The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Discover the best professional documents and content resources in AnyFlip Document Base.
Search
Published by abhishek54yt, 2021-10-27 05:08:47

math

math

Example 4: Find the number of bricks each of size 14cm × 8cm × 5cm to build a wall
Solution: 3m high and 30cm thick around the inside of a rectangular plot 15m ×
8m.

Here,

Length of the plot (l) = 15m

Breadth (b) = 8m

Height of the wall (h) = 3m

Thickness of the wall (t) = 30cm = 0.3m

 Volume of the surrounding wall (inside) the rectangular plot

(V) = 2t(l + b – 2t) × h

= 2 × 0.3m(15m + 8m – 2 × 0.3m) × 3m.
= 0.6m × 22.4m × 3cm = 40 .32m3.

Size of a brick = 14cm × 8cm × 5cm

 Volume of a brick (v) = 0.14m × 0.08m × 0.05m = 0.00056m3.

Number of bricks (N) = V = 40.32m3 = 72000.
v 0.00056m3

Therefore, 72000 bricks are required to build the wall.

Example 5: The external dimensions of a closed wooden box are 60cm × 50cm ×
Solution: 30cm and the thickness of the wood in 2cm. Find the volume of the
wood used to make the box.

Here,

External length (l) = 60cm, breadth (b) = 50cm and height 30cm.

External volume (Vext) =l×b×h
= 60cm × 50cm × 30cm

= 90000cn3

Thickness of the wood (t) = 2cm

Internal length (l1) = l – 2t
= 60cm – 2 × 2cm = 56cm

Internal breath (b1) = b – 2t
= 50cm – 2 × 2cm = 46cm

Internal height (h1) = h – 2t
= 30cm – 2 × 2cm = 26cm

Internal volume (Vint) = l1 × b1 × h1
= 56cm × 46cm × 26cm = 66976cm3

Now, volume of wood = Vext - Vint
= 90000cm3 - 66976cm3

= 23024cm3

94 | Mathematics - 9 Mensuration

Example 6: The internal dimensions of a wooden box are 36cm × 26cm × 16cm. Find
Solution: the external volume of the box, if the volume of wood used in 9024cm3.

Here,

Internal length (l) = 396cm, internal breadth (b) = 26cm and

Internal height (h) = 16cm

 Internal volume (Vint) =l×b×h
= 36cm × 26cm × 16cm

= 14976cm3

Volume of wood (V) = 9024cm3.

External volume of the box (Vext) = Vint + V
= 14976cm3 + 9024cm3
= 24000cm3

Therefore, the external volume of the box is 24000cm3.

Exercise 3.4

1. (a) Find the volume of a wall 10m long, 4m high and 25cm thick.

(b) Find the volume of a brick of size 25cm by 10cm by 5cm in cubic meter.
(c) A wall is 18m long and 3m high. If its volume is 10.8m3, find its thickness.
(d) A wall 3.5m high, 25cm thick. If its volume is 10.5m3, how long is the wall?

2. (a) A wall 20m long, 4m high and 25cm thick has two windows 2m × 1.5m each. Find the
volume of the wall excluding the windows.

(b) 1500 bricks 25cm by 10cm by 5cm each are required to build a wall. Find the volume of
the wall.

(c) Find the volume of each brick if 62500 bricks were used to construct a wall of size
18m × 4m × 1.5m.

(d) Find the number of bricks required to construct a wall 30m × 4m × 0.5m, if the size of
each brick is 25cm × 10cm × 5cm.

3. (a) A brick of volume 1250cm3 costs Rs. 3.50. A man spends Rs. 52,500 to build a wall.
Find the volume of the wall.

(b) A brick costs Rs. 3.50 and Tuyoocha Sahu had to spend Rs. 87,500 to build a wall of
volume 33.75m3. Find the volume of each brick.

(c) A wall 11m × 25cm × 3.5m is built by using bricks each of size 22cm × 10cm × 5cm. If
a brick costs Rs. 2.50, find the total expenditure.

(d) A wall 25m × 20cm × 3m built with 1250cm3 same sized bricks cost Rs. 40800. Find
the cost of each brick.

Mensuration Mathematics - 9 |95

4. (a) Find the volume of a wall 30cm thick to build a room 10m long, 8m wide and 3.5m
high leaving 1.06m3 of the volume for the doors and windows.

(b) A wall 40cm thick and 3.5m high is to be built all around, inside of a rectangular garden
40m × 30.8m with an open gate 3m wide. If 10% of the wall is occupied by the cement,
how many bricks of size 21cm × 13.7cm × 5cm are required to build the wall? Also, if a
brick costs Rs. 3.5, calculate the total cost of bricks.

(c) A rectangular plot 36m × 24.3m is divided into equal quadrants by building walls 30cm
thick and 3m high across the plot. If 10% of the wall is occupied by cement, how many
bricks of size 24cm × 12cm × 6cm are used in the wall?

(d) A cuboidal reservoir 32m × 24m × 4m is to be separated into 4 equal chambers by
constructing cross-walls 40cm thick up to its height. Find the volume of the cross wall.

5. (a) A lid less metal box has external dimension 25cm × 20cm × 12cm. If the thickness of
the material is uniformly 1cm, find the volume of the metal used in the box.

(b) A closed cubical metal box of side 25cm and 0.5cm thick is to be constructed.
(i) Find the volume of the metal required to construct the box.
(ii) The cost of the metal required, if 1cm3 of metal cost is Rs. 0.50.

(c) A cubical tin box has external dimensions 60cm × 50cm × 40cm. The thickness of the
tin is 0.2cm. Find the weight of the box, if 1cm3 of tin weighs 10gm.

Project Work
6. Take the measurements of the dimensions of your class room.

(a) Find the area of 4 walls of the room.
(b) Find the area and total cost of plastering at Rs. 200 per sq.m. floor of the room.

(c) Find the total cost of painting on the ceiling of the class room at the rate of Rs. 250 per
square meter.

(d) Find the area of 4 walls excluding the window and the door.

(e) Calculate the total cost of papering on its 4 walls excluding door and window at the rate
of Rs. 30 per square meter.

7. Measure the length and breadth of the basket ball court of your School. Calculate the number
of tiles of size 2ft. x 1ft. needed to pave on it at the rate of Rs. 315 per piece.


96 | Mathematics - 9 Mensuration

Unit Test

Time: 40 minutes F.M.- 24

Group-A (3×1=3)

1. What is the area of a square when the diagonal (d) is given?

2. Find the area of four walls of the room having length 10m, breath 8m and height 5m.
3. If the area of cross-section of a prism is A cm2 and height is h cm, find its volume.

Group- B (4 × 2 = 8)

4. The perimeter of a square garden is 100m. Find its area.

5. Find the perimeter of a semi-circle having diameter 14cm.

6. A room has length 9ft and breadth 6ft. What is the total cost of carpeting on its floor at the
rate of Rs. 150 per sq. ft?

7. The wall has to be made of length 11m, breadth 1m and height 5m. How many bricks of
dimensions 22 cm×10cm×5 cm are needed.

Group- C (2×4=8)

8. Find the total surface area of given prism.

9 cm

4 cm
7 cm

16 cm

9. A room of length 7m & breadth 5m has a door of side 2m×1.5m & two windows of side
1m×1.5m each. Total cost of colouring on it's four walls at the rate of Rs 7.5 per m2 is Rs
495. Find the height of the room.

Group- D (1× 5=5)
10. The volume of the room having length double of its breadth is 396m3. Total cost of

colouring its celling at the rate of Rs 30 per m2 is Rs 2160 what is the cost of colouring its
four walls at the rate of Rs60 per m2?



