9 Green C Math Class

EXAMPLE 2 The volume of a rectangular room of height 4m is 1600m³. Find the
cost of plastering its floor at the rate of Rs. 100 per sq. meter?

Solution : Volume of the room (V) = 1600m³

Height of the room (h) = 4m
Now, Area of the floor of room (A) = V

h
= 1600m³

4m
= 400m²

Rate of plastering (C) = Rs. 100 per sq. meter

Then, Total cost of plastering (T) = A × C

= 400 × Rs. 100

= Rs. 40000

EXAMPLE 3 The length of a room is 2 times its breadth and five times its height. If
the volume of the room be 800m³, find the cost of papering its walls at
Rs. 6 per m².

Solution : Let length of the room (l) = x m

Breadth of the room (b) = x m and
Now, x2
or, Height of the room (h) = 5 m
or,
Volume of the room (V) = 800m3
x× x × x
25 = 800m3
x³
= 8000m3

∴ x = 20m

Thus, length (l) = 20m
20m
breadth (b) = 2 = 10m

height (h) = 20m = 4m
2

Area of four walls (A) = 2h (l + b)

= 2 × 4(20 + 10)

= 240m²

Rate of papering the walls (C) = Rs. 6 per m²

∴ Total cost of papering the walls (T) = A × C

= 240 × Rs. 6

= Rs. 1440

GREEN Mathematics Book-9 95

EXAMPLE 4 A wall is 30m long, 4m wide and 80cm thick. It contains three square
windows having each side 2m and a door of dimension 2.5m × 1m.

i. Find the number of bricks with size 20cm × 16cm × 10cm to
construct the wall if 1 part of the wall is occupied by cement
4
joints.

ii. Find the total cost of bricks if cost per brick is Rs. 10.

Solution : Volume of wall (V1) = 30m × 4m × 80cm
= 30m × 4m × 0.8m

= 96m³

Volume of the wall occupied by a window and 3 doors (x)

= (2.5m × 1m × 80cm) + 3 (2m × 2m × 0.8m)

= 2.5m × 1m × 0.8m + 3(2m × 2m × 0.8m)

= 2m³ + 9.6m³

= 11.6m³

V olume of the wall occupied by cement = 1 of 96 m³
4

= 24m³

Now, V olume of wall excluding door, windows and cement (V)

= 96m³ – 11.6m³ – 24m³

= 60.4m³

= 60400000cm³

i. Volume a brick (v) = 20cm × 16cm × 10cm

= 3200cm³

Number of bricks (N) = V
v

= 603402000000
= 18875 pieces

ii. Cost of a brick (C) = Rs. 10

Total cost (T) = N × C

= 18875 × Rs. 10

= Rs. 188750

96 GREEN Mathematics Book-9

EXAMPLE 5 The cost of carpeting the floor of a room whose breadth is one-
fourth of its length is Rs. 16640 at the rate of Rs. 260 per sq m and

the cost of plastering the four walls of the room is Rs. 3000 at Rs. 25

per sq m. Calculate the height of the room.

Solution : Let, the length of the room(l) = xm
∴ the breadth of the room (b) = x m

4
Cost of carpeting (C) = Rs. 260 per sq. m.

Total cost of carpeting (T) = Rs. 16640
Now, Area of room (A) = T = 16640
or, C 260
or, l × b = 64
or, x × x = 64
4
x2 = 256

∴ x = 16m
So, length (l) = 16 m and breadth (b) = 16 = 4m

4
Again,

Cost of plastering the walls (C) = Rs. 25 per m²

Total cost of plastering the walls (T) = Rs. 3000

Now, Area of four walls (A) = T = 3000
C 25
or,
2h (l + b) = 120

or, 2h(16 + 4) = 120

or, h = 120
40

∴ h = 3m

So, the height of the room (h) = 3m

GREEN Mathematics Book-9 97

EXERCISE 5.4

A. Very Short Questions
1. a. The length, breadth and height of a room are 8m, 6m and 5m respectively. Find

the volume of the wall.

b. The area of a room is 30m2 and its height is 5m. Find the volume of the room.

2. a. If the volume of a brick is 0.01cm3, how many bricks are required to construct
a wall of 100cm3?

b. How many bricks each of volume 750cm3 will be required to construct a wall
of 10m × 5m × 30cm?

3. a. How many bricks each of volume 20cm3 will be required to construct a wall of
10m × 5m × 30cm?

b. If 100000 bricks each of volume 600cm3 are required to construct the wall, find
the volume of the wall.

B. Short Questions
4. a. The height of a wall is 1 of the length and thickness is 50cm. If the volume of

3
the wall is 18m³. Find the height of the wall.

b. A wall is of dimension 25m × 4m × 30cm. It contains a door of size 3m × 2m and
two windows each of 2m × 1.5m. Find the volume of the wall excluding door
and windows.

5. a. How many bricks of each dimension 25cm × 10cm × 55cm are required to pave
the bricks for the wall of dimension 11m × 11m × 3m?

b. If 35,000 bricks of the same size are needed to build a wall of dimensions 14.5m
× 6.3m × 30cm, what will be the volume of each brick?

6. a. 15,000 bricks of size 15cm × 8cm × 5cm are required to construct a wall. Find the
volume of the wall.

b. Find the cost of the stones each of volume 0.015m³, which are required to
construct the wall of volume 120m³ if the cost per stone is Rs. 20.

c. A lidless wooden box of external dimensions 52cm × 42 cm × 30cm has thickness
1cm. Find the volume of wood used in making the box.

C. Long Questions

7. The length, breadth and height of a room are 8m, 5m and 4m respectively. It has
a door of size 2.2m × 1.6m and 2 windows each of size 1.9m × 1.4m. Find the cost
painting its walls at the rate of Rs. 20 per sq meter.

8. The length, breadth and height of a wall are 50m, 30cm and 3m respectively. There
are 3 windows each of size 3m × 1.5m and two doors each of size 1m × 1.5 m. Find
the cost of constructing the wall at the rate of Rs. 5 per brick if the size of the a brick
is 12cm × 10cm × 8cm.

98 GREEN Mathematics Book-9

9. 23,500 bricks each of size 25cm × 16cm × 10cm are required to construct a wall 40m
long and 50cm wide containing four windows each of size 2m × 1.5 m each . Find the
height of the wall.

10. The length and breadth of a room are 12m and 8m respectively. If the volume of the
room is 576 cubic metre, find the cost of painting its walls and ceiling at the rate of
Rs. 30 per sq metre.

11. The length of a room is two times its breadth and the breadth is two times its height.
If the volume of the room is 512m³, find the cost of plastering its walls at Rs. 5.50 per
m².

12. A square room contains 256m³ of air. If the cost of carpeting its floor at Rs. 55 per m²
is Rs. 3520, find the cost painting its walls at Rs. 12.50 per sq metre.

13. The volume of a room is 480 m³ and height is 4m. Find the cost of carpeting its floor
at the rate of Rs 50 per square metre.

14. A square room contains 1800m³ of air. If the cost of painting its 4 walls at Rs. 20 per
m² is Rs. 7200, find the height of the room.

15. The cost of carpeting a square room at Rs. 36 per m² is Rs. 1764. If the cost of papering
its walls at Rs. 16 per m² is Rs. 2240, find the height of the room.

Project Work:
Measure dimension of your classroom, bedroom, studyroom and school compound
walls, then calculate the following:
a. Total surface area
b. Find the cost of painting its walls at Rs. 3.50 per sq. m.
c. Estimate the number of bricks to construct each types of walls.

GREEN Mathematics Book-9 99

6

Algebraic Expressions

Estimated Teaching Periods : 6

Muhammad ibn Mūsā al-Khwārizmī, formerly Latinized as Algoritmi,
was a Persian mathematician, astronomer, and geographer during the
Abbasid Caliphate, a scholar in the House of Wisdom in Baghdad.

Contents

6.1 Factorization
6.2 Factorization of a4 + a2b2 + b4
6.3 Introduction / Revision

Objectives

At the end of this unit, students will be able to:
introduce factorization, factorise algebraic expressions in the form of a2 - b2, a3 + b3,
a3 - b3, (a - b)3, ax2 ± bx + c, a4 + a2b2 + b4.
know the laws of indices and solve the problems related to indices
introduce exponential equations and solve the problems related to exponential
equations.
introduce ratio, proportion and the properties of proportion and solve the problems
related to ratio and proportion

Materials

Models of (a + b)2, (a - b)2, a2 - b2, a3 + b3 and a3 - b3, card board, thread, chart paper,
gum, panel board, chart of laws of indices, etc.

100 GREEN Mathematics Book-9

6.1 Factorization

When two or more than two algebraic expressions are multiplied, the result is called the
product and each expression is called factor of the product. The process of finding out
factors of an algebraic expression is known as factorisation.

