010th rectangle would have n = 10Therefore 8 10 + 10 = 90


Answer


3. Steven pays $20 a month for the telephone account. Phone calls to his girlfriend in another town cost $2.75 each.

Complete the table to show the monthly cost for different numbers of phone calls. Write a rule to show how much he would pay for n calls.

Calls 1 2 3 4 5

Cost22.75

25.50

Complete table with 28.25, 31, 33.75Difference = 2.75This gives 2.75, 5.5, 8.25, …Needs 20 added

Rule is 2.75n + 20


Answer


4. A rectangular courtyard is being tiled. Fifteen tiles form the rectangular centre as shown and the tiles are being placed on each side.Find a rule to find the number of tiles after n edges have been added.

ANSWER


Number of tiles is 21, 27, 33, …Difference is 66 x table gives: 6, 12, 18, …Need to add 15 to eachTherefore rule is 6n + 15

Answer


5. The table shows the number of savouries allowed by a catering company for different numbers of guests.Find an expression for the number of savouries allowed for n guests.

Guests 10 20 30 40

Savouries 25 45 65 85


Guests go up by 10Savouries go up by 20, so 2 savouries per person.But for 10, 20, 30, 40, this would give: 20, 40, 60, 80, so need to add 5Therefore rule is 2n + 5


6. The amount of material needed for a sewing pattern is given for different sizes as shown in the table.Write a general rule to give the amount of material for the width (w).

Waist (cm)

70 75 80 85 90

Material (m)

2.7 2.9 3.1 3.3 3.5

Answer


Waist goes up by 5.Material goes up by 0.2, so 0.04 metres per cm.But for 70, 75, 80, 85, this would give: 2.8, 3, 3.2, 3.4,so need to take 0.1Therefore rule is 0.04w – 0.1

Answer


7. The number of doors needed to be supplied for a block of flats depends on the number of storeys high the flats are. For one storey 27 doors are needed. If the building is two-storied there are 51 doors needed; three-storied, 75 doors; four-storied, 99 doors and so on. If a block of flats is to have 17 storeys, how many doors are needed?


Number of doors is 27, 51, 75, 99, …

Difference is 24

24 x table gives 24, 48, 72, 96, … Need to add 3 to each

Therefore rule is 24n + 3 For 17 storeys, n = 17 so 24 17 + 3 = 411 doors. Next Menu

Exercise A13 Menu

Solve linear equations

5 28

x Q7.

Q5.

Q3.

Q1.

Q8.

Q6.

Q4.

Q2.5x + 7 = 42

2(x + 7) = 28

26 = 3x – 2

3x – 8 = 82

5x + 19 = 32.4

4 – x = 16

3 278

x

Next

Solve the following equations showing working, then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution.



5x + 7 = 42 – 7 – 7

5x = 35 5 5

x = 7

7

42

7

35 + 7

1. Solve the following equation showing working: 5x + 7 = 42



2. Solve the following equation showing working:3x – 8 = 82

882

x x x

+ 882

x x x

3x – 8 = 82 +8 +8 3x = 90 3 3 x = 30



3. Solve the following equation showing working:2(x + 7) = 28

Note: Can divide by 2 or expand brackets as first step. This example shows expanding brackets.

2x + 14 = 28 – 14 – 14

2x = 14 2 2 x = 7



4. Solve the following equation showing working:5x + 19 = 32.4

5x + 19 = 32.4 – 19 – 19

5x = 13.4 5 5

x = 2.68



5. Solve the following equation showing working:26 = 3x – 226 = 3x – 2

+ 2 + 228 = 3x3x = 28 Usual to have x term on left side so swap.

3 3 Can swap sides at any time since sides equal

x = 9.333333 so 9.33 (to 2 d.p.)



6. Solve the following equation showing working:4 – x = 16

4 – x = 16 – 4 – 4

-x = 12 -1 -1 x = -12



7. Solve the following equation showing working:

3 278

x

248

x

8 8 x = 192

– 3 – 3

3 278

x




38

x

5 28

x

5 28

x


+ 5 + 5

x 8 x 8 x = 24

Next Menu

Exercise A14 MenuSolve more linear equations

33 2

5

xx Q7.

Q5.

Q3.

Q1.

Q8.

Q6.

Q4.

Q2.5x + 7x = 42

2.5(x + 7) = 27.5

5.6x = 1.8x + 7

3.6x – 8 = 2.4x

5x + 19 = 30x + 104

14 – 2x = 16x

3 56

xx

Next




1. Solve the following equation showing working:5x + 7x = 42

12x = 42 12 12 x = 3.5



2. Solve the following equation showing working:3.6x – 8 = 2.4x

3.6x – 8 = 2.4x – 3.6x – 3.6x -8 = -1.2x -1.2x = -8 Can swap sides at any time

-1.2 -1.2 x = 6.67 (to 2 d.p.)



3. Solve the following equation showing working:2.5(x + 7) = 27.5

Can be done by expanding bracket or dividing by 2.5This example is done by dividing by 2.5.

2.5(x + 7) = 27.5 2.5 2.5

x + 7 = 11 – 7 – 7 x = 4



4. Solve the following equation showing working:5x + 19 = 30x + 104

– 19 – 19 Could take any term from both sides

as a correct first step.

5x = 30x + 85 – 30x – 30x -25x = 85 -25 -25

x = -3.4




5.6x = 1.8x + 7

5.6x = 1.8x + 7 – 1.8x – 1.8x 3.8x = 7 3.8 3.8

x = 1.84 (to 2 d.p.)



6. Solve the following equation showing working: 14 – 2x = 16x

14 – 2x= 16x + 2x + 2x

14 = 18x 18x = 14

18 18 x = 0.78 (to 2 d.p.)



7. Solve the following equation showing working:3

3 25

xx Multiply all terms by 5

3x – 15 = 10x Note that each term has been 5

– 3x – 3x-15 = 7x

7x = -15 Can swap sides at any time

7 7 x = -2.14 (to 2 d.p.)




3 56

xx Multiply all terms by 6

x + 18 = -30x Note that each term has been 6

– x – x 18 = -31x -31x= 18 Can swap sides at any time

-31 -31 x = -0.58 (to 2 d.p.)


Exercise A15 MenuSolve factorised quadratic equations

(x – 3)(x – 2) = 0

(x + 1)(x + 4) = 0

(3x + 2)(x – 2) = 0

x(2x + 7) = 0

Q5.

Q7.

Q3.

Q1. (x – 5)(x + 3) = 0

(2x – 5)(x + 4) = 0

(3x + 2)(2x – 5) = 0

5x (3x – 7) = 0

Q6.

Q4.

Q2.

Q8.

Next




1. Solve the following equation showing working:(x – 3)(x – 2) = 0

x – 3 = 0 or x – 2 = 0 +3 +3 +2 +2

x = 3 x = 2



2. Solve the following equation showing working:(x – 5)(x + 3) = 0

x – 5 = 0 or x + 3 = 0 + 5 + 5 – 3 – 3

x = 5 x = -3



3. Solve the following equation showing working: (x + 1)(x + 4) = 0

x + 1 = 0 or x + 4 = 0 – 1 – 1 – 4 – 4 x = -1 x = -4



4. Solve the following equation showing working:(2x – 5)(x + 4) = 0

2x – 5 = 0 or x + 4 = 0

+ 5 + 5 – 4 – 4 2x = 5 x = -4 2 2

x = 2.5



5. Solve the following equation showing working: (3x + 2)(x – 2) = 0

3x + 2 = 0 or x – 2 = 0 – 2 – 2 + 2 + 2

3x = -2 x = 2

3 3 x = -0.67



6. Solve the following equation showing working:(3x + 2)(2x – 5) = 0

3x + 2 = 0 or 2x – 5 = 0 – 2 – 2 + 5 + 5

3x = -2 2x = 5 3 3 2 2

x = -0.67(2 d.p.) or x = 2.5



7. Solve the following equation showing working:x(2x + 7) = 0

x = 0 or 2x + 7 = 0 – 7 – 7

2x = -7 2 2

x = -3.5



8. Solve the following equation showing working:5x (3x – 7) = 0

5x = 0 or 3x – 7 = 0 5 5 + 7 + 7 x = 0 3x = 7 3 3

x = 2.33 (2 dp)


Click on the buttons to go to the review questions, solve each one, then press the answer button. Or scroll to each question using the next arrow (below). Click to move from step to step in the solution.