Mensuration Mathematics - 9 |97

Answers ____________________________________________________________

Exercise 3.1 (b) 144m2 (c) 192m2 (d) 252m2
(b) 43.12m2 (c) 128m2 (d) 258m2
1. (a) 352m2 (b) 924m2
2. (a) 29.26m2 (b) 2944cm2 (c) 234m2 (d) 144m2
3. (a) 1350m2 (b) 5544m2 (c) 140m2 (d) 120m2
4. (a) 876m2 (b) 576m2 (c) 19.25m2 (d) 2m
5. (a) 556m2 (b) 36m (c) 3.5m (d) 1056cm2
6. (a) 52m (b) Rs. 1,00,000 (c) Rs. 432 (d) Rs. 3,00,000
7. (a) 4m (b) Rs. 15030.4 (c) Rs. 38016 (d) Rs. 2,00,000
8. (a) Rs. 25344
9. (a) Rs. 52360

Exercise 3.2

1. (a) 660cm2, 600cm3 (b) 600cm2, 864cm3 (c) 247.18cm2, 187.08cm3 (d) 864cm2, 1728cm3
(d) 24cm
2. (a) 216cm2 (b) 512cm3 (c) 5cm (d) 452cm2, 504cm3
(c) 322cm2, 246cm3 (d) 324cm3
3. (a) 462cm2, 540cm3 (b) 60cm2, 72cm3 (c) 156cm3 (d) 12,00,000 litres.
(c) 69600cm3
4. (a) 102cm3 (b) 144cm3

5. (a) 152000cm3, 16800cm2 (b) 556cm3

6. 8 cm

Exercise 3.3 (b) 188m2 (c) 194m2 (d) 112m2
(b) 176m (c) 14m (d) 12m
1. (a) 300m2 (b) 24m (c) 189m2 (d) 8m, 6m
2. (a) 9m (b) 2.5m (c) 370 (d) 134
3. (a) 50m2 (b) Rs. 11375 (c) Rs. 2700 (d) Rs. 2880
4. (a) 20m (b) 4m (c) Rs. 11340 (d) 4.5m
5. (a) Rs. 2480 (b) Rs. 1800 (c) Rs. 10125 (d) Rs. 9600
6. (a) 7m (b) Rs. 2376
7. (a) Rs. 14400 (d)12m
8. (a) 4.4m (b) 0.00125m3 (c) 20cm (d) 48000
(b) 1.875m3 (c) 1728cm3 (d) Rs. 3.4
Exercise 3.4 (b) 1350cm3 (c) Rs. 21875 (d) 222.4m3
(c) 28125
1. (a) 10m3 (b) 1,20,000, Rs. 42,000 (c) 29.36 kg
2. (a) 18.5m3 (b) 1951cm3, Rs. 975.5
3. (a) 18.75m3
4. (a) 38m3
5. (a) 1446cm3
6. Show your teacher
7. Show your teacher.

n

98 | Mathematics - 9 Mensuration

4Chapter

Algebra

Objectives:

At the end of this chapter, the
students will be able to:
 factorize the different algebraic

expression.
 simplify the problem related to

laws of indices.
 solve the exponential equation.
 simplify the surds and rationalize

them.
 solve the problems related to

ratio and proportion.

Teaching Materials:

Chart of algebraic formula, chart of
laws of indices and surds, flash cards,
different color board markers, etc.

4.1 Factorisation

Introduction

Let us take two algebraic expressions 2x and (x – 3y). When these two expressions are multiplied
together, then the product of 2x and (x – 3y) is 2x × (x – 3y) = 2x2 – 6xy.

2x2 – 6xy is a single algebraic expression which is the product of the two expressions 2x and x – 3y.
So, 2x and x – 3y are called the factors of the expression 2x2 – 6xy.

Thus, the factorization is the process by which the given algebraic expression can be expressed as the
product of two or more algebraic expression or terms.

When we factorise an algebraic expression, we write the given expression as the product of its factors.
So, for finding the factors of the given expression, we apply the selected method of factorization for
the particular type of expression. Therefore, it is very important and useful to know about the types of
expression which are to be factorised.

1. (i) Factorisation of the expression by taking common only:

To factorise such as expression, the common factor is taken out and each term of the given
expression should be divided by the common factor to get another factor.

For example: 3x2y – 6xy2 is an expression containing two terms. The both terms have the
common factor is 3xy. Then, 3x2y – 6xy2 = 3xy (x – 2y)

(ii) Factorisation of the expression by making groups before taking common:

To factorise such an expression, the terms of the given expression are to be arranged for
making a suitable groups such that each group has a common factor.

For example: x2 + xy + zx + yz is an expression. In this expression, the first two terms are
taken in a group. Then by taking common in each group, the common factor is (x + y).

Now, x2 + xy + zx + yz

= x(x + y) + z(x + y)

= (x + y) (x + z)

(x + y) and (x + z) are the factors of the given expression x2 + xy + zx + yz.

Worked Out Examples

Example 1: Factorise: x2y + xy2 + 3x + 3y.
Solution: x2y + xy2 + 3x + 3y
= xy(x + y) + 3(x + y)
= (x + y)(xy + 3)

Example 2: Factorise:

(a) 2a (x + y) – b (x + y)

(b) 6x (a – b) + 3xy (b – a)

100 | Mathematics – 9 Algebra

Solution: (a) 2a (x + y) – b (x + y)
let x + y = p, then the given expression is
2ap – bp = p(2a – b)
= (x + y) (2a – b)
 2a(x + y) – b(x + y) = (x + y) (2a – b)

(b) 6x (a – b) + 3xy (b – a)
= 2 × 3x (a – b) + 3xy {– (a–b)}
= 2 × 3x (a – b) – 3xy (a – b)
= 3x (a – b) (2 – y)

Example 3: Factorise: (b) x2 – x(2m – n) – 2mn
Solution: (b) x2 – x(2m – n) – 2mn
(a) x2 – xy + y2 – xy = x2 – 2mx + nx – 2mn
(a) x2 – xy + y2 – xy
= x(x – 2m) + n(x – 2m)
= x2 – xy – xy +y2
= x(x – y) – y(x – y) = (x – 2m)(x + n)
= (x – y)(x – y)

2. Factorisation of the expression of the form a2 – b2: a

To factorise an expression a2 – b2, we should follow the following steps.
a. Let's take a square chart paper of length a units, where the area of it is a2 sq. a2 sq. unit a

units.

b. Cut a square piece of length b units from the one corner of the given square
paper as shown in the following given figure. The area of such square piece is b2 sq. units.
Then the area of the remaining part of the square paper is (a2 – b2) sq. units.

aa

a2 a2 - b2 Remaining part
a a
b

b b2

c. Cut the remaining part whose area is (a2 – b2) sq. units along a a-b
the diagonal as shown in the figure alongside.
ab
d. Make a rectangle by rearranging two parts after cutting, where b
the length and the breadth of the rectangle are (a + b) units and
(a – b) units respectively. a-b

ba

(a-b) (a-b)

ab

Algebra Mathematics – 9 |101

Now, area of the rectangle = length × breadth
= (a + b) × (a – b)
= a2 – ab + ab – b2
= a2 – b2

 Area of the rectangle = (a2 – b2) sq. units

a2 – b2 = (a + b)(a – b).

Here, the expression a2 – b2 is the difference of the square of two terms a and b. From the above
illustration, it is clear that (a + b) and (a – b) are the factors of a2 – b2. Thus, we can use the formula
a2 – b2 = (a + b)(a –b) to factorize the algebraic expression of the form a2 – b2.

Important formulae:

 (a + b)2 = a2 + 2ab + b2 or, (a – b)2 + 4ab
 (a – b)2 = a2 – 2ab + b2 or, (a + b)2 – 4ab
 a2 – b2 = (a + b)(a – b)
 a2 + b2 = (a + b)2 – 2ab or, (a – b)2 + 2ab

Worked Out Examples

Example 1: Factorize:
Solution:
(a) 16a2 – 49b2 (b) (2x + y)2 – (x – 2y)2
(a) 16a2 – 49b2

= (4a)2 – (7b)2

= (4a + 7b)(4a – 7b)
(b) (2x + y)2 – (x – 2y)2

= {(2x + y) + (x – 2y)}{(2x + y) – (x – 2y)}

= (2x + y + x – 2y)(2x + y – x + 2y)

= (3x – y)(x + 3y)

Example 2: Factorize:
Solution:
(a) 3a2b – 27b3 (b) m2 – 4n2 – m3 + 2m2n
(a) 3a2b – 27b3 = 3b(a2 – 9b2)
= 3b{(a)2 – (3b)2}

= 3b(a + 3b)(a – 3b)
(b) m2 – 4n2 – m3 + 2m2n = m2 – (2n)2 – m2(m – 2n)

= (m + 2n) (m – 2n) – m2 (m – 2n)
= (m – 2n) (m + 2n – m2)

102 | Mathematics – 9 Algebra

Example 3: Factorize: 4x2 – y2 + 2y – 1.
Solution: Here, 4x2 – y2 + 2y – 1

= (2x)2 – (y2 – 2y + 1) [ (a – b)2 = a2 – 2ab + b2]
= (2x)2 – (y – 1)2

= (2x + y – 1){2x – (y – 1)}

= (2x + y – 1)(2x – y + 1)

3. Factorization of the expression of the form of a3  b3:

We know that,

(a + b)3 = (a + b)1 (a + b)2

= (a + b) (a2 + 2ab + b2)

= a3 + 2a2b + ab2 + a2b + 2ab2 + b3

= a3 + b3 + 3a2b + 3ab2

= 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)

 a3 + b3 = (a + b)(a2 – ab + b2)
So, (a + b) and (a2 – ab + b2) are the factors of a3 + b3.