Worked Out EXAMPLES

EXAMPLE 1 Factorize: 3x + 3y
Solution : 3x + 3y
= 3(x + y)

EXAMPLE 2 Factorize: x² – 3a + 3x – ax
Solution : x² + 3x – ax – 3a
= x(x + 3) – a(x + 3)
= (x + 3) (x – a)

EXAMPLE 3 Factorize: x³ – 4x
Solution : x3 – 4x
= x(x² – 4)
= x(x² – 2²) a2 – b2 = (a + b) (a – b)
= x(x – 2) (x + 2)

EXAMPLE 4 Factorize: a³ + 1 – a² – a
Solution : a3 + 1 – a2 – a
= a³ – a² – a + 1
= a² (a – 1) – 1 (a – 1) Arranging in order
= (a – 1) (a² – 1)
= (a – 1) (a – 1) (a + 1) GREEN Mathematics Book-9 101
= (a – 1)² (a + 1)

EXAMPLE 5 Factorize: x³ + x² + x + 1

Solution : x3 + x² + x + 1
= x² (x + 1) + 1 (x + 1)
= (x + 1) (x² + 1)

EXAMPLE 6 Factorize: x³y² – 1
Solution : 4x

x³y² – 1
4x

= 1 ox4y² – 1 p
x4

=1 s(x² y)² – 12

o pt
x2

= 1 (x² y + 1)(x² y – 1)
x 22

EXERCISE: 6.1

A. Very Short Questions

Factorize the following algebraic expressions.

1. a. 2x + 4y b. 3ax – 3ay c. x²y + xy²

2. a. a(x – y) + b(x – y) b. 3ax – bx – 3ay + by
c. x²yz + z²x – xy² – yz
B. Short Questions d. 25x² – 1 e. x4 – x–4
36y²

Factorize the following algebraic expressions.

3. a. 64b² – 9c² b. 16p4 – 81q4 c. 3x3 – 6x²y + 3xy²
e. x² – 4a + 4x – xa
d. 2a² (3x – 2y) – 5b² (3x – 2y)

f. x² a² + x4 g. 9x²a² – 36 h. 3a² – 2a + 27a – 18

i. 3x³y – 243xy³ j. 2a8 – 18a²b²

4. a. (3a + 2b)² – (3a – 2b)² b. 5x + y + 25x² – y² c. a² + b² – 2ab – c²
f. 2 – 72
d. p4 + 2p³q + q²p² e. x² – 2x + xy – 2y
49x²
g. a³b² – 1 h. 441 – x²
4a a²b² y²z² i. 25a² – 9b
16b

102 GREEN Mathematics Book-9

5. a. mn²(m – p) – p + m b. a² + 2a + 1 – b² c. a² – 9 – 6b – b²
d. (2x + 5)² – 16y² e. x³ – x² – x + 1 f. 4 – (12a – 4)²
g. 400(m + n)² – (m – n)²
C. Long Questions b. 4x2 – 16z2 + 12xy + 9y2
d. p8 – q8
6. a. 64a6b – b7

c. x2 – 8x – 9 – y2 + 10y

e. yx44 – 44x2 + 16
y2

6.2 Factorization of a4 + a2b2 + b4

For the factorization of expression a4 + a2b2 + b4, we should reduce the algebraic expression
into the difference of two squares then we should apply the formulae of a2–b2 = (a+b)(a–b).
Look at the following workout examples to be more clear about it.

Worked Out EXAMPLES

EXAMPLE 1 Factorize: a4 + 4b4 a2 + b2 = (a + b)2 – 2ab
Solution : a4 + 4b4
= (a2)2 + (2b2)2
= (a2 + 2b2)2 – 2 . a2 . 2b2
= (a2 + 2b2)2 – (2ab)2
= (a2 + 2b2 + 2ab) (a2 + 2b2 – 2ab)
= (a2 + 2ab + 2b2) (a2 – 2ab + 2b2)

EXAMPLE 2 Factorize: x2 – 4xy + 4y2 – z2
Solution :
x2 – 4xy + 4y2 – z2
= x2 – 2 . x . 2y + (2y)2 – z2
= (x – 2y)2 – z2
= (x – 2y)2 – z2
= (z – 2y + z) (x – 2y – z)

GREEN Mathematics Book-9 103

EXAMPLE 3 Factorize: a4 + a2b2 + b4

Solution : a4 + a2b2 + b4

= (a2­)2 + (b2)2 + a2b2 a2 + b2 = (a + b)2 – 2ab
= (a2 + b2)2– 2a2b2 + a2b2 Divide 4ab = 2ab + 2ab
= (a2 + b­2)2 – a2b2
= (a2 + b­2)2 – (ab)2
= (a2 + b2 + ab) (a2 + b2 – ab)
= (a2 + ab + b2) (a2 – ab + b2)
Alternative Method :

a4 + a2b2 + b4

= (a2­)2 + 2 . a2 . b2 + (b2)2 – a2b2
= (a2 + b2)2 – (ab)2
= (a2 + b2 + ab) (a2 + b2 – ab)
= (a2 + ab + b2) (a2 – ab + b2)

EXAMPLE 4 Factorize: x2 – 3 + 1
Solution : x2

x2 – 3 + 1
x2

( )= x2 – 2 . x . 1 + 1 2 – 1
xx

( )= x – 1 2 – 12
x

( ) ( )= x – 1 + 1 x – 1 – 1
xx

( ) ( )= x + 1 – 1 x – 1 – 1
xx

EXAMPLE 5 Factorize: (1 – a2) (1 – b2) + 4ab
Solution : (1 – a2) (1 – b2) + 4ab

= 1 – a2 – b2 + a2b2 + 4ab
= 1 + 2ab + a2b2 – a2 + 2ab – b2
= {12 + 2 . 1 . ab + (ab)2} – {a2 – 2 . a . b + b2}
= (1 + ab)2 – (a – b)2
= (1 + ab + a – b) {1 + ab – (a – b)}
= (1 + ab + a – b) (1 + ab – a + b)

104 GREEN Mathematics Book-9

EXAMPLE 6 Factorize: x2 – 6x – 7 – 8y – y2
Solution : x2 – 6x – 7 – 8y – y2

= x2 – 2 . x . 3 + 32 – 16 – 8y – y2
= (x – 3)2 – (16 + 8y + y2)
= (x – 3)2 – (42 + 2 . 4 . y + y2)
= (x – 3)2 – (4 + y)2
= {(x – 3 + (4 + y)} {(x –3) – (4 + y)}
= (x – 3 + 4 + y) (x –3 – 4 – y)
= (x + y + 1) (x – y – 7)

EXAMPLE 7 If ab – 1 = 4, find the value of a2b2 + 1 .
Solution : ab a2b2

Here, ab – 1 = 4
ab

Now, a2b2 + 1
a2b2

( )= (ab)2 + 1 2
ab

( )= ab – 1 2 + 2 . ab . 1
ab ab

= (4)2 + 2

= 16 + 2

= 18

EXAMPLE 8 Factorize : 4x4 – 37x2y2 + 81y4

Solution : 4x4 – 37x2y2 + 81y4

= (2x2)2 – 2 . 2x2 . 9y2 + (9y2)2 – x2y2
= (2x2 – 9y2)2 – x2y2
= (2x2 – 9y2)2 – (xy)2
= (2x2 – 9y2 + xy) (2x2 – 9y2 – xy)
= (2x2 + xy – 9y2) (2x2 – xy – 9y2)

GREEN Mathematics Book-9 105

EXERCISE: 6.2

A. Very Short Questions

1. Factorize the following: b. a2 – 2ab + b2 – c2 c. a4 – b4
a. a2 – (b – c)2

2. Simplify :

a. x2 – 3x – 4 b. x2(–x 2–x1+) 1 c. 4x(24+x 5+x1+) 1
x –4

B. Short Questions

3. Factorize: b. 4a4 + b4 c. x4 + 4
e. a2 + 2a + 1 – b2 f. a2 – 9 – 6b – b2
a. x4 + 4y4 h. a2 + b2 – 2ab – c2
d. a2 – b2 + 2b – 1
g. 25(m + n)2 – (m – n)2

C. Long Questions b. x2 + 8x – 9 – a2 – 10a
4. Factorize the following expressions: d. (p2 – q2) (r2 – s2) + 4pqrs
a. (b2 – bc + c2)2 – (b2 + bc + c2)2

c. x2 – 6x – 7 – 8y – y2

e. (1 – x2) (1 – y2) + 4xy f. 4x2 – 16z2 + 12xy + 9y2

5. Simplify the following:

a. a2 – 2bc – c2 – b2 b. x2 – 2xy + y2 – z2
(a – b – c) x2 – y2 + 2yz – z2

c. a4 + a2b2 + b4 d. (1 – x2) (1 – y2) – 4xy
a3 + b3 (1 – x – xy + y)
6. a. If x – 1 = 4, find the value of x2 + 1 .
x x2
b. If 2x + 1 = 5, find the value of 4x2 + 1 .
2x 4x2

c. If ab = 10, a – b = 5, find the value of a2 + b2.

d. If x + y = 6, xy = 8, find the value of x2 + y2.

e. If a2 + 1 = 18, find the value of a – 1 .
a2 a

f. If m2 + 1 = 23, find the value of m – 1 .
m2 m

g. If a2 + b2 = 45 and ab = 10, find the value of a – b.