Exercise A16 Menu

Achievement review

Expand the equations

Expand the equations

Find the rule

Solve the equation

Q5.

Q7.

Q3.

Q1. Factorise the equations

Find the surface area

Solve the equationsQ6.

Q4.

Q2.

NextBack to Achievement Menu

Answer b


1. Expand and simplify: (a) 3xy (2x – 7) (b) (2x + 3)(x – 6)

Answer a


(a) 3xy 2x – 3xy 7 = 6x2y – 21xy

(b) First Outside Inside Last 2x x – 2x 6 + 3 x – 3 6 = 2x2 – 12x + 3x – 18 = 2x2 – 9x – 18

Answer a


2. Factorise fully:(a) 5x2y – 20x3y4 (b) x2 – 8x + 15

Answer b

(b) Pairs that multiply to give 15 are: 1, 15; 3, 5; -1, -15 and -3, -5

-3 and -5 add to give -8x2 – 8x + 15 = (x – 3)(x – 5) or (x – 5)(x – 3)


(a) = 5 x x y – 5 4 x x x y y y y = 5x2y(1 – 4xy3)


3. Simplify: (a) (b)2 3 2(5 ) (2 )x xy2 2(3 )

9

p q

p

Answer a

Answer b

2 2 9(3 )

9

pp q

p

3 2

9

p q

p

3 2p q


(b) Top = 9 p p p p q q Bottom = 9 p

Common factor is 9p

(a) = 53 (x2 )3 22 x2 y2 = 125 4 x6 x2 y2 = 500 x8y2


4. S.A. = gives the surface area of a sphere for given radius r. What is the surface area of a sphere with radius 8 cm?

24 r

Answer


S. A. = 4 8 8 = 804.2477193 so 804 (3 sig fig)


5. Hedge plants are to be placed around the edge of rectangular fields. The diagram shows the spacing for different 2 m wide rectangular fields.Find a rule to give the number of plants needed for a 2 m rectangular field that is n metres long.

0.5m length10 plants 1m long

12 plants 1.5m long14 plants

Answer

For n values 0.5, 1, 1.5,… the number of plants is 10, 12, 14 so that the values 1, 2, 3, … would give 12, 16, 20With difference 4, the 4 x table gives 4, 8, 12, … Therefore need to add 8.The rule for length n is 4n + 8


Answer bAnswer a


6. Solve: (a) 5x – 7 = 32 (b) 6x – 19 = 3.6x + 21


(a) 5x – 7 = 32

+ 7 + 7 5x =

39 5 5 x = 7.8

(b) 6x – 19 = 3.6x + 21 + 19 + 19 6x = 3.6x + 40 – 3.6x – 3.6x

2.4x = 402.4 2.4 x = 16.67

(2 d.p.)


7. Solve: (3x + 5)(6 – x) = 0

3x + 5 = 0 or 6 – x = 0 – 5 – 5 x = 6

3x = -5 3 3

x = -1.67 (2 d.p.)

Previous Merit MenuBack to Exercise A16 Menu

Back to Main Menu

Merit MenuPress the button to go to the menu page for each exercise group.

Exercise M9.

Exercise M11.

Exercise M7.

Exercise M5.

Exercise M3.

Exercise M1.

Exercise M10.

Exercise M8.

Exercise M6.

Exercise M4.

Exercise M2. Simplifying

Change the subject

Equations and inequations

Harder quadratic equations

Substitution in equations

Context equation solving

Rational expressions

Quadratic patterning

Solving inequations

Simple quadratic equations

Elimination in equations

Back to Main Menu

Q5. Q6.

Q7.

Q4.

Q3.Q2.Q1.

Exercise M1 MenuAdding and subtracting rational expressionsAnswer each question as a single expression. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution.

Next

Q8. Q9.

3 5

x x

4 7

x x

2

5 10

x x

3 2

4 5

x x

4

5

x

x 2

5 3

x x

5

3

x

x 7

4 8

x

x 2

2

5 10

p

p

Back to Merit Menu

Exercise M1 Question 1

3 and 5 have common

multiple of 15 Rewrite x/3 as 5x/15 by

multiplying top and bottom by 5

Rewrite x/5 as 3x/15 by multiplying top and bottom by 3

Write common bottom line and combine the top lines.

5 3

3 5 15 15

x x x x

5 3

15

8

15

xx x

1. Write as a single expression:

3 5

x x

NextBack to Exercise M1 Menu


4 and 7 have common

multiple of 28 Rewrite x/4 as 7x/28 by


Rewrite x/7 as 4x/28 by multiplying top and bottom by 4


7 4

4 7 28 28

x x x x

7 4

28

3

28

xx x


4 7

x x

NextPrevious Back to Exercise M1 Menu


4 and 5 have common

multiple of 20 Rewrite 2x/5 as 4x/10 by



2 4

5 10 10 10

x x x x

4 5

10 1 20

x x x x

3. Write as a single expression:2

5 10

x x



4 and 5 have common

multiple of 20 Rewrite 3x/4 as 15x/20 by


Rewrite 2x/5 as 8x/20 by multiplying top and bottom by 4


3 2 15 8

4 5 20 20

x x x x

15 8

20

7

20

xx x

4. Write as a single expression:3 2

4 5

x x



x and 5 multiply to give 5x Rewrite 4/x as 20/5x by


Rewrite x/5 as by multiplying top and bottom by x


24 20

5 5 5

x x

x x x

220

5

x

x


5

x

x

2

5

x

x



x2 and x have common

multiple of x2

Rewrite as by multiplying top and bottom by x


2 2 2

5 3 5 3x

x x x x

2

5 3x

x


2

5 3

x x

3

x 2

3x

x



3 and x multiply to give 3x

Rewrite x/3 as by multiplying top and bottom by x

Rewrite 5/x as 15/x by multiplying top and bottom by 3


25 15

3 3 3

x x

x x x

2 15

3

x

x


3

x

x

2

3

x

x



4 and 8x both divide

into 8x Rewrite x/4 as by

multiplying top and bottom by 2x


27 2 7

4 8 8 8

x x

x x x

22 7

8

x

x


4 8

x

x

22

8

x

x



Common denominator

10p2

First term multiply top and bottom by 2

Second term multiply top and bottom by p2


3

2 2 2

2 4

5 10 10 10

p p

p p p

3

2

4

10

p

p


2

2

5 10

p

p

Next MenuPrevious Back to Exercise M1 Menu

Exercise M2 MenuSimplifying rational expressions

Answer each question as a single expression. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution.

Next

2 3

3

x x

x

3 6

3

x 22 4

10

x x

x

2 3 2

2

x x

x

2

4

2 8

x

x x

2

5

3 40

x

x x

2 16

2 8

x

x

2

3 2

5

5 25

x x

x x

2

2

2 15

9

p p

p

Q5. Q6.

Q7.

Q4.

Q3.Q2.Q1.

Q8. Q9.

Back to Merit Menu


x2 – 3x = x(x – 3) 3x = x 3 Cancel out the x

x

( 3)x

x

33

3

x

1. Simplify fully: 2 3

3

x x

x



3x + 6 = 3(x + 2)

3 = 3 1 Cancel out the 33

( 2)

3

x

2

12

xx

2. Simplify fully: 3 6

3

x



2x2 – 4x = 2x(x – 2)

10x = 2x 5 Cancel out the 2x2x

( 2)

2

x

x

5

2

5

x

3. Simplify fully:22 4

10

x x

x



x2 + 3x + 2 = (x + 1)(x + 2)

x + 2 = (x + 2) 1 Cancel out the (x + 2)( 2)x

( 1)

( 2)

x

x

1

11

xx

4. Simplify fully:2 3 2

2

x x

x



x – 4 = (x – 4) 1 x2 – 2x – 8 = (x – 4)(x + 2) Cancel out the (x – 4)

( 4)x ( 4)x ( 2)x

1

2x

5. Simplify fully: 2

4

2 8

x

x x



x – 5 = (x – 5) 1 x2 + 3x – 40 = (x + 8)(x – 5) Cancel out the (x – 5)

( 5)x ( 5)x ( 8)x

1

8x

6. Simplify fully:25

3 40

x

x x



x2 – 16 = (x – 4)(x + 4) 2x – 8 = 2(x – 4) Cancel out the (x – 4)

( 4)x

( 4)

2 ( 4)

x

x

4

2

x

7. Simplify fully:2 16

2 8

x

x



x2 + 5x = x(x + 5) 5x3 + 25x2 = 5x (x)(x +

5) Cancel out the x(x + 5)x

( 5)x

5 ( )x x ( 5)x 1

5x

8. Simplify fully:2

3 2

5

5 25

x x

x x



p2 + 2p – 15 = (p + 5)(p – 3)

p2 – 9 = (p + 3)(p – 3) Cancel out the (p – 3)( 5) ( 3)p p

( 3) ( 3)p p

5

3

p

p

9. Simplify fully:2

2

2 15

9

p p

p

Previous Back to Exercise M2 Menu Next Menu

Q3.