Similarly,

a3 – b3 = (a – b)(a2 + ab + b2)

So, (a – b) and (a2 + ab + b2) are the factors of a3 – b3.

Thus, to factorize the expression of the form a3 + b3 and a3 – b3, we should use the following
formulae.

a3 + b3 = (a + b)(a2 – ab + b2)

a3 – b3 = (a – b)(a2 + ab + b2).

Worked Out Examples

Example 1: Factorize: (b) 8ax3 – ay3
Solution: = (a2)3 – (1)3
(a) a6 – 1 = (a2 – 1){(a2)2 + a2.1 + 12}
(a) a6 – 1 = (a2 –1){(a2 + 1)2 – 2a2 + a2}

Algebra Mathematics – 9 |103

= (a + 1)(a – 1){(a2 + 1)2 – a2)}
= (a + 1)(a – 1)(a2 + a + 1)(a2 – a + 1)
(b) 8ax3 – ay3 = a(8x3 – y3)

= a {(2x)3 – (y)3}
= a (2x – y) {(2x)2 + 2x.y + y2}
= a (2x – y) (4x2 + 2xy + y2)

Example 2: Factorize:
Solution:
(a) x6– y6 (b) (2x + y)3 – (2x – y)3

Here,

(a) x6– y6 = (x3)2 – (y3)2

= (x3 + y3) (x3 – y3)

= (x + y) (x2 – xy + y2) (x – y) (x2 +xy + y2)

= (x + y) (x – y) (x2 – xy + y2) (x2 +xy + y2)

(b) (2x + y)3 – (2x – y)3

= {(2x + y) – (2x – y)}{ (2x + y)2 + (2x + y)(2x – y) + (2x – y)2}

= (2x + y – 2x + y){(2x)2 + 2.2x.y + y2 + (2x)2 – y2 + (2x)2 – 2.2x.y + y2}

= (2y)(4x2 + 4xy + y2 +4x2 – y2 + 4x2 – 4xy + y2)

= 2y(12x2 + y2)

Example 3: Factorize: x3 – 2x2 + 8 – 4x
Solution: Here, x3 – 2x2 + 8 – 4x
= (x)3 + (2)3 – 2x2 – 4x
= (x + 2)(x2 – x.2 + 4) – 2x(x + 2)
= (x + 2)(x2 –2x + 4 – 2x)
= (x + 2)(x2 – 4x + 4)
= (x + 2)(x – 2)2

= (x + 2)(x – 2)(x – 2).

4. Factorization of the expression of the form ax2 + bx  c:

To factorize the trinomial expressions of the form ax2 + bx  c, we need to find the two numbers
whose product is ac and the sum or difference is b. Then the given trinomial expression is expanded to
four terms by breaking the middle term and factorization process is performed by grouping.

Let the two numbers be p and q. Then, pq = ac and p q = b. The expression ax2 + bx + c is written as
ax2 + (p  q)x + c.

104 | Mathematics – 9 Algebra

Worked Out Examples

Example 1: Factorize:
Solution:
(a) x2 + 7x + 12 (b) 4a2 – 16a – 9

(a) x2 + 7x + 12

Comparing this expression with ax2 + bx + c, then, a = 1, b = 7 and c = 12

Now, a × c = 1 × 12 = 12

The pairs of the possible product are, 1× 12, 2 × 6, 3 × 4,

Among these pairs, the sum of pairs 3 and 4 is 7.

3+4=7

Therefore, the two required numbers are 3 and 4.

Now,
x2 + 7x + 12
= x2 + (3 + 4)x + 12
= x2 + 3x + 4x + 12

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

= (x + 3)(x + 4)
(b) 4a2 – 16a – 9

Comparing this expression with ax2 + bx + c, then, a = 4, b = –16 and c = –9

Now, a × c = 4 × –9 = –36

The pairs of the possible product are, 1 × – 36, 2 × –18, 3 × –12, 6 × –6, 4 × –9

or, –1 × 36, –2 × 18, –3 × 12, –6 × 6, –4 × 9

Among these pairs, the required pairs whose sum is –16 are 2 and –18 since

2 + (–18) = 2 – 18 = –16
Now, 4a2 – 16a – 9

= 4a2 – (18 – 2)a – 9
= 4a2 – 18a + 2a – 9

= 2a(2a – 9) + 1(2a – 9)

= (2a – 9)(2a + 1)

Example 2: Factorize:
Solution:
(a) 12yx22 + x – 20 (b) 10ba22 – 1 – 2ba22
y

(a) 12yx22 + x – 20
y

= 12yx22 + (16 – 15) x – 20
y

= 12yx22 + 16 x – 15 x – 20
y y

Algebra Mathematics – 9 |105

= 4 x 3 x + 4 – 53 x + 4
y y y

= 3 x + 4 4 x – 5
y y

(b) 10ba22 – 1 – 2ba22

= 10ba22 – (5 – 4)ba . b – 2ba22
a

= 10ba22 – 5ba . b + 4ba . b – 2ba22
a a

= 5 a 2ba – b  + 2 b 2ba – b 
b a a a

= 5ba + 2ba  2ba – b 
a

Example 3: Factorize:
Solution:
(a) 8x6 – 7x3 – 1

(b) 2(2x + x2)2 – 2(2x + x2) – 12
(a) 8x6 – 7x3 – 1

= 8x6 – 8x3 + x3 – 1
= 8x3(x3 – 1) + 1(x3 – 1)
= (x3 – 1)(8x3 + 1)
= {(x)3 – (1)3}{(2x)3 + (1)3}
= (x – 1)(x2 + x.1 + 12)(2x + 1){(2x)2 – 2x.1 + (1)2)
= (x – 1)(2x + 1)(x2 + x + 1)(4x2 – 2x + 1)
(b) 2(2x + x2)2 – 2(2x + x2) – 12
Put 2x + x2 = y,
then, 2y2 – 2y – 12
= 2y2 – 6y + 4y – 12

= 2y(y – 3) + 4(y – 3)

= (y – 3)(2y + 4)
Putting the value of y, we get,(x2 + 2x – 3)[2(x2 +2x) + 4]
= (x2 + 2x – 3)(2x2 + 4x + 4)
= (x2 + 3x – x – 3). 2(x2 + 2x + 2)
= {x(x + 3) – 1(x + 3)}.2(x2 + 2x + 2)
= 2 (x + 3) (x – 1) (x2 + 2x + 2)

106 | Mathematics – 9 Algebra

Exercise 4.1.1

1. Factorize the following.

(a) 3x – 4xy (b) 2x(a – b) + 4y(a – b)