106 GREEN Mathematics Book-9

6.3 Introduction / Revision

Factorisation of the expression a³ + b³ and a³ – b³.

Here, we have,

(a + b)³ = a³ + b³ + 3ab (a + b)

∴ a³ + b³ = (a + b)³ – 3ab (a + b)

= (a + b) [(a + b)² – 3ab]

= (a + b) (a² + 2ab + b² – 3ab)

= (a + b) (a² – ab + b²)

Similarly,

(a – b)³ = a³ – b³ – 3ab (a – b)

∴ a³ – b³ = (a – b)³ + 3ab (a – b)

= (a – b) [(a – b)² + 3ab]

= (a – b) (a² – 2ab + b² + 3ab)

= (a – b) (a² + ab + b²)

Worked Out EXAMPLES

EXAMPLE 1 Factorize : 27a³ – 125b³

Solution : Here,
27a³ – 125b³
= (3a)³ – (5b)³ ∵ a³ – b³ = (a – b) (a² + ab + b²)
= (3a – 5b) {(3a)² + 3a. 5b + (5b)²}
= (3a – 5b) (9a² + 15ab + 25b²)

EXAMPLE 2 Factorize : 64x6 – 1
Solution : 64y6

Here, 64x6 – 1
64y6

= (8x3)2 – o 1 3
8y3 p

= o8x3 + 1 p o8x3 – 1 p
8y3 8y3

= s(2x)3 + o 1 p3ts(2x)3 – o 1 3
2y 2y
pt

= o2x + 1 po4x2 – x + 1 o2x – 1 po4x2 + x + 1
2y y 4y2 p 2y y 4y2 p

GREEN Mathematics Book-9 107

EXAMPLE 3 Factorize : (x – 5)³ + (x + 5)³

Solution : Here,
= {(x – 5) + (x + 5)} {(x – 5)² – (x – 5) (x + 5) + (x + 5)²}
= (x – 5 + x + 5)(x² – 10x + 25 – x² + 25 + x² + 10x + 25)
= 2x(x² + 75)

EXAMPLE 4 Factorize : 125x³ – 15x + 3 – 1
Solution : 5x 125x³

Here,

= (5x)³ – 3.(5x)² o 51xp . + 3. 5x o 1 p² – o 51xp³ ∵ a³ – 3a²b + 3ab² – b³ = (a – b)³
5x

= o5x – 1 p³
5x

EXAMPLE 5 If x = 1 – –1 prove that x(x² + 3) = a – 1 .
Solution : a
a3 a 3,

11

Here, x = a3 – a– 3

L.H.S. = x(x² + 3)

= x³ + 3x

11 11
= oa3 – a– 3 p3 + 3 oa3 – a– 3 p

11 1 11 1
= oa3 – a– 3 p3 + 3 a3 . a– 3 oa3 – a– 3 p


13 –1 3
= oa3 p – oa 3 p


= a – a– 1

= a – 1 = R.H.S. proved.
a

EXERCISE: 6.3

A. Very Short Questions

Factorize the following algebraic expressions.

1. a. 8x³ + 27y³ b. o a p³ – o b p³ c. 2(x + 2)³ – 16
2. a. x4y² – xy5 b a

b. (m + n)³ – n³ c. 8 – (x + 1)³ d. (a + b)³ + (a – b)³

3. a. a³ + 9a²b + 27ab² + 27b³ b. x³ + 3x + 3 + 1 c. 8a³ + 36a²b + 54ab² + 27b³
x x³

108 GREEN Mathematics Book-9

B. Short Questions b. o pq p³ – o q p³ c. 8m³ + n³
4. a. 27a³ + 125b³ p n³ 8m³

d. (a – b)³ + 125 e. (x – y)³ + 27 (x + y)³ f. 1 + 512 x
x2
16
g. (x + y)³ + 2 h. (x + y)³ + 64 i. a³ – 8x³ + 2ay – 4xy

j. 8(a + b)³ – (a – b )³ k. 8a³ – 4a²b + 2ab² – b³ l. (a – b)³ + 8

5. a. 16xy4 + 54x4y b. (x³ – y³) + (x² – y²) c. 8(p + q)³ – (p – q)³

d. (x + 1)³ – 125 e. (a – 1)³ – 1 f. 2x4 –16 xy³

g. 16a³ – 1 h. acb³ + 125ab i. (a x³ – 729
32b³ – b)³ (a – b)³

6. a. m³ + 9m²n + 27mn² + 27n³ b. 8a³ – 60a² + 150a – 125

c. (m³ – n³) + (m² – n²) d. (a³ – b³) – (a² – b²)

e. 125 – 150x + 60x² – 8x³ f. 8x³ + 6x + 3 + 1 g. x³ + y³ + x + y
2x 8x³ j. m³ – 8n³ + 2mp – 4np
125
h. a²b² – ab i. x9 – 512y9

B. Long Questions

7. a. If a + b = 6 and ab = 7, find the value of a³ + b³

b. If x – 1 = 9, find the value of x³ – 1 .
x x³

c. If a + 1 = 6, find the value of a³ + 1 .
a a³

d. If x + 1 = a, prove that x³ + 1 = a³ – 3a
x x³

e. 1 1 show that x³ + 3x + 1 = mn
mn
If x = (mn)3 – (mn)– 3 ,

Worked Out EXAMPLES

EXAMPLE 1 Factorize: 9y3 – 144x2y
Solution : 9y3 – 144x2y
= 9y (y2 – 16x2)
= 9y {y2 – (4x)2}
= 9y (y + 4x) (y – 4x)

GREEN Mathematics Book-9 109

EXAMPLE 2 Factorize: a4 – 7a2 + 1
b4 b2

Solution : a4 – 7a2 + 1
b4 b2

= o a2 2 a2 + 12 – 9a2
b2 b2 b2
p +2.

= o a2 2 3a 2
b2 b
+ 1p – o p

= o a2 +1– 3a po a2 +1+ 3a p
b2 b b2 b

EXAMPLE 3 Factorize: (x2 – 3x)2 – 11(x2 – 3x) + 28

Solution : (x2 – 3x)2 – 11(x2 – 3x) + 28

= (x2 – 3x)2 – 7(x2 – 3x) – 4(x2 – 3x) + 28
= (x2 – 3x) (x2 – 3x – 7) – 4(x2 – 3x – 7)
= (x2 – 3x – 7) (x2 – 3x – 4)

EXAMPLE 4 Factorize: a4 + (2y2 – x2) a2 + y4

Solution : a4 + (2y2 – x2) a2 + y4

= a4 + 2a2y2 – x2a2 + y4
= a4 + 2a2y2 + y4 – x2a2
= (a2 + y2)2 – (xa)2
= (a2 + y2)2 – (xa)2
= (a2 + y2 + xa) (a2 + y2 – xa)
= (a2 + xa + y2) (a2 – xa – y2)

EXAMPLE 5 Factorize: x2 + 10x + 24 – 2y – y2

Solution : x2 + 10x + 24 – 2y – y2
= x2 + 2 . x . 5 + 52 – 1 – 2y – y2
= (x + 5)2 – (1 + 2y + y2)
= (x + 5)2 – (1 + y)2
= (x + 5 + 1 + y) (x + 5 – 1 – y)
= (x + y + 6) (x – y + 4)

110 GREEN Mathematics Book-9

EXERCISE: 6.4

A. Very Short Questions
Factorize following algebraic expression.

1. a. x³ – 49xy² b. 3x4 – 6x² + 3 c. 2a² + 9ab + 4b²

d. 3a² + 7ab – 20b² e. p6 – 13p3 + 42 f. a4 – 5a2 + 4 g. x4 – y4

B. Short Questions

2. a. a5 +­ 4ab4 b. x4 + 4y4 c. x4 + x² + 1

d. yx44 – 7x² +1 e. a² – 12a – 28 + 16b – b² f. x4 – 9
y²

g. 4 – m4 – 2m²n² – n4 h. p4 – 6p²q² + q4 i. m4 – 6m² – 12n – 4n²

j. b4 – 2b²c² + c4 – 36 k. 3(x – y)² – 12z² l. 4(x4 + y4) – 17x²y²

C. Long Questions b. a² + 1 + 1
3. a. 4p4 – 35p²q² + 121q4 a²

c. mn44 – 44m² + 16 d. 169m² – 52m – 196n² – 56n
n²

e. t² – 10t + 24 + 6r – 9r² f. x10 – 10x6 + 9x²

g. x4 – 8x² – 33 – 14y – y4 h. x4 + 2 oy² – 9z² px ² + y4
2

i. a4 + (2y² – x²) a² + y4 j. a2 + 10a + 24 + 6b – 9b²

k. a4 + 1 – 7 l. x6 + x4 + x²
a4 y4 y4 y4

m. (x2 – 3x)2 – 11(x2 – 3x) + 28 n. (a2 – 5a)2 + 10(a2 – 5a) + 24

o. ba44 + 8a2 + 36
b2

GREEN Mathematics Book-9 111

7

Laws of Indices

Estimated Teaching Periods : 8

Body Mass Index, derived from a simple math formula, was devised in
the 1830s by Lambert Adolphe Jacques Quetelet (1796-1874), a Belgian
astronomer, mathematician, statistician and sociologist. BMI is said to
estimate how fat you are by dividing your weight in kilograms by your
height in meters square. However, as mentioned earlier, the measurement
is flawed, especially if the person carries a lot of muscle.