Q1.

Exercise M3 Menu

Quadratic patterningSolve the quadratic pattern in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution.

Next

Q4.

Q2.

Complete the table and write a rule for when n metres has been added.

Write a rule for the number of triangles needed for Design n.

Find a rule to give the number of plants needed if n rounds are to be added.

How many tins would there be in x layers?

Back to Merit Menu

Answer


1. A garden area is a rectangle 4 m by 3 m. It is being added to by extending 1 metre at a time along 2 adjacent sides. After 1 metre is added it has area 5 m by 4 m = 20 m2. After a second metre is added it has area 6 m by 5 m = 30 m2. Complete the table and write a rule for when n metres has been added.

Length added

1 2 3 4 5 6

Area 5x4=20

6x5=30

7x6=42

Next QuestionBack to Exercise M3 Menu

3 gives 7 6 = 424 gives 8 7 = 56 5 gives 9 8 = 726 gives 10 9 = 90n gives (n + 4)(n + 3)

Alternative Method

Exercise M3 Q1 (Answer b)

Length added, n

Area Diff 2nd diff

1 20

10

2 30 2

12

3 42 2

14

4 56 2

16

5 72

Question 1: Alternative methodHalf second difference = 1, so rule is to do with (1) n2.n2 gives 1, 4, 9, 16, 25, …But Area is 20, 30, 42, 56, 72, …

Need an extra 19, 26, 33, 40, 47,…This is a linear pattern with difference 7, so can be found by 7n + 12

Therefore rule is n2 + 7n + 12

Back to the question

Next QuestionBack to Exercise M3 Menu

Answer


2. A pattern is made from the basic design as shown.Write a rule for the number of triangles needed for Design n.

Design 1Triangles: 6

Design 2Triangles: 16 Design 3

Triangles: 30

Next QuestionPrevious

One way of seeing this pattern is:Design 1 = 6 triangles(2 by 2 square doubled to get triangles – 2 corner triangles) = 2 x (2 x 2) – 2 x 1

Design 2 = 2 x (3 x 4) – 2 x (2 x 2) doubled rectangle – 2 triangles which form a square

Design 3 = 2 x (4 x 6) – 2 x (3 x 3) Design 4 would have rectangle 5 by 8 = 2 x (5 x 8) – 2 x (4 x 4) Design n would be = 2(n + 1)(2n) – 2(n x n)

= 4n2 + 4n – 2n2 = 2n2 + 4n

Back to Exercise M3 Menu

Alternative Method


Design, n Triangles Diff 2nd diff

1 6

10

2 16 4

14

3 30 4

18

4 48 4

22

5 70

Question 2: Alternative methodThere are several different ways of seeing the design which lead to the same answer.This approach is the mathematical one:Half second difference = 2, so rule is to do with 2n2.n2 gives 1, 4, 9, 16, 25, …2n2 gives 2, 8, 18, 32, 50Triangles is 6, 16, 30, 48, 70, …

Need an extra 4, 8, 12, 16, 20,…This is a linear pattern with difference 4, so can be found by 4n

Therefore rule is: 2n2 + 4n




Answer


3. Plants are to be spaced out equally in a rectangular field. From a centre rectangular area with 8 plants, the plants are set out in rounds as shown in the diagram.Find a rule to give the number of plants needed if n rounds are to be added.

After 1 round 24 plants After 2 rounds

48 plants

After 3 rounds 80 plants


1 round 4 by 6 = 24 2 rounds 6 by 8 = 48 3 rounds 8 by 10 = 80 Dimensions are going up in 2s

so linear patterns.n rounds (2n + 2) by (2n + 4)So (2n + 2)(2n + 4) plants


Alternative Method


Rounds added, n

Plants Diff 2nd diff

1 24

24

2 48 8

32

3 80 8

40

4 120 8

48

5 168

Question 3: Alternative methodThere are several different ways of seeing the design which lead to the same answer.This approach is the mathematical one:Half second difference = 4, so rule is to do

with 4n2.n2 gives 1, 4, 9, 16, 25, …4n2 gives 4, 16, 36, 64, 100Triangles is 24, 48, 80, 120, 168, …

Need an extra 20, 32, 44, 56, 68,…This is a linear pattern with difference 12, so can be found by 12n + 8

Therefore rule is: 4n2 + 12n + 8




Answer


4. Cans are stacked on a shelf as shown in the diagrams. There are 4 cans in 1 layer, 11 cans if there are 2 layers, 21 cans if there are 3 layers and so on. How many cans would there be in x layers?

Previous

1 layer = 4 cans 2 layers = 0.5 x (2 x 11) 3 layers = 0.5 x (3 x 14) 4 layers = 0.5 x (4 x 17) Linear pattern for values in brackets, sox layers has 0.5 x x x (3x+5)

=0.5x(3x+5)

Back to Exercise M3 Menu Next Menu

Alternative Method


Layers, x Cans Diff 2nd diff

1 4

7

2 11 3

10

3 21 3

13

4 34 3

16

5 50

Question 4: Alternative methodThere are several different ways of seeing the design which lead to the same answer.This approach is the mathematical one:Half second difference = 3, so rule is to do with 1.5x2.x2 gives 1, 4, 9, 16, 25, …1.5x2 gives 1.5, 6, 13.5, 24, 37.5Cans is 4, 11, 21, 34, 50, …

Need an extra 2.5, 5, 7.5, 10, 12.5,…This is a linear pattern with difference 2.5, so can be found by 2.5x

Therefore rule is: 1.5x2 + 2.5x

Previous


Back to Exercise M3 Menu Next Menu

Q5.

Q3.

Q1.

Exercise M4 Menu

Change the subject

4q

k

Make k the subject in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution.

Next

x = 15k + 7

m = k2 – 5

z = Q6.

Q4.

Q2. P = 3(k – 2)

B =

g = 4a2k – 2p

5

3

k

Back to Merit Menu


1. Make k the subject in: x = 15k + 7

15k + 7 = x – 7 – 7

15k = x – 7 15 15

k = (x – 7) 15 or7

15

x



2. Make k the subject in: P = 3(k – 2)

3(k – 2) = P 3 3

k – 2 = P/3

+ 2 + 2 k = P/3 + 2 or 23

P



3. Make k the subject in: m = k2 – 5

k2 – 5 = m + 5 + 5

k2 = m + 5 take square root

k = m 5



4. Make k the subject in B =5

3

k


x 3 x 3 3B = k – 5 k – 5 = 3B + 5 +5 k = 3B + 5


5. Make k the subject in z = + q

NextPrevious

4z – q

4k


– q – q z – q = 4/k x k x k k(z – q) = 4 (z – q) (z – q)

k =


6. Make k the subject in g = 4a2k – 2p

Previous

g + 2p4a2


4a2k – 2p = g + 2p + 2p

4a2k = g + 2p 4a2 4a2

k =

Next Menu

58

3

k

3 115

k 6 11

6

k

Exercise M5 Menu

Solving Inequations

Next

Q5.

Q3.

Q1.

Q6.

Q4.

Q2.4x – 8 < 20

4(2 – x) > 16

3x + 8 > 5x

Solve the inequations in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution.