(c) 4ax – 6a2x + 8ay (d) 14p2q + 6pq2 – 12pq

(e) 4a2 – b2 (f) 3x3 – 48x

(g) 9a2 – 1 (h 9x2 – 16
4 25y2

(i) x4 – y4 (j) 27x3 – 12x

(k) 3x2 – 147 (l) 2a2 – 242

(m) a3 + 8b3 (n) x3 – 27y3

(o) m4n2 + mn5 (p) 16 p4 – 2p

(q) 64m6 + n6 (r) x3 + 1
x3

(s) x6 + 1 (t) (y + 1)2 – 9
x6

(u) (x – 3)2 – 4(x – 3) (v) 16 – (2a – b)2

(w) (2x–3)2–(3x – 2)2

2. Factorize the following.

(a) 3x2 + 8x + 4 (b) x2 + 7x + 12

(c) a2 + 5a + 4 (d) a2 + 5a – 6

(e) 4a2 – 16a – 9 (f) 3x2 – 7x – 6

(g) 2a4 – 5a2 – 12 (h) 2x6 – x3 – 3

(i) 7x2 – 4 – 3y2 (j) 2x2 – 3x – 5
y2 x2 y2 y

(k) 10p2 – 1 – 3q2 (l) 3a2 – 11 + 10b2
q2 p2 b2 a2

(m) 9 (a + b)2 + (a + b) – 8 (n) 3 (x + y)2 – 10 (x + y) – 8

(o) 4 (x + y)2 – 5 (x + y)z – 6z2 (p) 7a – 1a2 + 5a – 1a –2

(q) –a2 – 12a – 27 (r) –a2 + 9a – 14

3. Factorize: (b) a2 – b2 – ax – bx
(a) x2 – y2 – ax + ay (d) a2 + 6ab – ca – 6bc
(c) a2 + ab + ac + bc (f) b2 – ab + a2 – ab
(e) x2 – 2 (x+ y) – y2 (h) x3 – x2 – x + 1
(g) ab2 – b(a – c) – c (j) y3 + 1 + 2y2 + 2y
(i) x2y + 2xy – 3x – 6 (l) 8 – 36x – 2x2 + 9x3
(k) x3 + 2x2 + 8 + 4x

Algebra Mathematics – 9 |107

(m) a3 – a + b3 – b (n) x3 – 8y3 + 2xa – 4ya
(o) 8a3 + b3 + 12a2b + 6ab2 (p) x3 – x – y3 + y
(q) 4x2 – y2 + 2y – 1 (r) 4x2 – 9y2 + 4x + 1
(s) 9a2 – 4b2 + 6a + 1 (t) 1 + 4p + 4p2 – 36p4

Factorization of the expression of the form a4 + a2b2 + b4

To factorize the expression of the form a4 + a2b2 + b4 by using the similar method of factorizing the
expression of the form a2 – b2.

Now, a4 + a2b2 + b4

= a4 + b4 + a2b2
= (a2)2 + (b2)2 + a2b2

= (a2 + b2)2 – 2a2b2 + a2b2 [a2 + b2 = (a + b)2 – 2ab]

= (a2 + b2)2 – a2b2

= (a2 + b2)2 – (ab)2

= (a2 + b2 + ab) (a2 + b2 – ab) [a2 – b2 = (a + b)(a – b)]

= (a2 + ab + b2) (a2 – ab + b2)

Worked Out Examples

Example 1: Factorize: x4 + 6x2y2 + 25y4.
Solution: Here, x4 + 6x2y2 + 25y4
= (x2)2 + (5y2)2 + 6x2y2
Example 2: = (x2 + 5y2)2 – 2.x2.5y2 + 6x2y2
Solution: = (x2 + 5y2)2 – 10x2y2 + 6x2y2
= (x2 + 5y2)2 – 4x2y2
= (x2 + 5y2)2 – (2xy)2 [a2 + b2 = (a – b)2 + 2ab]
= (x2 + 5y2 + 2xy)(x2 + 5y2 – 2xy)
= (x2 + 2xy + 5y2) (x2 – 2xy + 5y2)

Resolve into factors: m4 – 6m2n2 + n4.
Here, m4 – 6m2n2 + n4
= m4 + n4 – 6m2n2
= (m2)2 + (n2)2 – 6m2n2
= (m2 – n2)2 + 2m2n2 – 6m2n2
= (m2 – n2)2 – 4m2n2
= (m2 – n2)2 – (2mn)2
= (m2 – n2 + 2mn) (m2 – n2 – 2mn)
= (m2 + 2mn – n2) (m2 – 2mn – n2)

108 | Mathematics – 9 Algebra

Example 3: Factorize: 4x4 + y4.
Solution: Here, 4x4 + y4
= (2x2)2 + (y2)2
Example 4: = (2x2 + y2)2 – 2.2x2.y2
Solution: = (2x2 + y2)2 – 4x2y2
= (2x2 + y2)2 – (2xy)2
Example 5: = (2x2 + y2 + 2xy)(2x2 + y2 – 2xy)
Solution: = (2x2 + 2xy + y2)(2x2 – 2xy + y2)

Factorize: a4 – 6a2 – 7 – 8x – x2.
Here, a4 – 6a2 – 7 – 8x – x2
= (a2)2 – 2.a2.3 + 32 – 16 – 8x – x2
= (a2 – 3)2 – (16 + 8x + x2)
= (a2 – 3)2 – (x2 + 2.x.4 + 42)
= (a2 – 3)2 – (x + 4)2
= [(a2 – 3) + (x + 4)] [(a2 – 3) – (x + 4)]
= (a2 – 3 + x + 4)(a2 – 3 – x – 4)
= (a2 + x + 1)(a2 – x – 7)

Factorize: x2 – y2 – 6x + 14y –40
Here, x2 – y2 – 6x + 14y – 40
= (x)2 – 2.x.3 + (3)2 – y2 + 14y – 40 – 9
= (x – 3)2 – (y2 – 14y + 49)
= (x – 3)2 – (y – 7)2

= (x – 3 + y – 7)(x – 3 – y + 7)

= (x + y – 10)(x – y + 4)

Exercise 4.1.2

1. Factorize the following.

(a) x 4 +4y4 (b) x4 + 4 (c) a4 + 64b4

(d) 324p4 + q4 (e) (7a)4 + 4b4 (f) x4 + 1
4x4

(g) 4x4 + 1 (h) 4x4 + 1 y4
81y4 625

2. Factorize following.

(a) x4 + x2y2 + y4 (b) x4 + x2 + 1 (c) x4 – 7x2 + 1

(d) 4p4 + 3p2 + 1 (e) m4 – 3m2 + 1 (f) 9a4 – 4a2 + 4

(g) m4 – 14m2 n2 + n4 (h) 4a4 + 35a2 b2 + 121 b4 (i) x4 – 17x2 + 256

Algebra Mathematics – 9 |109

(j) x4 – 11x2 + 1 (k) x4 – 6x2 + 1 (l) a4 – 14a2 b2 + b4
(n) 25x4 – 9x2y2 + 16y4
(m) x2 + 1 + y2 (o) p4 – 7 p2 + 1
y2 x2 q4 q2

(p) x4 + 4y4 – 7 (q) x4 – 7 + 1 (r) x4 – 11x2 + 1
4y4 x4 x4 y4 y2

(s) 4x2 – 27 + 1 (t) x2 – 3 + 1 (u) x4 + 1 + y4
4x2 x2 y4 x4

3. Factorize the following:

(a) x2 – 10x + 24 + 6y – 9y2 (b) p2 – 6p – 40 + 14q – q2

(c) a2 – 6a + 8 – b2 + 2b (d) x4 + 9 – 7x2 + 2xy – y2

(e) a2 – 12a – 28 + 16b – b2 (f) x2 + 3y2 – z2 + 2yz – 4xy

(g) x2 – 90xy + 2000y2 – 550yz – 3025z2

(h) a2 – 2ab + b2 – c2 – 2cd – d2

4.2 Indices

Introduction

Any algebraic term has three parts. The three parts are coefficient, base and power. Let’s take an
algebraic term 3a2. In this term 3a2, 3 is the coefficient of a2, a is the base and 2 is the power of a. The
power of the base is called the index. It is also called the exponent. So, in a term – 4x5. 5 is called the
index of x. Indices is the plural form of index.

Laws of Indices

There are certain rules which are used to solve the problems of indices. These rules are called the laws
of indices.

(i) Product Law of Indices
Let xm and xn are the two algebraic terms, where m and n are assumed to be positive integers,
then xm × xn = xm + n.

Proof:
We know that,
x2 = x × x
x3 = x × x × x
x4 = x × x × x × x
xm = x × x × x × x × x ... ... ... ... to ‘m’ factors
xn = x × x × x × x × x ... ... ... to ‘n’ factors
 xm × xn = (x × x × x × x × x ... ... ... to ‘m’ factors) × (x × x × x × x × x ... ... ... to ‘n’ factors)

= x × x × x × x × x ... ... ... to (m + n) factors
= xm + n

110 | Mathematics – 9 Algebra

Thus, xm × xn = xm + n, where ‘m’ and ‘n’ are positive integers.

Similarly, If a, b, c, d are the positive integers, then

xa × xb × xc × xd = xa + b × xc × xd

= xa+b+c × xd

= xa + b + c + d

So, in general,

xa × xb × xc × xd × xe × ... ... ... = xa + b + c + d + e + ... ... ...