Contents

7.1 Indices
7.2 Exponential equation

Objectives

At the end of this unit, students will be able to:
identify the base, index or exponent or power.
identify the laws of indices.
solve the problems involving the laws of indices

Objectives

Blocks, chart papers and geo-board.
Rule of indices in chart paper.

112 GREEN Mathematics Book-9

7.1 Indices

Index is the power of the base of an algebraic 3x5 Index
term. It contains coefficient, base and index. Base
For example in 3x5, x is base, 5 is index and 3 is Coefficient
its coefficient.

Laws of indices:

1. xa × xb = xa + b 2. xa ÷ xb = xa – b 3. (xa)b = xab

11 b xa a
4. x– a = xa 5. x– a = xa 6.
= xb

7. x a = y – a xa 9. If xa = ya then x = y
y x 8. xa = xa–a = x0 = 1 12. xn = m i.e. n m = x
o p o p

1

10. If xa = xb then a = b 11. a xb = xab

Worked Out EXAMPLES

4

EXAMPLE 1 Find the value of : 27 3

Solution : 4

27 3

4

= (3)3 × 3

= 34

= 81

–1

EXAMPLE 2 Find the value of : (2401) 4

Solution : –1

(2401) 4

–1

= (74) 4

= (7)4 ×–4 1

= 7– 1

= 1
7

GREEN Mathematics Book-9 113

EXAMPLE 3 Find the value of : q 3 64 ÷ o 1 –1 2
125 3

pr

Solution : q 3 64 ÷ o 1 –1 2
125 3

pr

3× 1 13 × –1 2
3
= q4 3 ÷ –1 r

53× 3

= q4 ÷ 1–1 2
5–1
r

= [4 ÷ 5]2

= q 4 2
5
r

= 16
25

–3 –3

EXAMPLE 4 Find the value of : 81 4 × 25 2 ÷ 5 –3
16 9 2
o p qo p o pr

–3 –3

Solution : 81 4 × 25 2 ÷ 5 –3
16 9 2
o p qo p o pr

= 3 4× –3 5 2 × – 3 5 –3
2
o p 4 × qo p ÷ o pr
2 32

= o 3 –3 qo 5 –3 5 –3

p× p ÷o pr
2 32

= o 2 3 qo 3 3 o 2 3

p× p÷ pr
3 55

= 8 qo 27 p× 125

o pr
27 125 8

= 8 × 27 × 125
27 125 8

= 1

114 GREEN Mathematics Book-9

EXAMPLE 5 Find the value of : o xa a – b × o xb b – c × o xc p c – a
Solution : x–b x–c x–a
p p

o xa a – b × o xb b – c × o xc p c – a
x–b x–c x–a
p p

(a + b) (a – b) (b + c)(b – c) (c + a) ( c – a)
= x × x × x

= xa² – b² × xb² – c² × xc² – a²

= xa² – b² + b² – c² + c² – a²

= x0

= 1

EXAMPLE 6 Find the value of : (xa ÷ xb)a² + ab + b². (xb ÷ xc)b² + bc + c² . (xc ÷ xa)c² + ca + a²

Solution : (xa ÷ xb)a² + ab + b². (xb ÷ xc)b² + bc + c² . (xc ÷ xa)c² + ca + a²
= (xa–b)a² + ab + b². (xb–c)b² + bc + c² . (xc–a)c² + ca + a²
= x­a³–b³ . xb³–c³ . xc³ – a³
= x­a³ – b³ + b³ – c³ + c³ – a³
= x0
= 1

EXAMPLE 7 Simplify: ab xa × bc xb × ca xc
xb xc xa

Solution : ab xa × bc xb × ca xc
xb xc xa

a bc

= x ab × x bc × x ca

b c a

x ab x bc x ca

1 11

= x b × x c × x a
1 11
xa xc xc

1+1+1–1–1

= x b c a b c

= x0

= 1

GREEN Mathematics Book-9 115

EXAMPLE 8 Find the value of : 1 + 1 + 1
Solution : 1 + xa–b + xa–c 1 + xb–c + xb–a 1 + xc–a + xc–b

L.H.S. = 1 + 1 + 1

1 + xa–b + xa–c 1 + xb–c + xb–a 1 + xc–a + xc–b

= 1 × x–a + 1 × x–b + 1 × x–c

1 + xa–b + xa–c x–a 1 + xb–c + xb–a x–b 1 + xc–a + xc–b x–c

= x–a + x–b + x–c
x–a + x–b + x–c x–b + x–c + x–a x–c + x–a + x–b

= (x–a + x–b + x–c)
(x–a + x–b + x–c)

= 1

EXAMPLE 9 Find the value of : o xxa+cbpa – b × o xb+c b – c × o xc+a p c – a
Solution : xa xb
p

o xa+b a – b × o xb+c b – c × oxxc+bap c – a
xc xa
p p

x­ x­ x­(a + b) (a – b)
(b + c) (b – c) (c + a) (c – a)

= x x x×
c(a – b)
× b(c – a)
a(b – c)

xa­² – b² x­b² – c² x­c² – a²
= ××
ab – ca bc – ab
x x xca – bc

= xa² – b² + b² – c² + c² – a²
xca – bc + ab – ca + bc – ab

= x0
x0

= 1
1

= 1

EXAMPLE 10 Find the value of : 2 × 2n+ 1 (n – 1) (n + 1)
Solution : 2n­ (n – 1) × 4(n + 1)

2 × 2n+ 1 (n – 1) (n + 1)
2­n(n – 1) × 4(n + 1)

= 2n+ 1 × 2n2 – 1
2­n2 – n × 22n + 2

= 2n + 1 + n2 – 1
2­n2 – n + 2n + 2

= 2n + n2 – n2 – n – 2

= 1
22

= 1
4

116 GREEN Mathematics Book-9

EXERCISE: 7.1

A. Very Short Questions

1. Find the value of:

1 b. (216)– 1 c. 51 20143
3
a. 273
3
d. (2401)– 1 e. 49 – 2 f. (81)– 3
4 o 81 p 2

2. Simplify :

6 ÷ x– 1 b. 3a 4 ÷ a– 4 c. x a–3 × 2x a–7
5 5
a. x7 3

3. Simplify:

a. 3 × 3x + 1 (x – 1) (x + 1) xa a² + ab + b²× xb b² + bc + c² × xc c² + ca + a²
xb xc xa
3x(x–1) × 9x+1 b. o p o p o p

c. x2m+3n × x3m–6n d. x2a + 3b × x3a – 6b
xm+2 × x4m – 4n xa + 2b x4a – 4b

B. Short Questions

4. Find the value of: 1

a. 27– 2 ÷ 16 – 3 b. qo 25 – 3 ÷ o 5 –3 c. 1 d. 5 x5y10
3 2 243
4 p p r
92 5

2 3 216 – 1 23
o8p
e. (8a³ ÷ 27b– 3)­ 3 f. g. (23013)2

h. o 3 5 × o 3 12 i. 27 – 2 × – 2
o 64 p 3
625 p3 3125p 3 8 3

5. Simplify: oa + 1 px × 1 – apx

o
bb
a. 3 8a² ÷ 8 a–5 4 x– 3 o8 – 12 1 px 1
b. ­ ÷ c. ob + × – bpx
xp o

24 1 1 3 1 aa
2 2p
d. a 3 oa3 ÷ a3p e. a– oa2 – a– f. m–1 (m–1 – 2m)

1– 14 ÷ 4x– 1 y–23 h. o 297––­84 –3 3– 3 1 m5 × 3 n²
3 3–­6 6 n–² × 4 m10
g. o8x2y 2p3 p 4×o 3 i.

p

j. 6 a5 4 a³ a² k. 3 × 3n+ 1 (n – 1) (n + 1) l. x2a + 3b × x3a – 6b
3n­ (n – 1) × 9 (n + 1) xa + 2b × x4a – 4b

m. aa45mm + 3n × am – 9n n. o xm m + n × o xn n + p ÷ 3(xm . xp) m– p
– 6n xn xp
p p