Back to Merit Menu


1. Solve: 4x – 8 < 20

+ 8 +8 4x < 28 4 4 x < 7



2. Solve: 3x + 8 > 5x

Swap sides, reversing inequation 5x < 3x + 8

– 3x – 3x 2x < 8 2 2 x < 4



3. Solve: 4(2 – x) > 16

4 4

2 – x > 4 – 2 – 2 -x > 2

-x -1 < 2 -1 Note: Change of sign direction

x < -2



4. Solve:

x 3 x 3 k – 5 < 24 + 5 + 5 k < 29

58

3

k



5. Solve:

+ 3 + 3 k/5 > 14

x 5 x 5 k > 70

k3 11

5



6. Solve:

– 6 – 6 -k/6 > 5

x 6 x 6 -k > 30

-k -1 < 30 -1 Note: Change of sign direction k < -30

6 116

k


Exercise M6 MenuForm and solve equations and inequations

Five more than three times a number is forty seven. What is the number?

How many items of jewellery were sold on commission?

How many weeks will it be before the account is overdrawn?

What day will Marama be jogging more than 5 km?

Click on the buttons to go to each problem or scroll to each question/answer using the ‘next’ arrow (below). Write an equation or inequation to represent each problem, then solve it. Press the answer button for the correct answer. Click to move from step to step in the solution. After how many weeks will

Tyrone have more than $500?

What is the mechanic charging per hour for labour?

How many calls can be made, without spending more than $60 in a week?

How many glasses of drink can be poured?

Q1.

Q5.

Q3.

Q7.

Q2.

Q6.

Q4.

Q8.

NextBack to Merit Menu

Answer


1. Write an equation or inequation to represent:Five more than three times a number is forty seven. What is the number?

3x + 5 = 47 – 5 – 5 3x = 42 3 3 x = 14


Answer


2. Write an equation or inequation to represent:Tyrone saves $16 a week for a holiday.If he starts with $83, after how many weeks will he have more than $500?

16w + 83 > 500 w = number of weeks – 83 – 83

16w > 417 16 16 w > 26.0625

Therefore 27 weeks


Answer


3. Write an equation or inequation to represent:Trish sells jewellery on commission at a jewellery

evening. She gets $8 for each item she sells, but it costs her $45 to cover the costs of the jewellery evening.

If she makes $203 one evening, how many items of jewellery did she sell?

8j - 45 = 203 j = number of jewellery items + 45 + 45 8j = 248 8 8 j = 31 Therefore 31 items


Answer


4. Write an equation or inequation to represent: A mechanic charges $56 for parts and has worked for

3.5 hours on the car.If the total account was $248.50, what is the mechanic charging per hour for labour?

3.5h + 56 = 248.50 h = hourly rate – 56 – 56 3.5h = 192.50 3.5 3.5

h = 55 Therefore $55 per hr


Answer


5. Write an equation or inequation to represent:Jenny has $1364 in an account. If she pays her rent

from this account and it is $75 per week, how many weeks will it be before her account is overdrawn?

1364 – 75w < 0 w = number of weeks – 1364 – 1364

-75w < -1364 -75w -75 > -1364 -75

w > 18.1866667Therefore 19 weeks


Answer


6. Write an equation or inequation to represent: Steven pays $20 a month for the telephone account.

Phone calls to his girlfriend in another town cost $2.75 each.He wants to call his girlfriend as much as possible. How

many calls can he make but not spend more than $60 in a week?

20 + 2.75c < 60 c = number of calls

– 20 – 20 2.75c < 40

2.75 2.75 c < 14.54545455

Therefore 14 calls


Answer


7. Write an equation or inequation to represent:Marama is increasing the distance she jogs by one more telephone pole length, which means she adds 24 m each day she jogs.If on day 1 she jogged 2 km (2000 m), on what day’s jogging will she be jogging more than 5 km? 2000 + 24(d – 1) > 5000 d = number of days 1976 + 24d > 5000 – 1976 – 1976

24d > 3024 24 24 d > 126 Therefore Day 127


Answer


150g + 30 = 9500 g = number of glassesNote: Could be < instead of =

– 30 – 30 150g = 9470 150 150

g = 63.133333 Therefore 63 glasses

8. Write an equation or inequation to represent:From a 9.5 L (9500 ml) drink container, 150 ml

glasses are being poured.If 30 ml was spilt from the container, how many

glasses of drink could be poured?

Back to Exercise M6 Menu Next MenuPrevious

Exercise M7 MenuSolving simple quadratic equations

x2 + 6x + 5 = 0

x2 + 4x – 12 = 0

x2 + 6x = 16

x2 – 3x = 10

Q5.

Q3.

Q1.

Q7.

x2 + 9x + 18 = 0

x2 – 5x – 24 = 0

x2 + 10x = -24

x2 – 2x = 8

Q6.

Q4.

Q2.

Q8.

Next

Solve the simple quadratic equations in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution.

Back to Merit Menu


1. Solve: x2 + 6x + 5 = 0

Pairs of factors of 5 are 1, 5; -1, -5

Pair that add to give 6 are 1, 5Therefore (x + 1)(x + 5) = 0

x + 1 = 0 or x + 5 = 0 – 1 – 1 – 5 – 5

x = -1 or x = -5



2. Solve: x2 + 9x + 18 = 0

Pairs of factors of 18 are: 1, 18; 2, 9; 3, 6; -1, -18; -2, -9; -3, -6 Pair that add to 9 are 3, 6Therefore (x + 3)(x + 6) = 0 x + 3 = 0 or x + 6 = 0

– 3 – 3 – 6 – 6 x = -3 or x = -6



3. Solve: x2 + 4x – 12 = 0 Pairs of factors of -12 are:1, -12; 2, -6; 3, -4; 4, -3; 6, -2; 12, -1Pair that add to 4 are 6, -2Therefore (x + 6)(x – 2) = 0x + 6 = 0 or x – 2 = 0

– 6 – 6 + 2 + 2 x = -6 or x = 2



4. Solve: x2 – 5x – 24 = 0

Pairs of factors of -24 are:1, -24; 2, -12; 3, -8; 4, -6; 6, -4; 8, -3; 12, -2; 24, -1Pair that add to -5 are 3, -8Therefore (x + 3)(x – 8) = 0

x + 3 = 0 or x – 8 = 0 – 3 – 3 + 8 + 8 x = -3 or x = 8



5. Solve: x2 + 6x = 16

Get all terms on left by – 16 giving x2 + 6x – 16 = 0Pairs of factors of -16 are: 1, -16; 2, -8; 4, -4; 8, -2; 16, -1Pair that add to 6 are 8, -2Therefore (x + 8)(x – 2) = 0 x + 8 = 0 or x – 2 = 0

– 8 – 8 + 2 + 2 x = -8 or x = 2



6. Solve: x2 + 10x = -24

Get all terms on left by + 24 giving x2 + 10x + 24 = 0Pairs of factors of 24 include: 1, 24; 2,

12; 4, 6Pair that add to 10 are 4, 6Therefore (x + 4)(x + 6) = 0x + 4 = 0 or x + 6 = 0

– 4 – 4 – 6 – 6 x = -4 or x = -6



7. Solve: x2 – 3x = 10

Get all terms on left by – 10 giving x2 – 3x – 10 = 0Pairs of factors of -10 are: 1, -10; 2, -5; 5, -2;

10, -1Pair that add to -3 are 2, -5Therefore (x + 2)(x – 5) = 0x + 2 = 0 or x – 5 = 0

– 2 – 2 + 5 + 5 x = -2 or x = 5



8. Solve: x2 – 2x = 8

Get all terms on left by – 8 giving x2 – 6x – 8 = 0Pairs of factors of -8 are: 1, -8; 2, -4; 4, -2; 8,

-1Pair that add to -2 are 2, -4Therefore (x + 2)(x – 4) = 0 x + 2 = 0 or x – 4 = 0

– 2 – 2 + 4 + 4 x = -2 or x = 4


Exercise M8 MenuSolving harder quadratic equations

Next

2x2 + 6x – 56 = 0

4x2 + 4x = 288

5p2 + 32p + 12 = 0

4x2 + 20x + 25 = 0

Q5.

Q3.

Q1.

Q7.

3x2 + 12x + 9 = 0

3x2 – 19x + 20 = 0

4x2 – 7x + 3 = 0

3x2 + 19x = 14

Q6.