Thus, the index of the product of the two or more algebraic terms with the same base is the sum
of the indices of each term.

For examples:

32 × 35 = 32 + 5 = 37

(4x)2 × (4x)3 × (4x)5 = (4x)2 + 3 + 5 = (4x)10

(ii) Quotient Law of Indices

Let xm and xn are the two algebraic terms with the same base, where m and n are assumed to be
positive integers, then

xm ÷ xn = xm – n, where m > n.

Proof:

We have,

xm – n × xn = xm – n + n [by using the product law of indices]

or, xm – n × xn = xm

or, xm – n = xm
xn

or, xm = xm–n
xn

or, xm ÷ xn = xm – n, where m > n.

But, if m < n, then

xm ÷ xn = 1 = xm–n
xn–m

Thus, the index of the quotient of the two algebraic terms with the same base is the difference
of the indices of each term.

For examples:
74 ÷ 73 = 74 – 3 = 7
(3a)5 ÷ (3a)2 = (3a)5 – 2 = (3a)3

(4m)3 ÷ (4m)7 = 1 = 1
(4m)7 – 3 (4m)4

Algebra Mathematics – 9 |111

(iii) Power Law of Indices

If xm is an algebraic term with the base x and index m, then (xm)n = xmn, where m and n both are

the positive integers and n is the index of xm.
Proof:
We know,

(xm)n = xm × xm × xm × ... ... ... to n factors.

= xm + m + m + ... ... ... n terms
= xmn

(xm)n = xmn

Thus, the power of an index of any algebraic term is the product to two indices.

[ ]It can also be extended as { }(xm)n p q = xmnpq,

xymnp = (xm)p = xmp
(yn)p ynp

For examples:

(32)3 = 32 × 3 = 36

(4a)3 × 5 = (4a)15

 23 5 = (23)5 = 23 × 5 = 215
32 (32)5 32 × 5 310

(iv) Law of Negative Index

Let x–m is an algebraic term, where m is a negative integer, then x–m = 1
xm

Proof:

We have, x–m = xm – 2m

= xm ÷ x2m [Quotient law of index]
[Product law of indices]
= xm = xm m
x2m xm +

= xm = 1
xm × xm xm

 x–m = 1
xm

Similarly, xm = 1
x–m

112 | Mathematics – 9 Algebra

Thus, an algebraic term with negative index is the reciprocal of the term with the positive index.

For example: x–3 = 1 , 4–3x = 1 and 1 = 32x
x3 43x 3–2x

(v) Law of Zero Index
Let ao be an algebraic term with the base a and the index 0, then ao = 1.

Proof: OR x = 1
We have, ao = a2 – 2 x

= a2 × a–2 and x1
x1
= a2 × 1 [Law of negative index.] x = = x1–1 = x0
a2 x

x0 = 1

=1

 ao = 1
Thus, any algebraic term having the index 0 is always equal to 1. But 0o  1.
For examples, 2o = 1, (100)o = 1, (3x)o = 1 and so on.

(vi) Root Law of Index

m be an algebraic term with base x and index m , where m and n both are positive
n
Let xn

m

integers, then, xn = n xm

Proof:

We have,

( )m n m m m
xn = xn × xn × x n × ... ... ... to n factors

mmm

= x n + n + n ... ... ... n terms

= m × n = xm

xn

m

 x n = n xm

m

Thus, x n is equal to nth root of xm.

For examples, 1 2 x , but it is written as only x.

x2 =

1 = 3 a , 31 = 4 x3

a3 x4 = (x3)4

Algebra Mathematics – 9 |113

Now, we may summarize the above laws of indices as the following.

(i) xm × xn = xm + n (ii) (xy)m = xmym

(iii) xm÷xn = xm–n if m > n and 1 if n > m. (iv)  x m = xm
xn–m y  ym

[ ](v) { }(xm)n = xmn , (xm)n p q = xmnpq (vi) x–m = 1 , 1 = xm
xm x–m

(vii) p = q xP (viii) xo = 1

xq

Worked Out Examples

Example 1: Find the products in the exponential form:
Solution:
(a) 32 × 33 × 34 × 35 × 3 (b) 2x × 4x × 8x (c) (5x)3 × (5x)–4 × (5x)7
Example 2:
Solution: Here,

(a) 32 × 33 × 34 × 35 × 3 = 32 + 3 + 4 + 5 + 1 = 315
(b) 2x × 4x × 8x = 2x ×(22)x ×(23)x = 2x × 22x × 23x = 2x + 2x + 3x = 26x
(c) (5x)3 × (5x)–4 × (5x)7 = (5x)3 + (–4) + 7 = (5x)10 – 4 = (5x)6

Find the quotient in their exponential form:

(a) 74 ÷ 72 (b) (25)2 ÷ (125)2

Here,

(a) 74 ÷ 72 = 74 – 2 = 72

(b) (25)2 ÷ (125)2 = (52)2 ÷ (53)2= 54 ÷ 56 = 54 – 6 = 5–2 = 1
52

Example 3: Express 2xy 5 243x–10y–15 with the positive indices without radical sign.

Solution: Here, 2xy 5 243x–10y–15

11 1

= 2xy(243)5 (x–10)5 (y–15)5

1

= 2xy(35)5 x–2 y–3

= 2xy × 3 × 1 × 1
x2 y3

= 6xy = 6
x2 y3 x y2

Example 4: Evaluate:

2 (b) 3 729–1 3 4 1
144
(a) (125)3 (c) (144)4 ×

114 | Mathematics – 9 Algebra

Solution: Here,

22

(a) (125)3 = (53)3 = 52 = 25

31 1
3
729–1 = (729–1)2 =
{ }(b) (36)–16
3 (729)–12 = = 3–1 = 1
3

(c) 3 4 1 = 3 ×  1144 1
144 4
(144)4 × (144)4

( )3 1 3 (144)–41
4
= (144)4 × (144)4
–1 = ×

(144)

= 3 – 1 = 2
4
(144)4 (144)4

1

= [(12)2]2 = 12

Example 5: Simplify:

146 × 155 –2 7 3 9 –5
(a) 356 × 65 (b) a3b 3 c6 ÷ a2b–1c 4

Solution: Here, –2 7 3 9

Example 6: 146 × 155 –5
Solution: 356 × 65 (b) a3 b 3 c6 ÷ a2 b–1 c 4
(a)

9 –5 1
(2 × 7)6 × (3 × 5)5 ( ) ( )–2 7 1
= (5 × 7)6 × (2 × 3)5 = a3 b 3 c6 2 ÷ a2 b–1 c 4 3

= 26 × 76 × 35 × 55 3 –1 7 3 –1 –5
56 × 76 × 25 × 35
= a2 b 3 c12 ÷ a2 b 3 c12

= 26 × 2–5 × 76 × 7–6 × 35 × 3–5 ( ) ( )3 3 –1 –1 7 –5
56 × 5–5
= a2 – 2 b 3 – 3 c12 – 12

= 26–5 × 76–6 × 35–5 –1 1 7 5
56–5
= aob 3 + 3 c12 + 12

= 21 × 7o × 3o 12
51
= ao boc12

= 1×1×c

= 2 × 1 × 1 = 2 =c
5 5

Simplify: xxmn m + n × xxnpn + p × xx–pmp – m .