GREEN Mathematics Book-9 117

C. Long Questions

6. Simplify : 1a 1 b–a
y²p yp
ox² – × ox – 1 a–b
xp
a. ab xa × bc xb × ca xc b. oy² – x1²pb × oy +
xb xc xa

c. ox a² – b² 1 . ox 1 1 q+r 1 . r+p 1 . p+q 1
pa+b
. xb² – c² pb + c o c² – a² p c + a d. ox r – p pp – q xo p – q p q – r xo q – r p r – p

e. a + b xa² × b+c xb² × c+a xc² f. o xp ×p² + pq + q² o xq q² + qr + r² × xr r² + pr + p²
xb² xc² xa² xq xr o xp p
p p

g. ab a – b × bc b – c × ca c – a h. s(x)a – a1 t a 1 1
a b +
xb xc xa
c

op + 1 m × o 1 – ppm j. ooyx ++ y1x1 ppmm ×× ooxy––y1x1 n
q q
p p

i. + 1 m × o 1 – qpm n
oq p p
p p



k. (xa + b . yc)a – b­(xb + c .ya)b – c . (xc + a . yb)c – a l. o xa pa – b . o xb pb – c . o xc pc – a
x–b x–c x–a

7. Prove that:

a. 1 + 1 + 1 =1
1 + xb – a + xc – a 1 + xa – b + xc – b 1 + xa – c + xb – c

b. xa(a + b) . xb(b + c) . xc(c + a) . x–a² – b² – c² = 1
xab xbc xca

c. 1 + 1 – a + 1 1 = 1
xb + xa – b

d. If a³ + b³ + c³ = 1, prove that : o xa a² – ab + b²× xb b² – bc + c² × xc c² – ca + a² = x²
x–b ox–cp ox–ap
p

e. If x + y + z = 0, prove that a(xy)–1. a(­ yz)­ –1. a­(zx­)–1 = 1

f. If a = xq + r × yp, b = xr+P . yq and c = xp+p. yr, then prove that aq–r × br – p × cp – q = 1

118 GREEN Mathematics Book-9

7.2 Exponential equation

An algebraic equation, in which unknown variable appears in the exponent of base is
known as exponential equation.

Rules for exponential equation

1. If bases are equal, then powers of bases are also equal.
i.e. If xa = xb then a = b
2. If powers are equal, then bases are also equal.
i.e. If xa = ya then x = y.

1

3. If ax = b1, then a = b x

y

4. If ax = by­ then a = (b) x

Worked Out EXAMPLES

EXAMPLE 1 Solve the following exponential equations.
Solution :
x+5 x+1 c. 5x = 1
0.04
a. 2x+4 = 8x b. 9 2 = 81 2

Here, Here, Here,
8x = 2x+4
or, (2³)x = 2x + 4 x+5 x+1 5x = 0.104
or, 23x = 2x + 4 or, 5x = 1040
9 2 = 81 2
or, 5x = 25
x+5 = 9 2×x + 1
2
or, 9 2

x+5

or, 9 2 = 9 x + 1

or, 3x = x + 4 or, x + 5 = x + 1 or, 5x = 52
or, 3x – x = 4 2
or, 2x = 4 ∴ x = 2
∴ x = 2
or, x + 5 = 2(x + 1)

or, 2x + 2 = x + 5

or, 2x – x = 5 – 2

∴ x = 3

GREEN Mathematics Book-9 119

EXAMPLE 2 Solve : ax–2 . b = bx–2 . a

Solution : ax–2 . b = bx–2 . a

or, ax–2 . b1 = 1
bx–2 . b

or, ax–2–1 =1
bx–2–1

or, o a px – 3 = 1
b

or, o a px – 3 = o a p0
bb

or, x – 3 = 0

∴ x = 0

EXAMPLE 3 Solve : 32x+1 – 9x+1 + 54 = 0

Solution : 32x+1 – 9x+1 + 54 = 0
or, 32x . 31 – 32x . 32 = – 54

or, 32x(3 – 9) = – 54 Comparing both sides
or, 32x = – 54 2x = 2
x=1
–6
or, 32x = 9

or, 32x = 32

EXAMPLE 4 If xa = yb = zc = and y = 8 xz , prove that 2 = 1 + 1 .
Solution : b a c

Given xa = yb = zc

Then, x = b and z = ycb

ya

Now, y = xz

or, y = y b . y b
a c

or, y = y b , y b
2a 2c

or, y = y b + b
2a 2c

or, y = y b ( 1a + 1 )
2 c

or, 1= b ( 1a + 1 )
c

2
∴ 2 = 1 + 1
bac

120 GREEN Mathematics Book-9

EXAMPLE 5 If x = 1 3– 1 , prove that 3x³ – 9x – 10 = 0
3
33+

Solution : L.H.S. = 3x³ – 9x – 10

= 3(x³ – 3x) – 10

1 –1 1 –1
{ } = 3 o33 + 33 p³ – 3 o33 + 3 3 p – 10

1 –1 1 –1 1 –1
{ } = 3 o33 + 3 3 p³ – 3. 33 . 3 3 o33 + 3 3 p – 10

1 –1

{ } = 3 o33 p³ + o3 3 p³ – 10 [∵ (a + b)³ – 3ab (a + b) = a³ + b³]

= 3{3 + 3– 1­}– 10
= 3o3 + 1 p– 10

3
= 3 × 10– 10

3

= 10 – 10

= 0 ∴ L.H.S. = R.H.S. Proved.

EXERCISE: 7.2

A. Very Short Questions

1. Solve for 'x':

a. 3x + 1 = 1 b. 2x + 2 = 1 c. 52x + 1 = 0.04
4

d. 3 = 9x+1 e. b . ax – 1 = bx f. 4 2 = 16x
2. Solve for 'x':
a. 22x + 1 = 42x – 1 b. 5x + 2 – 5x = 600 c. 4x + 2 = 22x + 1 + 14
e. 3 × 81x = 9x+4 f. 2x+1 – 2x = 8
p2x + 1= o 5 h. 4x + 1 – 22­x = 12 i. 3 3x+1 = 3 1
d. 16 64o 3 px + 1

g. 3x + 1 – 3x = 54

B. Short Questions b. 4x + 2 – 4x = 240 c. 9x– 2 + 2 × 32x – 3 = 63
3. Solve: e. 22x + 1 ­– 4x+ 2 + 14 = 0
a. 32x + 1 = 92x – 1 f. 2x + 1 = 1
2–x 2
d. 3x– 2 + 3x = 10

GREEN Mathematics Book-9 121

g. 2x – 1 = 15 h. 3x + 3x – 2 = 3 1 i. 2c + 1 – 2c = 8
j. 2a – 2 + 2a = 5 3 l. 2x + 2 + 2x + 3 = 12
C. Long Questions 3
k. 3x + 1 + 31–x = 36

4. Solve : b. 23n – 5 × a n – 2 = 2 n – 3 × 2a1 – n
a. 22m –3 × 5m – 1 = 200 d. 7c + 3 – 7c + 5 + 336 = 0
c. 55p – 4 × a4p – 3 = 52p – 3 × ap – 2 f. 9x + 33 – 4 × 32x + 1 = 0
e. 9x – 10 × 32x + 9 = 0

5. Prove that xyz = 1 if x = yz, y = zx, z = xy

6. If am. an = (am)n, show that m(n – 2) + n(m – 2) = 0.

7. Ifa = 10x, b= 10y and axby = 100, prove that xy = 1.

8. If ax = by = cz and b² = ac, prove that 1 + 1 – 2 = 0.
x z y

9. If an – bn = 0, prove that m = n.
am bm

10. If x2 + 2 = 2 + 3– 2 , find the value of 3x3 – 9x.
3
33

11. If a2 – 2 = 2 + 2– 2 , find the value of 2a3 – 6a.
3
23

12. If m2 + 2 = 2 + 5– 2 , prove that 5m3 + 15m – 24 = 0.
3
53

122 GREEN Mathematics Book-9

8

Ratio and Proportion

Estimated Teaching Periods : 5

Body Mass Index, derived from a simple math formula, was devised in
the 1830s by Lambert Adolphe Jacques Quetelet (1796-1874), a Belgian
astronomer, mathematician, statistician and sociologist. BMI is said to
estimate how fat you are by dividing your weight in kilograms by your
height in meters square. However, as mentioned earlier, the measurement
is flawed, especially if the person carries a lot of muscle.

Contents

8.1 Ratio and proportion
8.2 Proportional (in proportion)
8.3 Properties of proportion

Objectives

At the end of this unit, students will be able to:
find the ratios of given problems.
find out the unknown values from the relation of proportion.
Compare two given ratios.
find out unknown values from given proportion.
solve the problems related on ratios and proportion.

Materials

Chart paper, balance, coloured sticks, etc.

GREEN Mathematics Book-9 123

8.1 Ratio and proportion

Ratio is the fraction which Ratio of a and b is a : b or a (Read as 'a' is to 'b')
is obtained by comparing b
two or more quantities of b
same kind. If 'a' and 'b' Ratio of b and a is b : a or a .
are two quantities, then.
In the ratio a : b . 'a' is antecedent and 'b' is consequent.