Q4.

Q2.

Q8.

Solve the quadratic equations in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution.

Back to Merit Menu


1. Solve: 2x2 + 6x – 56 = 0

Common factor of 2, so divide by this leaving x2 + 3x – 28 = 0

Factors of -28 are: 1, -28; 2, -14; 4, -7; 7, -4; 14, -2; 28, -1Pair that add to give 3 are 7, -4Therefore (x + 7)(x – 4) = 0 x + 7 = 0 or x – 4 = 0

– 7 – 7 + 4 + 4 x = -7 or x = 4



2. Solve: 3x2 + 12x + 9 = 0

Common factor of 3, so divide by this leaving

x2 + 4x + 3 = 0 Factors of 3 are 1, 3; -1, -3;Pair that add to give 4 are 1, 3Therefore (x + 1)(x + 3) = 0x + 1 = 0 or x + 3 = 0

– 1 – 1 – 3 – 3 x = -1 or x = -3



3. Solve: 4x2 + 4x = 288 Common factor of 4, so divide by this leaving

x2 + x = 72Subtract 72 from both sides, so x2 + x – 72 = 0Factors of -72 include 4, -18; 6, -12; 8, -9; 9, -8;Pair that add to give 1 are: 9, -8Therefore (x + 9)(x – 8) = 0x + 9 = 0 or x – 8 = 0 – 9 – 9 + 8 + 8 x = -9 or x = 8



4. Solve: 3x2 – 19x + 20 = 0

No common factor; x2 term and constant multiplied gives 60x2

Pairs that multiply to give 60x2 include: -x, -60x; -2x, -30x; -3x,-10x; -4x, -15x; -5x, -12x; -

6x, -10xPair that add to give -19x are -4x and -15x

3x2 – 4x – 15x + 20 = 0 x(3x – 4) – 5(3x – 4) = 0 Therefore (3x – 4)(x – 5) = 0 3x – 4 = 0 or x – 5 = 0

3x = 4 + 5 + 5 x = 4/3 or x = 5



5. Solve: 5p2 + 32p + 12 = 0 No common factor; x2 term and constant multiplied gives 60p2

Pairs that multiply to give 60x2 include: p, 60p; 2p, 30p; Pair that add to give to 32p are 2p and 30p

5p2 + 2p + 30p + 12 = 0 p(5p + 2) + 6(5p + 2) = 0 Therefore (5p + 2)(p + 6) = 0 5p + 2 = 0 or p + 6 = 0

5p = -2 – 6 – 6 p = -2/5 or x = -6



6. Solve: 4x2 – 7x + 3 = 0 No common factor; x2 term and constant multiplied gives 12x2

Pairs that multiply to give 12x2 include:-x, -12x; -2x, -6x; -3x, -4x;

Pair that add to give to -7x are -3x and -4x4x2 – 3x – 4x + 3 = 0

x(4x – 3) – 1(4x – -3) = 0Therefore (4x – 3)(x – 1) = 04x – 3 = 0 or x – 1 = 0

4x = 3 + 1 + 1 x = 3/4 or x = 1



7. Solve: 4x2 + 20x + 25 = 0No common factor; x2 term and constant multiplied gives 100x2

Pairs that multiply to give 100x2 include: 2x, 50x; 4x, 25x; 5x, 20x; 10x, 10x;Pair that add to give 20x are 10x and 10x

4x2 + 10x + 10x + 25 = 0 2x(2x + 5) + 5(2x + 5) = 0 Therefore (2x + 5)(2x + 5) = 0

2x + 5 = 0 Note: Other factor gives same result

2x = -5 x = -5/2 or -2.5



8. Solve: 3x2 + 19x = 14

No common factor; take 14 from both sides 3x2 + 19x – 14 = 0

x2 term and constant multiplied gives -42x2

Pairs that multiply to give -42x2 include: 7x, -6x; 14x, -3x; 21x,-2x;Pair that add to give 19x are 21x and -2x

3x2 + 21x – 2x – 14 = 0 3x(x + 7) – 2(x + 7) = 0

Therefore (x + 7)(3x – 2) = 0 x + 7 = 0 or 3x – 2 = 0

– 7 – 7 3x = 2 x = -7 or x = 2/3


Exercise M9 MenuSolve simultaneous equations using elimination

2x + 7y = 6812x – 7y = 16

4x + 9y = 11 -2x – 3y = -1

5f + 6g = 26 f + g = 4

4k + 7n = 413k – 2n = 9

3t + 4u = 255t – 6u = 11.9

3x – 4y = 20 -3x + 20y = 20

7x + 5y = 11 4x – 10y = 32

100d – 20e = 114 5d + 4e = 17.2

34p + 11q = 91 4p – 2q = 14

2x + 8y = 13 5x – 7y = -62

Q5.

Q3.

Q1.

Q7.

Q9.

Q6.

Q4.

Q2.

Q8.

Q10.

Next

Solve these equations. Press the button for the correct answer or scroll using the ‘next’ arrow (below). Click to move through the solution.

Back to Merit Menu


1. Solve: 2x + 7y = 6812x – 7y = 16

As 7y and -7y will cancel out, add the two equations to get:

14x = 84 gives x = 6Substitute into first equation to find y

12 + 7y = 68 gives 7y = 56 so y = 8Check second equation

12 6 – 7 8 = 72 – 56 = 16 x = 6, y = 8



2. Solve: 3x – 4y = 20-3x + 20y = 20

As 3x and -3x will cancel out, add the two equations to get: 16y = 40

gives y = 2.5Substitute into first equation to find x

3x – 10 = 20 gives 3x = 30 so x = 10Check second equation

-3 10 + 20 2.5 = -30 + 50 = 20 x = 10, y = 2.5



3. Solve: 4x + 9y = 11 -2x – 3y = -1

Need to multiply equation 2 by 3 so 9y and -9y will cancel out(Note: could have been by 2 to eliminate the x with 4x and -4x)

-6x – 9y = -3 Add this new equation to the first equation

-2x = 8 gives x = -4Substitute into first equation to find y

-16 + 9y = 11 gives 9y = 27 so y = 3

Check second equation -2 -4 – 3 3 = 8 – 9 = -1

x = -4, y = 3



4. Solve: 7x + 5y = 11 4x – 10y = 32

Need to multiply equation 1 by 2 so 10y and -10y will cancel out

14x + 10y = 22Add this new equation to the second equation

18x = 54 gives x = 3Substitute into first equation to find y

21 + 5y = 11 gives 5y = -10 so y = -2Check second equation

4 3 – 10 -2 = 12 – -20 = 32 x = 3, y = -2



5. Solve: 5f + 6g = 26f + g = 4

Need to multiply equation 2 by -5 so 5f and -5f will cancel out (Note could have been by -6 to eliminate using 6g and -6g)

-5f + -5g = -20Add this new equation to the second equation

gives g = 6Substitute into first equation to find y

5f + 36 = 26 gives 5f = -10 so f = -2

Check second equation -2 + 6 = 4

f = -2, g = 6



6. Solve: 100d – 20e = 114 5d + 4e = 17.2

Multiply equation 2 by 5 so -20e and 20e will cancel out (Note could have been by -20 to eliminate using 100d and -100d)

25d + 20e = 86Add this new equation to the first equation 125d = 200

gives d = 1.6Substitute into first equation to find e

160 – 20e = 114 gives -20e = -46 so e = 2.3

Check second equation 5 1.6 + 4 2.3 = 8 + 9.2 = 17.2

d = 1.6, e = 2.3



7. Solve: 4k + 7n = 41 3k – 2n = 9

Multiply eqn 1 by 2 and eqn 2 by 7 so 14n and -14n will cancel out. (Note could have been eqn 1 by 3 and eqn 2 by -4 to eliminate using 12k and -12k)

8k + 14n = 8221k – 14n = 63

Add these 29k = 145gives k = 5

Substitute into first equation to find n 20 + 7n = 41

gives 7n = 21 so n = 3Check second equation

3 5 – 2 3 = 15 – 6 = 9 k = 5, n = 3



8. Solve: 34p + 11q = 91 4p – 2q = 14

Multiply eqn 1 by 2 and eqn 2 by 11 so 22q and -22q will cancel out. (Note could have been eqn 1 by 2 and eqn 2 by -17 to eliminate using 68p and -68p)