Here, xxmn m + n × xxpnn + p × xx–pmp – m

= (xm–n)m+n × (xn–p)n+p × (xp+m)p–m
= xm2 – n2 × xn2 – p2 × xp2 – m2
= xm2 – n2 + n2 – p2 + p2 – m2
= xo

=1

Algebra Mathematics – 9 |115

Example 7: 12
Solution:
If x = 23 + 23 , find the value of x3 – 6x – 6.
Here, x3 – 6x – 6

( ) ( )1 2 3 12

= 23 + 23 – 6 23+ 23 – 6

( ) ( ) ( ) ( )1 3 2 3
12 1 2 12

= 23 + 23 + 3. 23 .23 22 + 23 – 6 22 + 23 – 6

[ (a + b)3 = a3 + b3 + 3ab(a + b)]

( ) ( )=
2 + 22+ 3.(2) 1 + 2 12 12 –6
3 3
22 + 23 – 6 22 + 23

( ) ( )1+2 1 2
12

= 2 + 4 + 3.(2) 3 22 + 23 – 6 22 + 23 – 6

( ) ( )1 2
12

= 6 + 3.2 22 + 23 – 6 22 + 23 – 6

( ) ( )1 2
12

= 6 – 6 + 6 22 + 23 – 6 22 + 23

=0+0

=0

Exercise 4.2

1. Find the products in their exponential forms. (b) 3a × 9a × 27a × 81–a
(a) 43 × 4–2 × 45 (d) (a + b)3 × (a + b)9 × (a + b)–7
(c) (7x)2 × (7x)–3 × (7x)5
(b) 83 ÷ 25
2. Find the quotients in their exponential forms. (d) (4a2)3 ÷ (8a)5
(a) 137 ÷ 134 (f) (4a3)2 ÷ (2a)7
(c) 163 ÷ 82
(e) (3p + q)7 ÷ (3p + q)2

3. Express the following with the positive indices without radical sign.

(a) 3 x3y–6 (b) 3 27a6 b–3 (c) x 5 x3 b–12

(d) a2 a–2 (e) 4 (81x–4 y4)3 (f) a3 a–4
4. Evaluate:
2 ( )(c) 4–34 –34
(a) 3 82
(d) (16)–43 (b) (64)3

(e) 2674–23

116 | Mathematics – 9 Algebra

5. Evaluate: (b) 1 (c) 3 8x–3
(a) 3 (64)–1 4 81–1

(d) 13–2 ÷ 31–3 (e) –3 1
64

6. Evaluate: (b) 3 729–1 (c) 6a4o –32

(a) ( 5)6

(d) 9 × 5 32 (e) 3 64 (f) 3 4 1
25 243 36
(36)4 ×

7. Simplify: (b) 3x m3 × x mx – 1 (c) (8x3 ÷ 27y–3)–32
(a) a–2b × 3 ab3

(d) 3 8x–3y6 ÷ 3 27z9 (e) 3 a6b–2c4 ÷ 4 a–4b4c2 (f) xx4–y1y–227 ÷ xx––32yy–35–2

8. Simplify:

(a) 534 × 5–31 34 (b) 83 × 272 ÷ 32–1 (c) 5–m × 252m–2
 1  93 162 81–1 53m–2 × 15–1

252

(d) 124 × 154 (e) (50)4 × (30)–4 × (20)–5 (f) (35)4 × (24)5
183 × 454 (40)3 × (60)–3 × (80)–2 (28)5 × (30)4

9. Simplify:

(a) ax–y × ay–z × az–x (b) (xa+b)a–b × (xb+c)b–c × (xa+c)a–c
(c) (ap)q–r × (aq)r–p × (ar)p–q
(d) x5b – 6c
xb + 2c . x3b – 8c

(e) x2m + 3n × x3m + 6n (f) xm + n × xm – n × xp–3m
xm + 2n × x4m – 4n xp–m

10. Simplify:

(a) 1 – 1 – b + 1 – 1 – a (b) (xa+ b)2 × (xb+ c)2 × (xc+ a)2
xa xb (xa xb xc)4

(c) aayxz . aayzx . aaxzy (d) 2x + 3 – 2x + 2
2x + 2

(e) (am – n)p × (an – p)m × (ap – m)n (f) xa – b + c. xb – c + a. xc – a + b
xa + b + c

Algebra Mathematics – 9 |117

11. Simplify:

(a) xxbca × xxacb × xxbac (b) xxbcb + c × xxcac + a × xxbaa + b

(c) aamnm + n × aa–npn – p × aa–pmp – m (d) xxbcb + c – a ×xxcac + a – b ×xxbaa + b – c

(e) aayxx2 + xy + y2 × aayzy2 + yz + z2 × aaxzz2 + zx + x2

12. Simplify:

111 (b) xxpn + mp m – n × xxpp + mn n – p × xxmn + pnp – m
– – –
(a) xxbcbc × xxcaca × xxbaab

(c) x ay × y az × z ax (d) ab xa bc xb ca xc
az ax ay xb × xc × xa

(e) x + y1m x – 1yn (f) a2 – b12m a – 1bn–m
y + 1xm y – 1xn b2 – a12n b + 1am–n

(g) 1 + 1 + 1 + ax–z + 1 + 1 + ay–x
1 + ax–y + az–y ay–z az–x

13. (a) If a = 3, b = –1 and c = 0, find the value of (ac)b + (ba)c .

(b) If m2 + n2 + p2 = mn + np + pm, prove that: xxmnm – n × xxnpn – p × xxmpp – m = 1
(c) If a3 + b3 + c3 = 1, show that: xx–aba2 – ab + b2 × xx–bcb2 – bc + c2 × xx–cac2 – ca + a2 = x2

(d) If x = 1 – a–13 , prove that: x3 + 3x = a – 1 .
a
a3

(e) If x = 1 + 3–31 , prove that: 3x3 – 9x – 10 = 0

33

(f) If x2 + 2 = 2 + 3–23 , prove that: 3x3 + 9x = 8.

33

4.3 Exponential Equation

Let’s take an equation 2x = 8. In this equation, the variable x is an unknown which is the power of the
base 2. So x is also called the index or exponent of 2. Such an equation in which the variable is
contained in exponent of the base is called the exponential equation. The more examples of the
exponential equations are 5x + 1 + 5x = 6, 2. 3x + 1 = 18, 2x + 4 × 3x + 4 = 144 etc.

118 | Mathematics – 9 Algebra

To solve an exponential equation, we should follow the following steps.
1. We should simplify the equation and reduce the single term on the both sides of the equation.
2. We should simplify the equation by using the laws of indices.
3. We should make the same base on the both sides of the equation.
4. We should compare the index or base and simplify for the required solution.
5. In solving such equations, we should be familiar with the following axioms:

If ax = ab, then x = b
If ax = 1, then ax = ao x = 0

Worked Out Examples

Example 1: Solve:
Solution:
(a) 2x = 16 (b) 27x = 3x + 4
Example 2: Here,
Solution: (a) 2x = 16

or, 2x = 24

x=4
(b) 27x = 3x + 4

or, (33)x = 3x + 4
or, 33x = 3x + 4

 3x = x + 4
or, 3x – x = 4

or, 2x = 4

or, x = 4
2

 x=2

Solve: 3 × 81x = 9x + 4
Here, 3 × 81x = 9x + 4
or, 3 × (34)x = (32)x + 4
or, 3 × 34x = 32x + 8
or, 31 + 4x = 32x + 8

 1 + 4x = 2x + 8

or, 4x – 2x = 8 – 1

or, 2x = 7

 x = 7 = 312
2

Algebra Mathematics – 9 |119

Example 3: Solve: 2x + 2 + 2x = 5
Solution:
Here, 2x + 2 + 2x = 5
Example 4:
Solution: or, 2x × 22 + 2x = 5

Example 5: or, 2x (22 + 1) = 5
Solution:
or, 2x (4 + 1) = 5

or, 2x × 5 = 5

or, 2x = 5
5

or, 2x = 1

or, 2x = 20

 x=0

Solve: 3x + 3 × 2x + 2 = 18

Here, 3x + 3 × 2x + 2 = 18

or, 3x × 33 × 2x × 22 = 18

or, 3x × 2x × 27 × 4 = 18

or, (3 × 2)x = 18 4
27 ×

or, 6x = 1
3×2

or, 6x = 1
6

or, 6x = 6–1

 x = –1

If ax = by and b = a2, prove that x – 2y = 0.

Here, ax = by

11 And, b = a2
or, a2 = b
or, (a x)x = (b y)x

y ... (i)

 a = bx

From equations (i) and (ii), we get 11

y1 or, (a2)2 = (b)2

bx = b2 1

y = 1 a = b2 ... (ii)
x 2


or, x = 2y

 x – 2y = 0 proved.