Remember

1. Duplicate ratio of a is a² .
b b²
1
2. Sub-duplicate ratio of a is a or
b b a2

1

3. Triplicate ratio of a is a³ b2
b b³
1
4. Sub - triplicate ratio of a is 3a
b 3b or a 3
1

b3

Worked Out EXAMPLES

EXAMPLE 1 Express as ratio of a to b in each of the following statements.

Solution : a. a is double of b

then a = 2b

∴ a = 2
b 1

∴ a : b = 2 :1

b. a is three – fourth of b

then, a = 3 b
4

a:b=3:4

124 GREEN Mathematics Book-9

EXAMPLE 2 If 5x – 2y = 3x + 4y, find : a. duplicate ratio of x : y
Solution :
b. triplicate ratio of x : y

Here,

5x – 2y = 3x + 4y

or, 5x – 3x = 2y + 4y

or, 2x = 6y

or, x 63
y = 2 1

∴ x = 3
y 1

a. The duplicate ratio of x = o 3 p2 =9:1
y 1

b. The triplicate ratio of x = o 3 p3 = 27 : 1
y 1

EXAMPLE 3 If a = 3 , find the value of 5a – 3b .
Solution : b 4 7a + 2b

We have, a = 3
b 4

3b
or, a = 4

Then, 5a – 3b
7a + 2b
Alternative Method:
3b
5 × 4 – 3b Given a = 3
b 4
= 3b
4
7× + 2b Let x be multiple of ratio, then

15b – 3b a = 3x, b = 4x
4
= Now, 5a – 3b
21b + 2b 7a + 2b

4 = 5.3x – 3.4x
7.3x + 2.4x
15b – 12b

4 15x – 12x
= = 21x + 8x

21b + 8b

4 = 3b × 4 = 3x
3b 4 29b 29x
4
= = 3 = 3
29b 29 29
4

GREEN Mathematics Book-9 125

EXAMPLE 4 If x:y = 7:9, find the ratio of 5x – 3y : 2x + y.
Solution :
x7 Alternate Method:
Here, y = 9
Give, x = 7
x y y 9
= 7 = 9 = k (let)
5x – 3y
= 2x + y

∴ x = 7k, y = 9k

Thus, 5x – 3y 5x – 3y x
2x + y = 5 y – 3
= y

= 5.7k – 3.9k 2x – y 2 x +1
2.7k + 9k y y

= 35k – 27k 5. 7 –3 35 – 27
14k + 9k 9 9
= =
8k 2. 7 +1 14 + 9
= 23k 9 9

= 8 = 8
23 23

EXAMPLE 5 If a:b:c = 2:3:5, find the value of 2a – 3b + 2c .
a – 2b + 4c

Solution : Here, a : b : c = 2 : 3 : 5
Let k be the multiple of ratio.

Then a = 2k, b = 3k, c = 5k

Now, 2a – 3b + 2c
a – 2b + 4c

= 2.2k – 3.3k + 2.5k
2k – 2.3k + 4.5k

= 4k – 9k + 10k
2k – 6k + 20k

= 14k – 9k
22k – 6k

= 5k
16k

= 5
16

126 GREEN Mathematics Book-9

EXAMPLE 6 If a : b = 2 : 3 and b : c = 4 : 5, find the value of a : b : c.

Solution : a = 2 , b = 4
b 3 c 5

or, a = 2b , c = 5b
3 4

Now, a : b : c = 2b : b : 5b Dividing all ratio by b.
3 4 Multiplying all ratio by 12.

= 2 :1: 5
3 4

= 2 :1: 5
3 4

= 2 × 12 : 1 × 12 : 5 × 12
3 4

= 8 : 12 : 15

∴ The ratio of a : b : c = 8 : 12 : 15

EXAMPLE 7 What common number must be subtracted from each term of the ratio
Solution : 19:16, so that the resulting ratio will be 7 : 4?

Let the number be k

Then, 19 – kk = 7
16 – 4

or, 76 – 4k = 112 – 7k

or, 7k – 4k = 112 – 76

or, 3k = 36

or, k = 36
3

or, k = 12

∴ The required number is 12.

EXAMPLE 8 Find the value of 2x² + 2xy + 3y² , if 2x = 3y.
6(4x² – 6xy + 5y²)

Solution : Here, 2x = 3y

or, x = y
3 2

Let, x = y =k
3 2

Then, x = 3k, y = 2k

GREEN Mathematics Book-9 127

Now, 2x² + 2xy + 3y² = 2.(3k)² + 2 × 3k × 2k + 3. (2k)²
6(4x² – 6xy + 5y²) 6{4.(3k)² – 6 × 3k × 2k + 5.(2k)²}

= 2 × 9k² + 12k² + 3 × 4k²
6{4 × 9k² – 36k² + 5 × 4k²}

= 18k² + 12k² + 12k²
6(36k² – 36k² + 20k²)

= 42k²
120k²

= 7
20

EXAMPLE 9 If 6x – y = 2 , find the value of x2 + y2 : x2 – y2.
5x + 4y 3

Solution : Here, 6x – y = 2
5x + 4y 3

or, 18x – 3y = 10x + 8y

or, 18x – 10x = 8y + 3y

or, 8x = 11y

or, x = 11
y 8

x2 + y2
Now, x2 – y2

x2 y2
= y2 + y2
x2 y2
y2 – y2

x2 121 + 64
o yp +1 121
=
=
x2 121 – 64
o yp –1 121

11 2 = 185
o 8p +1 57
=
o 181p2 – 1 ∴ x2 + y2 : x2 – y2 = 185 : 57

128 GREEN Mathematics Book-9

EXAMPLE 10 Twenty years ago, the ratio of ages of father and his son was 6:1. At
present, the ratio of their ages is only 8 : 3. Find their present ages.

Solution : Let

Father's present age be x years

and son's present age be y years

Twenty years ago,

Father's age was (x – 20) years

and son's age was (y – 20) years

According to the question

x – 2200 = 6 (i)
y – 1

yx = 8 (ii)
3

From (ii)

yx = 8
or, 3
Let
Then x = y
8 3

xy
8 = 3 = k

x = 8k

y = 3k

Substituting the values of x and y in (i)

8k – 20 = 6
3k – 20 1

or, 8k – 20 = 6(3k – 20)

or, 8k – 20 = 18k – 120

or, 18k – 120 = 8k – 20

or, 18k – 8k = 120 – 20

or, 10k = 100 Then, x = 8k = 8 × 10 = 80 years.
y = 3k = 3 × 10 = 30 years.
k = 100 ∴ Father's present age is 80 years.
10 son's present age is 30 years.

∴ k = 10

GREEN Mathematics Book-9 129

EXAMPLE 11 Two numbers are in the ratio of 3 : 7 and their sum is 120. Find the
numbers.

Solution : Here,

The ratio of two numbers = 3 : 7

Let k be the multiple of ratios.

∴ The first number = 3k

The second number = 7k

By the question:

7k + 3k = 120

or, 10k = 120

∴ k = 12

∴ The first number = 3k = 36

The second number = 7k = 84

EXAMPLE 12 Divide Rs. 3600 between Suvasha and Supreme in the ratio 5 : 4.

Solution : Given ratio = 5 : 4
Let x be the multiple of ratio
Then, = Rs. 2500
Suvasha's share = 5x = Rs. 1600
Supreme's share = 4x
Now,
5x + 4x = Rs. 3600
or, 9x = Rs. 3600
∴ x = Rs. 400
Suvasha's share = 5x = Rs. 5 × 500
Supreme's share = 4x = Rs. 4 × 400

130 GREEN Mathematics Book-9

EXAMPLE 13 Dipendra, Binesh and Rajesh invested Rs. 30,000, Rs. 25,000 and 20,000
respectively in a business. How should they divide profit of Rs. 15450,

if they agree to share the profit in the same ratios as their investments.

Solution : Total profit amount = Rs. 15450

The ratio of their investment = Rs. 30,000 : Rs. 25,000 : Rs. 20,000

= 30 : 25 : 20

=6:5:4

Let x be the multiple of ratios

Then Dipendra's profit = 6x

Binesh's profit = 5x

Rajesh's profit = 4x
Now,
6x + 5x + 4x = Rs. 15450

or, 15x = Rs. 15450

or, x = Rs. 15450
15

x = Rs. 1030

∴ Dipendra's profit amount = 6x = Rs. 6 × 1030 = Rs. 6180

Binesh's profit amount = 5x = Rs. 5 × 1030 = Rs. 5150

Rajesh's profit amount = 4x = Rs. 4 × 1030 = Rs. 4120

EXERCISE: 8.1

A. Very Short Questions

1. If x = 1 , find
y 8

a. duplicate ratio of x : y b. triplicate ration of x : y
d. sub – triplicate ratio
c. sub duplicate ratio

2. a. If 16x = 9y, find the ratio of x : y.

b. If a:b = 3:4, find the value of a in the terms of b.

c. If a : b = 3 : 4 and b : c = 2 : 5, find a : c.