68p + 22q = 18244p – 22q = 154

Add these 112p = 336gives p = 3

Substitute into first equation to find q 102 + 11q = 91

gives 11q = -11 so q = -1Check second equation

4 3 – 2 -1 = 12 – -2 =14 p = 3, q = -1



9. Solve: 3t + 4u = 255t – 6u = 11.9

Multiply eqn 1 by 3 and eqn 2 by 2 so 12u and -12u will cancel out.(Note could have been eqn 1 by 5 and eqn 2 by -3 to eliminate using 15t and -15t)

9t + 12u = 75 10t – 12u = 23.8

Add these 19t = 98.8gives t = 5.2

Substitute into first equation to find u 15.6 + 4u = 25

gives 4u = 9.4 so u = 2.35Check second equation

5 5.2 – 6 2.35 = 26 – 14.1 = 11.9 t = 5.2, u = 2.35



10. Solve: 2x + 8y = 135x – 7y = -62

Multiply eqn 1 by 7 and eqn 2 by 8 so 56y and -56y will cancel out.(Note could have been eqn 1 by 5 and eqn 2 by -2 to eliminate using 10x and -10x)

14x + 56y = 9140x – 56y = -496

Add these 54x = -405gives x = -7.5

Substitute into first equation to find y-15 + 8y = 13

gives 8y = 28 so y = 3.5Check second equation

5 -7.5 – 7 3.5 = -37.5 – 24.5 = -62x = -7.5, y = 3.5


Exercise M10 MenuSolve simultaneous equations using substitution

x = y – 20x + 3y = 40

y = 7 + 3x4x – 2y = 15

d = 5e – 17 2d + 6e = 18

q = 7 – 6p4p – 2q = 14

y = -3x – 74x – 2y = -12

Q5.

Q3.

Q1.

Q7.

Q9.

Q6.

Q4.

Q2.

Q8.

Q10.

Next

Solve these equations. Press the button for the correct answer or scroll using the ‘next’ arrow (below). Click to move from step to step in the solution.

y = 2x + 83x + y = 16

x = y – 112x – 3y = -1

f = 2g + 6 f + g = 4

k = 20 – n2n – 3k = 8

y = 3x + 17y = 2x – 3

Back to Merit Menu


1. Solve: y = 2x + 83x + y = 16

Replace y in equation 2 by what it equals in equation 1:3x + (2x + 8) = 16

Simplify5x + 8 = 16

gives 5x = 8so x = 1.6

Substitute into first equation to find y y = 2 1.6 + 8 so y = 11.2

Check second equation3 1.6 + 11.2 = 4.8 + 11.2 = 16

x = 1.6, y = 11.2

Back to Exercise M10 Menu Next


2. Solve: x = y – 20x + 3y = 40

Replace x in equation 2 by what it equals in equation 1: (y – 20) + 3 y = 40

Simplify4 y – 20 = 40

gives 4 y = 60so y = 15Substitute into first equation to find x

x = 15 – 20so x = -5

Check second equation -5 + 3 15 = -5 + 45 = 40

x = -5, y = 15



3. Solve: x = y – 112x – 3y = -1

Replace x in equation 2 by what it equals in equation 1: 2(y – 11) – 3y = -1

Simplify -y – 22 = -1

gives -y = 21so y = -21Substitute into first equation to find x

x = -21 – 11 so x = -32

Check second equation 2 -32 – 3 -21 = -64 + 63 = -1 x = -32, y = -21



4. Solve: y = 7 + 3x 4x – 2y = 15

Replace y in equation 2 by what it equals in equation 1: 4x – 2(7 + 3x) = 15

Simplify -2x – 14 = 15

gives -2x = 29so x = -14.5

Substitute into first equation to find y y = 7 + -43.5

so y = -36.5Check second equation 4 -14.5 – 2 -36.5 = -58 + 73 = 15

x = -14.5, y = -36.5



5. Solve: f = 2g + 6f + g = 24

Replace f in equation 2 by what it equals in equation 1: (2g + 6) + g = 24

Simplify 3g + 6 = 24

gives 3g = 18so g = 6

Substitute into first equation to find f f = 2 6 + 6 so f = 18

Check second equation 18 + 6 = 24 f = 18, g = 6



6. Solve: d = 5e – 172d + 6e = 18

Replace d in equation 2 by what it equals in equation 1: 2(5e – 17) + 6e = 18

Simplify 16e – 34 = 18

gives 16e = 52so e = 3.25

Substitute into first equation to find d d = 5 3.25 – 17 so d = -0.75

Check second equation 2 -0.75 + 6 3.25 = -1.5 + 19.5 = 18 d = -0.75, e = 3.25



7. Solve: k = 20 – n 2n – 3k = 8

Replace k in equation 2 by what it equals in equation 1: 2n – 3(20 – n) = 8

Simplify 5n – 60 = 8 Note: take care with the – (..-..) to get

+

gives 5n = 68so n = 13.6

Substitute into first equation to find k k = 20 – 13.6 so k = 6.4

Check second equation 2 13.6 – 3 6.4 = 27.2 – 19.2 = 8

k = 6.4, n = 13.6NextPrevious Back to Exercise M10 Menu


8. Solve: q = 7 – 6p4p – 2q = 14

Replace q in equation 2 by what it equals in equation 1: 4p – 2(7 – 6p) = 14

Simplify 16p – 14 = 14 Note: take care with the – (..-..) to get +

gives 16p = 28so p = 1.75

Substitute into first equation to find q q = 7 – 6 1.75 so q = 7 – 10.5 = -3.5

Check second equation 4 1.75 – 2 -3.5 = 7 + 7 = 14

p = 1.75, q = -3.5



9. Solve: y = 3x + 17y = 2x – 3

Replace y in equation 2 by what it equals in equation 1: 3x + 17 = 2x – 3

Simplify: – 17 from both sides 3x = 2x – 20

-2x gives x = -20Substitute into first equation to find y

y = 3 -20 + 17so y = -60 + 17 = -43

Check second equation y = 2 -20 – 3 = -43 as required x = -20, y = -43



10. Solve: y = -3x – 74x – 2y = -12

Replace y in equation 2 by what it equals in equation 1: 4x – 2(-3x – 7) = -12

Simplify 10x + 14 = -12

10x = -26 x = -2.6

Substitute into first equation to find y y = -3 -2.6 – 7 so y = 7.8 – 7 = 0.8

Check second equation 4 -2.6 – 2 0.8 = -10.4 – 1.6 = -12

x = -2.6, y = 0.8


Exercise M11 MenuSolve simultaneous equations in context

Q5.

Q3.

Q1.

Next Question

Find the price of a chicken pie.

Find the price of the drink and chips.

Find how many metres of wallpaper.

Find the number of 2-legged pets.

Q6.

Q4.

Q2. Find the adult price to swim at the pool.

Find the time taken to prepare each sandwich.

Find the size of the larger number.

Find how many trees Ricky pruned.

Q7. Q8.

Solve each of the simultaneous equations. Press the button for question details or scroll to each question using the ‘next question’ arrow (below). Click to move from step to step in the solution.

Back to Merit Menu

1. Chicken pies cost 50c more than mince pies. If 6 mince and 9 chicken pies cost a total of $40.50, solve the pair of simultaneous equations:

c = m + 0.5 6m + 9c = 40.5to find the price of a chicken pie.