120 | Mathematics – 9 Algebra

Exercise 4.3

1. Solve: (b) 2x + 2 = 8 (c) 4x = 1
(a) 3x = 27 (e) 52x + 3 = 1 128

(d) ay – 1 = 1 x

(f) 272 = 36

2. Solve: (b) 8x = 27 – 4x (c) 64x = 4 x + 6
(a) 92x = 33 – 2 x

(d) 52x + 1 = (25)2x x+5 x +1 5x + 3 2x – 9
(e) 3 2 = 9 2 (f) 11 4 = 121 2

3. Solve: (b) 2 × 8x = 2x + 5 (c) 2 × 83 = 2x – 4
(a) 3 × 27x = 9x – 1

(d) 5 × (125)x – 1 = 52x +3 (e) 4 × 64x – 1 = 1 (f) 36 × 62x–1 = 1

4. Solve:

(a) 2x + 1 + 2x = 3 (b) 3x + 1 + 3x = 108 (c) 32x – 3 + 32x = 913

(d) 2x + 1 – 2x – 8 = 0 (e) 3x + 2 + 3x + 1 = 131 2x + 2 2x + 3
2
(f) + = 1

5. Solve: (b) 3x – 2 × 2x + 1 = 8 (c) 2x + 3 × 3x + 4 = 18
(a) 4x + 2 × 5x + 3 = 100

(d) 72x + 1 × 52x – 1 = 7 (e) 22x – 3 × 52x – 1 = 25 (f) 23x – 5×bx – 2 = 2x – 2×b1 – x
5

(g) 33x –4 × a5–2x = 3x+2 × a11 –4x = 0 (h) m5–2x  n2x–1 = m11–5xn5x –7

6. Solve.

(a) 2x + 1 = 212 (b) 4x + 1 = 441 (c) 22x+3 + 1= 3×2x+2
2x 4x

(d) 5 × 4x+1 – 16x= 64 (e) 3x + 3–x = 919

7. (a) If xm = yn and y = x2, show that m – 2n = 0.
(b) If a = bc, b = ca and c = ab, prove that abc = 1.

11

(c) If am = b5 and ab = 1, prove that m + 5 = 0.

(d) If ax ay = (ax)y, show that x(y – 2) + y(x – 2) = 0.

(e) If a = 4x, b = 4y and aybx = 16, show that xy = 1.

Algebra Mathematics – 9 |121

4.4 Ratio and Proportion

4.4.1 Ratio

Let’s consider the ages of two boys. If the age of one boy is 10 years and the age of another boy is 14

years, the age of the first boy is 10 or 5 times the age of the second boy. In fact we say that the ages
14 7

of two boys are in the ratio 10 to 14 or 5 to 7.

The ratio of two quantities is a comparison of their magnitudes where the quantities are of same kind.

If x and y are two quantities of the same kind, we say that their ratio is x or x : y (read as x is to y). (It
y

means the ratio is a fraction of two quantities of the same kind by comparing that two quantities. So,

we can write the ratio of two quantities of the same kind conveniently in the form of a fraction.

It should be noted that the quantities must be of the same kind. So, we cannot compare between Rs. 6

and 10cm. Because Rs. 6 and 10cm. are the different quantities. But we can compare 40cm. with 3m.

after converting m. into cm. and their ratio is 40cm.: 300cm. or 2:15. From this example, we should

also be careful to express two quantities in the same unit before putting their ratio in the form of a

fraction. We should also remember that every ratio is simply a number. It has no unit. Thus, the ratio

of 15cm. to 20cm. is 15 or 3 only.
20 4

A ratio is said to be how many times a quantity is greater or smaller than another quantity of the same
kind. The ratio of a to b is usually written as a:b. The quantities a and b are called the terms of the
ratio. The first term of the ratio is called the antecedent and the second term the consequent. So, in the
ratio 4:7, 4 is the antecedent and 7 the consequent.

Compounded Ratio

Let’s suppose a:b and c:d are two ratios. Then a:b × c:d = a × c = ac = ac : bd is called the
b d bd

compounded ratio of these two given ratios.

Similarly, let’s suppose a:b, c:d and e:f are three ratios. Then, a:b × c:d × e:f i.e. ace: bdf is the

compounded ratio of these given ratios. Thus, the product of two or more ratio is called the

compounded ratio. For example, 4:5 and 7:9 are two ratios. The compounded ratio of 4:5 and

7:9 = 4:5 × 7:9 = 4 × 7 = 28 = 28:45.
5 9 45

Duplicate and Sub–Duplicate Ratios

Let’s suppose a:b is a ratio. The square of the ratio a:b is (a:b)2 i.e. a2:b2 is the compounded ratio of
a:b and a:b. Then a2:b2 is called the duplicate ratio of a:b and a:b is called the sub–duplicate ratio of
a2:b2. For example, let’s take a ratio 2:3.

The duplicate ratio of 2:3 = 22:32 = 4:9

The sub–duplicate ratio of 4:9 = 4 : 9 = 2:3

122 | Mathematics – 9 Algebra

Triplicate and Sub–Triplicate Ratios

Let’s suppose a:b is a ratio. The cube of the ratio a:b is (a:b)3 i.e. a3:b3. The ratio a3:b3 is the

compounded ratio of a:b, a:b and a:b. Thus, the ratio a3:b3 is called the triplicate ratio of a:b. The cube

root of a3:b3 = 3 a3:b3 1 11 = a:b
= (a3:b3)3 = (a3)3 : (b3)3

Then, a: b is called the sub–triplicate ratio of a3:b3.

For example, let’s take a ratio 4:5. Then, the triplicate ratio of 4:5 = 43:53 = 64:125.

The sub–triplicate ratio of 64:125 = 3 64:125 = 3 64 :3 125 = 4:5

Inverse Ratio

Let’s take a ratio a:b, where a is the antecedent and b is the consequent. When the antecedent and
consequent of the ratio a:b are interchanged, the ratio becomes b:a. Then the ratio b:a is called the
inverse ratio of a:b. For example, 2:3 be a ratio. Then the inverse of ratio 2:3 is 3:2.

Worked Out Examples

Example 1: Find the ratio of the following:

(a) 6 min to 3 hours (b) 3 kg to 750 gm
Here,
Solution: (a) 6 min to 3 hours

The ratio of 6min to 3 hours = 6 min : 3hours [... 1 hour = 60 min]

(b) 3 kg to 750gm. = 6 min : 3×60 min
The ratio of 3kg to 750gm = 1:30

= 3kg : 750gm [...1kg = 1000gm.]

= 3 × 1000gm : 750gm
= 3 × 4 : 3 = 4:1

Example 2: If x:y = 3:4 and y:z = 5:7, find x:z and x:y:z.
Solution:
Here, x:y = 3:4 and y:z = 5:7

Now, x:y × y:z = 3:4 × 5:7

or, x × y = 3 × 5
y z 4 7

or, x = 15
z 28

 x:z = 15:28.

Again, in the ratio x:y = 3:4, y is 4 and in the ratio y:z = 5:7, y is 5. The L.C.M of 4
and 5 is 20.

Algebra Mathematics – 9 |123

Now, x:y = 3:4 = 3×5:4×5 = 15:20
y:z = 5:7 = 5×4:7:×4 = 20:28
 x:y: z = 15:20:28

Example 3: If x:y = 3:4, find the value of 3x+4y:5x – 2y.
Solution:
Here, x:y = 3:4 i.e. x = 3
y 4

3x + 4y

Now, 3x + 4y = y [Dividing the numerator and denominator by y.]
5x – 2y 5x – 2y

y

3x + 4y 3 × 3 + 4 9 + 4 9 + 16
y y 4 4
= = = = 4 = 25
5x 2y 3 15 15 – 8 7
y – y 5 × 4 – 2 4 – 2
4

Example 4: If 5x + 2y:7x + 3y = 9:13, find the value of x:y.
Solution:
Here,

5x + 2y:7x + 3y = 9:13

or, 5x + 2y = 9
7x + 3y 13

or, 13(5x + 2y) = 9(7x + 3y) [by the cross – multiplication ]

or, 65x + 26y = 63x + 27y

or, 65 – 63x = 27y – 26y

or, 2x = y

or, x = 1
y 2

 x:y = 1:2

Example 5: The sum of two numbers which are in the ratio of 3:5 is 56, find the
Solution: numbers.

Here,

The ratio of two numbers is 3:5.

Let x be a common factor of two numbers.

Then, the two numbers are 3x and 5x.

According to question,

3x + 5x = 56

or, 8x = 56

or, x = 56 = 7
8

Now, 3x = 3 × 7 = 21

5x = 5 × 7 = 35

Hence, the two numbers are 21 and 35.

124 | Mathematics – 9 Algebra

Example 6: Divide Rs. 252 into two parts so that their ratio is 5:9.
Solution:
Here,

The ratio of two parts is 5:9

Let x be a common factor of two parts. Then, the two parts are 5x and 9x.