GREEN Mathematics Book-9 131

B. Short Questions

3. a. If a = 3 , find the value of 2a – 3b .
b 4 2a + 3b

b. If a = 2b , find i. the ratio of a : b ii. the value of a + 2b
2 5 3a – 2b

4. a. If 2x = 3y, find the value of 5x – 3y .
x + y

b. If 7(5x – 7y) = 5x + 7y, show that x = 28 .
y 15

3a – b + c
c. If a : b : c = 1 : 2 : 3, find the value of a + b + c .

d. If x : y = 2 : 3 and y : z = 4 : 5, find the x + 2y – z .
2x + y + z

C. Long Questions

5. Find a : b each of the following conditions:

a. 3a + 2b = 1 b. 2a – b = a + b c. a + 7b = 3
a – b 3 7 6 3a – b 4

d. 3a + 5b : a – 2b = 4 : 5 e. 42aa–+3bb = 1
4

6. If x – 2y = 2 , find the value of following:
x + 3y 3

x2 + y2 b. 35xx+–2yy c. x + 2y x3 + y3
a. x2 – y2 3x – y d. 2x2y + xy2

7. a. If two numbers are in the ratio 2 : 3 and their sum is 95, find the numbers.

b. If two numbers are in the ratio 11 : 9 and their differences is 10, find the number.

c. What common number must be added to each term of the ratio 19:29 so that new
ratio will be 3 : 4 ?

d. What common number must be subtracted from each term of the ratio 19 : 16, so
that the resulting ratio will be 7 : 4.

e. The ratio of present ages of two brothers is 4 : 5. After 10 years, ratio of their ages
will be in the ratio 6 : 7. Find their present ages.

f. The ratio of present ages of two sister is 3 : 4. Before 5 years, the ratio of their ages
was 7 : 11. Find their present age.

g. Two numbers are in the ratio 2:3. If 9 is added to each of the number, the result
will be in the ratio 3:4. Find the numbers.

h. Two numbers are in the ratio 3 : 4. If 10 is subtracted from each of the number,
the result will be in the ratio 2 : 3. Find the number.

i. If (3a + 2b) : (3a – 2b) = 3 : 1 find the value of (8a² + b²) : (5a² – b²).

132 GREEN Mathematics Book-9

8.2 Proportional (in proportion)

If two ratios are equal then they are said to be in proportion or Extreame

proportional. If a = c , it is written as a : b: : c: d and read as 'a' A : B = C : D
b d Mean proportion

is to 'b' is as 'c' is to 'd'. The first term 'a' and the fourth term 'd' are

extremes and second term 'b' and third term 'c' are means.

If 2 : 3 = 3 : 4 = 4 : 5 = ........... i.e. 2 = 3 = 4 = ............. the number 2, 3, 4, 5 ....... etc. are said
3 4 5

to be in continued proportion. If a = b then b2 = ac
b c

Thus three quantities are said to be in conti nued proportion if the product of extremes is

equal to the square of mean.

8.3 Properties of proportion

w, x, y, and z are four members then,

a. Invertendo :

If w = y then x = z
x z w y

Proof :

Here, w = y
or, x z
or,
or, 1÷ w = 1 ÷ y
x z

1× x = 1 × z
w y

x = z this property is called as invertendo.
w y

b. Componendo :

If w = y then w+x = y+z
x z x z

Proof :

Here, w = y
or, x z
or,
w + 1 = y +1 Addiing 1 in both side.
x z

w+x = y+z proved.
x z

GREEN Mathematics Book-9 133

c. Dividendo

If w = y then w–x = y–z
x z x z

Proof : wy
Here, x =z

or, w – 1 = y –1 Subtracting 1 from both side.
x z

or, w–x = y–z
x z

This property is called as dividendo.

d. Alternendo :

If w = y then w = x
x z y z

Proof : w = y
Here, x z

or, w × x = y × x
x y z y

or, wx Multiplying both side by x
y=z y

This property is called as alternendo.

e. Componendo and dividendo :

If w = y then w+x = y+z
x z w–x y–z

Proof : wy
Here, x =z

by componendo,

w+x = y+z ............. equation (i)
x z

By dividendo,

w–x = y–z ............. equation (ii)
x z

Dividing equation (i) by equation (ii)

w+x y+z

x=z
w–x y–z

xz

or, w+x = y+z . This property is called componendo and dividendo.
w–x y–z



134 GREEN Mathematics Book-9

f. Addendo :

If w = y = a then w = w + y + a + c ....
x z b x x + z + b + d + .......

Proof : w yac
Here, x = z = b = d = ....................... = K (let)

w = xK, y = zK, a = bK, c = dK ..............

Now, w + y + a + c ....
x + z + b + d + .......

= xK + zK + bK + dK + ....
x + z + b + d + .......

= K(w + y + a + c + d + ... =K
x + z + b + d + .......

∴ w = y = a = c = .......................
x z b d

This propertiy is called addendo.

Worked Out EXAMPLES

EXAMPLE 1 If 5, 10, 15 are in proportion, find the fourth proportion.
Solution :
Let a be the fourth proportion

Then 5, 10, 15 and a are in proportion.

So, 5 = 15
10 a

or, 5a = 150

or, a = 150
5

or, a = 30

∴ The fourth proportion is 30.

EXAMPLE 2 If 2, x, 6 and 12 are in proportion, find the value of x.
Solution :
Here, 2, x, 6 and 12 are in proportion.

Then, 2 = 6
x 12

or, 6x = 24

24
or, x = 6

or, x = 4

∴ The value of x is 4.

GREEN Mathematics Book-9 135

EXAMPLE 3 4, x and 9 are in continued proportion, find the value of x.
Solution :
Here, 4, x, 9 are in continued proportion.

Then, 4 = x
x 9

or, x2 = 36

or, x = 36

or, x = 6 Note: 6 is also called mean proportion.
∴ The value of x is 6.

EXAMPLE 4 What number should be added to each of the numbers 2, 4, 8 and 12 to
Solution : make them proportional?

Let the required number be a.

Then, 2+a = 8+a
4+a 12 + a

or, (2 + a) (12 + a) = (8 + a) (4 + a)

or, 24 + 2a + 12a + a2 = 32 + 8a + 4a + a2

or, 24 + 14a + a2 – 32 = 12a = a2 = 0

or, –8 + 2a =0

or, 2a =8

a =4

∴ The required number is 4.

EXAMPLE 5 If a = c then prove that a² + b² = o a + b 2
b d c² + d² c + d
p.

Solution : Let a = c = k , then
b d

a =k c = k
b d

a = bk c = dk

L.H.S. = a² + b²
c² + d²

= (bk)² + b²
(dk)² + d²

136 GREEN Mathematics Book-9

= b²k² + b²
d²k² + d²

= b²(k² + 1) = b²
d²(k² + 1) d²

Again,

a+b 2
R.H.S. = o c + d p

= o bk + b 2

dk + d p

b(k + 1) 2
= so d(k + 1) pt

= b²(k + 1)²
d²(k + 1)²

= b²
d²

∴ L.H.S. = R.H.S. proved.

EXAMPLE 6 If a = c = e , prove that a³ + b³ + e³ = a ce .
Solution : b d f b³ + d³ + f³ b df

We have, a = c = e = k (Say)
b d f

Then, a = k c =k e =k
b d f

a = bk c = dk e = fk

L.H.S. = a³ + c³ + e³
b³ + d³ + f³

= b³k³ + d³k³ + f³k³
b³ + d³ + f³

= k³(b³ + d³ + f³)
(b³ + d³ + f³)

= k³

GREEN Mathematics Book-9 137

R.H.S. = ace
ddf

bk . dk. fk
= bdf

= k³. b d f
bdf

= k³

∴ L.H.S. = R.H.S. proved.

EXAMPLE 7 If x + 3y + 2z + 6a = x – 3y + 2z – 6a , prove that x = z
x + 3y – 2z – 6a x – 3y – 2z + 6a y a

Solution : Given,

x + 3y + 2z + 6a = x – 3y + 2z – 6a
x + 3y – 2z – 6a x – 3y – 2z + 6a

By componendo and dividendo rule

or, (x+3y+2z+6a) + (x+3y–2z–6a) = (x–3y+2z–6a) + (x–3y–2z+6a)
(x+3y+2z+6a) – (x+3y–2z–6a) (x–3y+2z–6a) – (x–3y–2z+6a)

x+3y+2z+6a + x+3y–2z–6a x–3y+2z–6a + x–3y–2z+6a
or, x+3y+2z+6a – x–3y+2z+6a = x–3y+2z–6a + x+3y–2z–6a

or, 2x + 6y = 2x – 6y
4z + 12a 4z – 12 a

or, 2(x + 3y) = 2(x – 3y)
4(z + 3a) 4(z – 3a)

or, (x + 3y) (z – 3a) = (z + 3a) (x – 3y)

or, xz – 3ax + 3yz – 9ay = zx – 3yz + 3ax – 9ay

or, –6ax = – 6yz

or, ax = yz

or, x = z
y a

∴ L.H.S. = R.H.S. proved.