Answer


Next QuestionBack to the Exercise 11 Menu

Question: Find the cost of a chicken pieAnswer: c = m + 0.5

6m + 9c = 40.5Substitute for c in the second equation:

6m + 9(m + 0.5) = 40.5Simplifying: 15m + 4.5 = 40.5Giving: 15m = 36

m = 2.4Substitute into the first equation: c = m + 0.5

c = 2.4 + 0.5 = 2.9Read the problem and check the words for chicken pies at $2.90 and mince pies $2.40Chicken pies 50c more than mince, 6 mince and 9 chicken pies = 6 $2.40 + 9 $2.90 = $40.50Chicken pie costs $2.90

Exercise M11 Q1 AnswerBack to the question


2. A family with 2 adults and 3 children pay $14.80 to go swimming at the pool. Another group with 5 adults and 2 children pay $30.40 in total to swim at the pool. Solve the simultaneous equations

2a + 3c = 14.85a + 2c = 30.4

to find the price for an adult to go swimming at the pool. Answer


Next QuestionPrevious Back to the Exercise 11 Menu

Question: Find the adult price at a swimming pool.Answer: 2a + 3c = 14.8

5a + 2c = 30.4Multiply first equation by 2: 4a + 6c = 29.6Multiply second equation by -3:-15a – 6c = -91.2Adding gives: -11a = -61.6Dividing by -11 gives: a = 5.6This gives adult price $5.60, but best to check out by:Subsitute into the first equation 11.2 + 3c = 14.8Giving: 3c = 3.6 so c = 1.2Check wording of problem with 2 adults, 3 children $14.80, 5 adults and 2 children (5 $5.60 + 2 $1.20) = $30.40 Adults price is $5.60

Exercise M11 Q2 Answer



3. Six bags of chips and 4 bottles of drink cost $24.80. The money needed for the bottle of drink is 10 cents less than twice the price of the chips. Solve the simultaneous equations

6c + 4d = 24.8d = 2c – 0.1

and use the results to find the price of 1 bottle of drink and 1 bag of chips.

Answer



Question: Find the price of 1 bottle of drink and 1 bag of chips.Answer: 6c + 4d = 24.8

d = 2c – 0.1Substitute for d in the first equation:

6c + 4(2c – 0.1) = 24.8Simplifying: 14c – 0.4 = 24.8

14c = 25.2 c = 1.8

Substitute into the second equation: d = 2 1.8 – 0.1d = 3.5

Read the problem and check the words: 6 chips and 4 drinks = 6 $1.80 + 4 $3.50 = $24.80Drink ($3.50) is 10c less than twice chipsPrice of 1 drink and 1 chips = d + c = $3.50 + $1.80 Price is $5.30





4. A kitchen worker takes 17.6 minutes to prepare 10 sandwiches and 2 burgers. She takes 14.6 minutes to prepare 4 sandwiches and 5 burgers. Assuming the worker is working at a constant rate, solve the equations:

10x + 2y = 17.64x + 5y = 14.6

to find the time taken to prepare each sandwich.

Answer


Question: Find the time taken to prepare each sandwich.Answer: 10x + 2y = 17.6

4x + 5y = 14.6 x = time for sandwich, y = time for burgerWant x, so eliminate y by multiplying first equation by 5:

50x + 10y = 88Multiply second equation by -2:

-8x – 10y = -29.2Adding gives: 42x = 58.8Dividing by 42: x = 1.4This gives time for sandwich 1.4 min, but best to check out by:Substitute into the first equation 14 + 2y = 17.6 giving: 2y = 3.6 so y = 1.8Check wording of problem with 10 sandwiches, 2 burgers 17.6 min, 4 sandwiches and 5 burgers (4 1.4 + 5 1.8) = 14.6 Time to prepare sandwich = 1.4 min




Answer


5. A formula is used to work out how many metres of wallpaper is needed for a standard room, using the distance around the room and the total width of the windows. The formula gives 123.84 m for a room with perimeter 28 m and with windows with total width 3.6 m. The formula gives 71.1 m for a room with perimeter 16 m and windows with total length 1.5 m. Solve the equations:

28x + 3.6y = 123.8416x + 1.5y = 71.1

to find out how many metres of wallpaper is allowed per metre of perimeter and per metre of total window width.


Question: Find out how many metres of wallpaper.Answer: 28x + 3.6y = 123.84

16x + 1.5y = 71.1 x = perimeter of room, y = width of windowEliminate y by multiplying first equation by 1.5:

42x + 5.4y = 185.76Multiply second equation by -3.6:

-57.6x – 5.4y = -255.96Adding: -15.6x = -70.2Dividing by -15.6: x = 4.5Substitute into first equation 126 + 3.6y = 123.84 giving: 3.6y = -2.16 so y = -0.6Check wording of problem: with 28 m perimeter, 3.6 m window = 123.84, 16 m perimeter, 1.5 m windows (16 4.5 + 1.5 -0.6) = 71.1 Formula allows 4.5 m per metre perimeter and takes 0.6 m per window width metre.





6. The sum of two numbers is 98 and the larger number is five more than twice the smaller number. Solve the equations:

x + y = 98y = 2x + 5

to find the size of the larger number.

Answer


Question: Find the size of the larger number.Answer: x + y = 98

y = 2x + 5 Substitute for y in the first equation: x + (2x + 5) = 98Simplifying 3x + 5 = 98Giving 3x = 93

x = 31Substitute into the second equation:

y = 2 31 + 5 y = 67

Read the problem and check the words Sum of 31 and 67 is 9867 is twice 31 plus 5Question asks for larger numberLarger number is 67


Next Question


Back to the Exercise 11 Menu


7. There are 48 pets in a pet parade. The pets either have 4 legs or 2 legs. If there were 150 legs on the pets, solve the equations:

x + y = 484x + 2y = 150

to find how many 2-legged pets are in the parade.

Answer


Question: Find the number of 2-legged pets.

Answer: x + y = 484x + 2y = 150

y = number of 2-legged pets. Multiply the first equation by -4 -4x – 4y = -192

Adding this to the second equation gives -2y = -42 y = 21

As a check, substitute into the first equation, x + 21 = 48 so x = 27

Check legs: 4 27 + 2 21 = 108 + 42 = 150

Number of 2-legged pets is 21





8. If Ricky had pruned 6 more trees he would have pruned twice as many as Steve. Together they pruned 96 trees. Solve the equations:

x + 6 = 2yx + y = 96

to find how many trees Ricky pruned.

Answer

Previous Back to the Exercise 11 Menu Excellence Menu

Question: Find how many trees Ricky pruned.Answer: x + 6 = 2y can be rearranged to give

x – 2 y = -6 (1)We also know that x + y = 96 (2)x = number of trees Ricky prunedMultiplying equation (2) by 2 gives2x + 2y = 192 (3), Adding equation (3) to equation (1) gives 3x = 186Therefore x, the number of trees Ricky pruned, = 62As a check, substitute into the first equation,

62 + 6 = 2yTherefore y = 34This also fits the second equation x + y = 96



Excellence MenuBack to Merit Menu

Back to Main Menu

Excellence Menu: Exercise E1Solve equations in context by modelling

How many movies were shown each week 5 years ago?Estimate the charge to erect a fence 50 m long.

Calculate the worth when the cabinet is full.

Calculate is the area of the section.

How much do they each earn?

Give the dimensions of each of the rectangular regions.Give the maximum length of the garden.

Give the side lengths of the original field.

Calculate how many tile layers there are.

Calculate how many weeks.

Q1.

Q8.

Q3.

Q5.

Q7.

Q2.

Q4.

Q6.

Q9. Q10.

Solve each of the equations. Press the button for question details or scroll to each question using the ‘next question’ arrow (below). Click to move from step to step in the solution.

Next QuestionBack to Main Menu

Exercise E1 Question 1

1. The average number of movies shown in a movie complex each week this year has increased by 8 since 5 years ago. The average number of viewers per movie was 4 times the number of weekly movies shown 5 years ago, but is now half that number. If there are now 480 viewers per week, how many movies were shown each week 5 years ago?

Back to Excellence Menu

Answer

Next Question

Exercise E1 Answer 1

1. How many movies were shown each week 5 years ago?Let number of movies per week 5 years ago = x.The number of movies shown each week has increased by 8 since 5 years ago.Number of movies per week now = x + 8The average number of viewers per movie was 4 times the number of weekly movies 5 years ago so: Average number of viewers per movie was 4x but is now half that number so: Average number of viewers per movie is 2xThere are now 480 viewers per weekNumber of viewers per week now = average number per movie x number of movies per week = 2x(x + 8)

2x(x + 8) = 480 2x2 + 16x – 480 = 0

Dividing by 2 x2 + 8x – 240 = 0Factors of -400 include: -10, 24; -12, 20

(x – 12)(x + 20) = 0x = -20 is not realistic. Therefore x = 1212 movies shown each week 5 years ago.

Back to question

Next QuestionBack to Excellence Menu


2. A building floor plan has 2 rectangular regions. The left region has one wall completely in common with the right region. The left region has length 5 m longer than its width. The left region has the shared wall ¾ of the width of the right region and the left region is ½ of the length of the right region. If the total perimeter around the building is 186 m, give the dimensions of each of the rectangular regions.

Answer

Previous Next QuestionBack to Excellence Menu


2. Give the dimensions of each of the rectangular regions. Let the width of the left region = xThe left region has length 5m longer than its width.

Length of left region = x + 5The left region has the shared wall ¾ of the width of the right regionWidth of right region = 4x 3 and it is ½ of the length of the right region. Length of right region = 2(x + 5)The total perimeter around the building is 186 m 6(x + 5) + 2(4x 3) = 186 26x 3 + 30 = 186

x = 156 x 3 26 = 18

Therefore the smaller region is 18 m by 23 m and the larger region is 46 m by 24 m.

Back to question



3. A fencing firm have a standard fee that they charge for labour and for the number of metres of fencing they build. They charged one customer $2477.40 for a 165 m long fence that takes 3 hours to erect. Another customer is charged $645.79 for a fence 42 m long that takes1.3 hours to erect. What will they estimate as a charge to erect a fence 50 m long that they expect to take 1.5 hours to erect?

Next Question

Answer

Previous Back to Excellence Menu

Exercise E1 Answer 33. Fencers have a standard fee that they charge for labour and for the

number of metres of fencing they build Let charge rate for labour = $x per hour and cost per metre of fence = y.They charge one customer $2477.40 for a 165 m long fence that takes 3 hours to erect.

3x + 165y = 2477.4Another customer is charged $645.79 for a fence 42 m long that takes 1.3 hours to erect.

1.3x + 42y = 645.79To solve these, multiply equation 1 by 1.3: 3.9x + 214.5y = 3220.62Multiply equation 2 by -3:

-3.9x – 126y = -1937.37 Adding gives:88.5y = 1283.25 y = 14.5

Substitute into equation 13x + 2392.5 = 2477.79 leading to 3x = 85.9

x = 28.3Therefore cost for labour is $28.30 per hour and cost per metre of fence is $14.50.What will they estimate as a charge to erect a fence 50 m long that they expect to take1.5 hours to erect?Charge = (1.5 x $28.30) + (50 x $14.50) = $767.45

Back to question



4. Concrete is needed for a path that is to surround four sides of a rectangular garden. It is to be 0.85 m wide and 0.05 m thick. The garden is 2 metres longer than it is wide. If 0.8 cubic metres of concrete is needed, what is the maximum length the garden can be?

Answer



4. What is the length of the garden? Let the length of the garden = xThe garden is 2 metres longer than it is

wide..

Width of garden = x – 2Concrete is to be 0.85 m wide and 0.05 m

thick.Area of concrete =

2 x 0.85 x x + 2 x 0.85 x (x – 2) + 4 x 0.852

= 1.7x + 1.7x – 3.4 + 2.89 = 3.4x – 0.51Volume of concrete =

(3.4x – 0.51) x 0.05 = 0.17x – 0.02550.8 cubic metres of concrete is needed0.17x – 0.0255 < 0.8

0.17x < 0.8255 x < 4.856

Therefore maximum length is 4.85 m

x - 2

x

Back to question



5. A jeweller displays watches in a cabinet. The watches are all the same price. When the cabinet is ¾ full, the value of the cabinet and watches is $5518.65. When the cabinet is 1/3 full, the combined value is $3119.40. What is the cabinet and watches worth when it is full?

Answer


Exercise E1 Answer 55. What is the cabinet and watches worth when it is full?Let value of cabinet = $c and value of watches = $w.When the cabinet is ¾ full, the value of the cabinet and watches is valued at $5518.65.

c + 0.75w = 5518.65When the cabinet is 1/3 full, the combined value is $3119.40.

c + w = 3119.4To solve these, multiply equation 2 by -1

-c – w = -3119.4Adding gives:

w = 2399.25 w = 5758.2

Substitute into equation 1 c + 4318.65 = 5518.65

c = 1200Therefore cost for cabinet full of watches is $1200 + $5758.20 = $6958.20

Back to question

1

3

1

3

5

12



6. A square field has a right-angled isosceles triangle area cut from one corner. The sides of the rectangular field where the corner has been cut off are reduced to 2/3 of their original length.

If the resulting field has a perimeter of 96 m, what was the original side lengths of the field?

Answer



6. What was the original side lengths of the field?

Let the length of the original sides = xA square field has a right-angled isosceles triangle area cut from one cornerThe sides off the rectangular field where the corner has been cut off are reduced to 2/3 of their original length.

The resulting field has a perimeter of 96 m Diagonal length = =

= 0.4714x Perimeter = 2x + 2(2/3x) + 0.4714x = 3.8047x 3.8047x = 96 x = 25.23 mOriginal side = 25.23 m

x

x2 21 1

3 3( ) ( )x x 29 x

Back to question



7. A right-angled triangle section has the longest side 4 m longer than the shortest side. The other side is 2 m longer than the shortest side. What is the area of the section?

Answer



7. A right-angled triangle section has the longest side 4 m longer than the shortest side.

Let the shortest side = xLongest side = x + 4The other side is 2 m longer than the shortest

side. Length of other side = x + 2What is the area of the section? Using Pythagoras’ theorem, (x + 4)2 = (x + 2)2 + x2 x2 + 8x + 16 = x2 + 4x + 4 + x2 x2 + 8x + 16 = 2x2 + 4x + 4 x2 – 4x – 12 = 0 (x – 6)(x + 2) = 0x = 6 (since x = -2 not sensible)Area = 0.5 6 8 = 24 m2

x

x + 2

x + 4

Back to question



8. A pattern of tiles is shown in the diagram.

If 754 tiles are used, how many layers are there?

1 layer4 tiles

2 layers10 tiles

3 layers18 tiles

Answer



8. Number of tiles 4, 10, 18, 28, 40, …Difference between terms is 6, 8, 10, 12, …This is increasing at a constant rate of 2Therefore quadratic pattern involving 1n2

Creating table shows difference between 1n2 and number of tiles is 3, 6, 9, 12, …

Therefore pattern is n2 + 3n

If 754 tiles are used, how many layers are there? n2 + 3n = 754 n2 + 3n – 754 = 0

Factors of -754 include -26, 29 (n – 26)(n + 29) = 0

-29 not sensible so 26 layers.

Layers, n

Tiles

n2 diff

1 4 1 3

2 10 4 6

3 18 9 9

4 28 16 12

5 40 25 15

Back to question



9. If Susan was paid $30 more a week, she would earn half as much as Sophie. Together they are paid $510. How much do they each earn?

Previous

Let m = money that Susan earns.Sophie earns 2 (m + 30) = 2m + 60Together they earn m + (2m + 60) = 510

3m + 60 = 510 3m = 450 (– 60 from both sides) m = 150 ( 3 on both sides)

Susan earns $150 and Sophie earns $360



10. Nathan decides to exercise each week by jogging. He starts in week 1 by jogging for 20 minutes. In week 2 he jogs for 30 minutes; in week 3 he jogs for 40 minutes, and so on. He keeps a record of the total time he spends jogging. After how many weeks is his total 18 290 minutes?

Answer

Previous Back to Excellence MenuEXIT


10. Total number of minutes is 20, 50, 90 140, …Difference between terms is 30, 40, 50, …This is increasing at a constant rate of 10Therefore quadratic pattern involving 5n2

Creating table shows difference between 5n2 and number of minutes is 15, 30, 45, 60, …Therefore pattern is 5n2 + 15nAfter how many weeks is his total 18 290 minutes?

5n2 + 15n = 18 290 n2 + 3n – 3654 = 0

Factors of -3658 include -59, 62 (n – 59)(n + 62) = 0-62 not sensible so Week 59.

Weeks, n

Total 5n2 diff

1 20 5 15

2 50 20 30

3 90 45 45

4 140 80 60

Back to question

EXITBack to Excellence Menu

Back to MAIN Menu

press the ‘esc’ key to exit at any time pass with pearson maths achievement standard 90147 use...

Documents

exercise a1 menu slide

exercise a1 question

section menu exercise

exercise groups

exercise a9

exercise a11

exercise a13

exercise a7