138 GREEN Mathematics Book-9

EXAMPLE 8 1 1 1 a3 + b3 + c3
If a, b, c, are in continued proportion, prove a3 + b3 + c3 = a2b2c2 .

Solution : a = b =k
b c

Here, a = bk b = ck

= ck . k

ck2

111 1
L.H.S. = a3 + b3 + c3 + c3

11
= (ck2)3 + (ck)3

= 1 + k3 + k6

c3k6

= (k6 + k3 + 1)
c3k6

a3 + b3 + c3
R.H.S. = a2b2c2

(ck2)3– (ck)3 + c3
= (ck2)2 (ck)2 . c2

c3k6– c3k3 + c3
= c2k4 . c2k2 . c2

= c3(k6+ k3 + 1)

c6k6

= k6+ k3 + 1

c3k6

∴ L.H.S. = R.H.S. Proved.

Alternate Method:

Give, a = b
b c

or, b2 = ac

Now, 1 + 1 + 1
a3 b3 c3

or, c3 + a3 + 1 = c3 + a3 + 1 = c3 + a3 + b3
a3c3 b3 (b2)3 b3 b6

c3 + a3 + b3 c3 + a3 + b3 a3 + b3 + c3
= (b2) b2 . b2 = b2 (c2a2) = a2b2c2 Proved.

GREEN Mathematics Book-9 139

EXAMPLE 9 If a, b, c and d are in continued proportion, prove that:
Solution :
a. ab – bc + cd = (a – b + c) (b – c + d)

a a3 + b3 + c3
b. d = b3 + c3 + d3

Given a = b = c = k (let)
b c d

a = bk, b = ck, c = dk

= dk2 . k = dk . k
= dk3 = dk2

a. L.H.S. = ab – bc + cd

= dk3 . dk2 – dk2 . dk + dk . d

= d2k4 . k – d2k2 . k + d2k
= dk2 k – dk k + d k
= d k (k2 – k + 1)

R.H.S. = (a – b + c) (b – c + d)

= (dk3 – dk2 + dk) (dk2 – dk + d)

= d2k (k2 – k + 1)2

= d k (k2 – k + 1)
∴ R.H.S. = R.H.S. Proved.

a a3 + b3 + c3 R.H.S. a3 + b3 + c3
b. d = b3 + c3 + d3 = b3 + c3 + d3

L.H.S. = a = (dk3)3 + (dk2)3 + (dk)3
d (dk2)3 + (dk)3 + d3

dk3 d3k9 + d3k6 + d3k3
= d = d3k6 + d3k3 + d3

= k3 d3k3 (k6 + k3 + 1)
= d3(k6 + k3 + 1)

∴ R.H.S. = R.H.S. Proved. = k3

140 GREEN Mathematics Book-9

EXERCISE: 8.2

A. Very Short Questions

1. a. If a : 3 = 4 : 9, find the value of a.

b. If a = c , prove that ad – bc = 0
b d

c. If a = c , write alternendo proportion.
b d

d. If w = y , write ratio in dividendo form.
x z
a c
e. If b = d , what invertendo form of it.

B. Short Questions

2. a. If (x + y), y and (x – y) are in proportion show that x2 – 2y2 = 0.

b. If 14, x, 56 and 40 are in proportion, find x.

c. If a + b, b and a – b are in proportion, show that a2 – 2b2 = 0.

d. If 9, x, 4 are continued proportion, find the value of x.

e. If 2, a, 4 and 8 are proportion, find the value of a.

f. Find the fourth proportion if 10, 15 and 20 are in proportion.

3. a. What number should be subtracted from each of the numbers 15, 19, 21 and 27
make them proportion?

b. What number should be added to each of the number 2, 4, 7 so that the sum will
be in continued in proportion?

c. What number must be added to each of numbers, 3, 5, 7 and 10 to make four
number in proportion?

C. Long Questions

4. If a = dc , prove that
b

a² + c² ac xa + yb b b² + d² d²
a. b² + d² = bd b. xc + yd = d c. a² + c² = c²

d. o a + c p² = a² – ac + c² e. aacc + bd = a² + b²
b + d b² – bd + d² – bd a² – b²

f. (ma + nb) (mc – nd) = (ma – nb) (mc + nd) g. a(a + c) = c
b(b + d) d

h. (a + c) (a + b) = 1 + 1 + 1 + 1 i. a² + ac + c² = b² + bd + d²
abc a b c d a² – ac + c² b² – bd + d²

GREEN Mathematics Book-9 141

5. If a, b, c and d are in continued proportion then prove that:

a. a³ + b³ + c³ = da b. ab + bc + ad = (a + b + c) (b + c + d)
b³ + c³ + d³

c. (a + c) (c + d) = (b + c) (b + d) d. (a + b)2 : ab = (c + d)2 : cd

e. a³ + c³ – b² c = a³ f. ma³ + nb³ + pc³ = a
abc – bd² + d³ b³ mb³ + nc³ + pd³ d

g. (a + b)² = (b + c)² + (c + d)² h. (ab + bc + cd)2 = (a2 + b2 + c2) (b2 + c2 + d2)
ab bc cd

i. (a + b) (b + c) – (a + c) (b + d) = (b – c)2

j. (b – c)2 + (c – a)2 + (d – b)2 = (a – d)2

a+b c+d 2
b d
[ ]k. bd + = 4(a + b) (c + d)

l. b2c2 + c2a2 + a2b2 = a3 + b3 + c3
a b c

m. (a – b + c) (c2 + b2 + c2) = (a + b + c)3

n. (a2 + c2) (b2 + d2) = (ab + cd)2

a2 + b2 – c² a2 + bc
o. b2 – c² + d2 = b2 + cd

6. If a = c = e then show that each of the following ratios are equal.
b d f

a. pa – qc – re = xa – yc – ze b. a³ + c³ + e³ = aec
pb – qd – rf xb – yd – zf b³ + c³ + f³ bdf

a³ – c³ c – e 3 d. ma + nc + pe = 3 ace
c. b³ – d³ = o d – f p mb + nd + pf bdf

e. 3 a3c3 + c3e3 +a3c3 = ace
b3d3 + d3b3 + b3f3 bdf

7. a. If x = y = z , prove that x³ + y³ + z³ = (x + y + z)3
a b c a³ b³ c³ (a + b + c)2

b. If a = c = e , prove that a² + c² + e² = ca
b d f b² + d² + f² bd

c. a = p = r = d , prove that a+p+r+d = a
b q s e b+q+s+r b

142 GREEN Mathematics Book-9

8. If a = b = c , prove that:
b c d

a. a2 – b2 = a
b2 – c2 c

b. a2 + b2 = a2 + ab + b2
b2 + c2 b2 – bc + c2

c. (a2 + b2) (b2 + c2) = b2 (a + b)2

d. (ab + bc + ca)3 = abc (a + b + c)3

e. a3b3 + b3c3 + c3a3 = abc (a3 + b3 + c3)

1 1 + 1 1 (a2 + b2 + c2)
f. o a2 + b2 c2 p= ab2c

g. a³ + b³ + c³ = a²b + b²c + c²a
b³ + c³ + d³ b²c + c²d + b²d

9. a. If a, b, c, d and e be in continued proportion, prove that a : e = a4 : b4.

b. If y a z = z b x = x c y , find the value of a + b + c.
– – –

c. If 4a + 9b = 4c + 9d , prove that a = c
4a – 9b 4c – 9d b d

d. If x = y = z , find the value of x + y – z
b+c–a c+a–b 2c

e. If p, q, r and s are in continued proportion then prove that (p + q + r) (q + r + s)
= pq + qr + rs

f. If (a2 + b2) (x2 + y2) = (ax + by)2, show that x = y .
ab

g. If a = b then prove that 1 + 1 + 1 = a + b + c
b c a³ b³ b³ b²c² c²a² a²b²

h. If (4a + 9b) (4c – 9d) = (4a – 9b) (4c + 9d), prove a = c .
bd

GREEN Mathematics Book-9 143

9

Simultaneous linear equations

Estimated Teaching Periods : 5

In the context where algebra is identified with the theory of equations,
the Greek mathematician Diophantus has traditionally been known as
the "father of algebra" but in more recent times there is much debate over
whether al-Khwarizmi, who founded the discipline of al-jabr, deserves
that title instead.

Contents

9.1 Equation
9.2 Solving equation by graphical method

Objectives

At the end of this unit, students will be able to:
know about the simultaneous equations
solve the simultaneous equations by substitution method, elimination method and
graphical method

Materials

Pencil, scale, graph papers, graph chart, etc.

144 GREEN Mathematics Book-9