Now, 5x + 9x = Rs. 252

or, 14x = Rs. 252

or, x = Rs. 252
14

x = Rs. 18

Therefore, the first part = Rs. 5x = Rs. 5 × 18 = Rs. 90

The second part = Rs. 9x = Rs. 9 × 18 = Rs. 162

Hence, the two parts of Rs. 252 which are in the ratio 5:9 are Rs. 90 and Rs. 162.

Example 7: What number must be subtracted from each term of a ratio 27:35 to
Solution: make it equal to 7:11?

Let the required number be x.

Then, 27 – x = 7
35 – x 11

or, 11(27 – x) = 7(35 – x)

or, 297 – 11x = 245 – 7x

or, –11x + 7x = 245 – 297

or, – 4x = – 52

or, x = 52
4

or, x = 13

Hence, the required number is 13.

Example 8: The ratio of two numbers is 5:7. If 15 is added to each number, the ratio
Solution: becomes 3:4. Find the numbers.

Here, The ratio of two numbers is 5:7.

Let x be a common factor of two numbers.

Then, two numbers are 5x and 7x.

According to question,

5x +15 = 3
7x + 15 4

or, 4(5x + 15) = 3(7x + 15)

or, 20x + 60 = 21x + 45

or, 20x – 21x = 45 – 60

or, – x = –15

 x = 15

The first number = 5x = 5 × 15 = 75

Algebra Mathematics – 9 |125

The second number = 7x = 7 × 15 = 105
Hence, the required two numbers are 75 and 105.

Example 9: The ratio of the present ages of a daughter and her mother is 5:12. If the
Solution:: ratio of their ages after 6 years will be 1:2, find their present ages.

Here, The ratio of the present ages of a daughter and her mother is 5:12.

Let x be a common factor of their ages.

Then, the present ages of the daughter and her mother are 5x years and 12x years

respectively.

After 6 years, their ages will be (5x + 6) years and (12x + 6) years respectively.

According to question,

(5x + 6) : (12x + 6) = 1:2

or, 5x + 6 = 1
12x + 6 2

or, 2(5x + 6) = 1(12x + 6)

or, 10x + 12 = 12x + 6

or, 10x – 12x = 6 – 12

or, – 2x = – 6

or, x = 6
2

 x=3

 The present age of the daughter = 5x = 5 × 3 = 15 years.
The present age of the mother = 12x = 12 × 3 = 36 years.

Exercise 4.4.1

1. Find the ratios of the following:

(a) 9min to 8hours (b) 70cm to 1.4m (c) Rs. 4.50 to 75 paisa

(d) 7km to 500m. (e) 2.5kg to 950gm. (f) 12days to 3weeks.

2. Find the compounded ratio of the following ratios: –

(a) 4:5 and 10:18 (b) 7:9 and 27:21 (c) 2:3, 4:7 and 1:5

3. Find the duplicate ratio of the following ratios: –

(a) 2:5 (b) 2x:7y (c) 12:7 (d) 14a:9b

4. Find the sub – duplicate ratio of the following ratios:

(a) 9:16 (b) 49:36 (c) 169a2:144b2 (d) 4:9

5. Find the triplicate ratio of the following ratios:

(a) 4:7 (b) 2a:b (c) 5:2 (d) 6x:11y

6. Find the sub – triplicate ratio of the following ratios:

(a) 8:27 (b) 125x3:1 (c) 216a3:64b3 (d) 729x6 : 512y9

126 | Mathematics – 9 Algebra

7. Find the inverse ratio of the following ratios:

(a) 12:19 (b) 16x2:21y2 (c) (2x + 3):(9y – 5) (d) 18a:7b

8. (a) If x:y = 4:9 and y:z = 12:11, find x:z and x:y:z.

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

(c) If p:q = 8:11 and q:r = 33:16, find p:r and p:q:r.

9. (a) If a:b = 3:4, find the value of 4a + 3b:12a – b.

(b) If x:y = 6:7, find the value of 4x – y:5x + y.

(c) If p:q = 2:3, find the value of 5p – 2q:2p + q.

(d) If a:b = 2:3, find the value of a – 2b
3

(e) If x:y = 3:4, what will be the value of x – 3y ?
4

10. (a) If 7a – 11b = 0, what will be the ratio of a:b?

(b) If 6a – 8b = 0, find the value of a:b.

(c) If 3m – 5n:3m + 5n = 1:4, find the value of m:n.

(d) If 5a + b:5a – b = 3:1, find the value of a:b

(e) If 3x – 5y:3x + 5y = 2:5, find the value of x:y.

11. Find the value of x:y in the following each cases.

(a) 2x – 3y:2x + 3y = – 1:3 (b) 2x – 3y = 2y – x

(c) 10x – 8y = 0 (d) 3x – y:2x + y = 1:2

(e) 7x – 4y:3x + y = 5:13 (f) x + 3y:3x + 2y = 5:7

(g) 5x + 3y:4x + 5y = 2:3 (h) 2x + 3y : 23x + y2 = 12: 17

12. (a) If a:b:c = 2:5:7, find the value of 2a:3b + c.

(b) If a:b:c = 3:4:7, find the value of (3a + 2b – c):(6a – 2c).

(c) If x:y:z = 4:5:6, find the value of (2x – 3y):(x + 2y – 3z).

13. (a) The sum of two numbers which are in the ratio 5:9 is 98. Find the numbers.

(b) Two numbers are in the ratio of 5:7 and their sum is 168. Find the numbers.

(c) The difference of two numbers whose ratio is 3:5 is 50. Find the numbers.

(d) Two numbers are in the ratio 4:7 and their difference is 57. Find the numbers.

(e) If the angles of a triangle are in the ratio 2:3:4, find the angles of the triangle.

14. (a) Divide Rs. 68 into two parts so that their ratio is 8:9.

(b) Divide 72 into two parts such that their ratio is 5:7.

(c) Divide Rs. 21 into two parts such that their ratio is 3:4.

(d) When Rs. 105 is divided into two parts in the ratio of 2:5, how much money will be in
the first part?

(e) 99 students are divided in two rooms in the ratio of 5:6. How many students will there
be in the second room?

15. (a) What number must be subtracted from each term of a ratio 4:9 so that it may become
equal to 9:4?

Algebra Mathematics – 9 |127

(b) What number must be subtracted from each term of a ratio 5:7 so that it may become
equal to 6:5?

(c) What number must be added to each term of the ratio 4:7 to make it equal to 2:3?

(d) What number must be added to each term of the ratio 29:36 to make it equal to 3:4?

16. (a) The ratio of two numbers is 3:5. If 8 is added to each number, the ratio becomes 7:9.
Find the numbers.

(b) The ratio of two numbers is 2:3. If 17 is added to each number, the ratio becomes 3:4.
Find the numbers.

(c) The ratio of two numbers is 25:37. If 14 is subtracted from each, the ratio of the
remainder is 3:5. Find the numbers.

(d) Two numbers are in the ratio of 4:5. If 7 is subtracted from each, the ratio of the
remainder is 7:9. Find the numbers.

17. (a) The ratio of the present ages of two men is 3:4. If the ratio of their ages after 4 years
will be 7:9, find their present ages.

(b) The ratio of the present ages of two persons A and B is 3:4. If the ratio of their ages
before 7 years was 2:3, what are their present ages be?

(c) The ratio of the present ages of a son and his father is 2:7. After 5 years, the ratio of
their ages will be 3:8. Find their present ages.

18. (a) The ratio of milk and water in a container is 7:4 by volume. If it contains 55 litres of the
mixtures, find the quantities of milk and water in it.

(b) Two numbers are in the ratio of 5:9. If the antecedent is 30, find the consequent.

(c) If the ratio of two complementary angles is 2:7, find the angles.

(d) On a map with a scale of 1cm:1km, what actual distance on the ground is represented on
the map by 8cm?

4.4.2 Proportion

Let a, b, c and d be four quantities. The ratio of a and b is a:b and the ratio of c and d is c:d. If
a:b = c:d, then these two equal ratios are called the proportional and the four quantities a, b, c and d
are said to be in proportion. For example, 4, 5, 16 and 20 are four quantities. The ratio of 4 and 5 is
4:5 and the ratio of 16 and 20 is 16:20. The ratio 16:20 is equivalent to the ratio 4:5. Therefore,
4:5 = 16:20. So, these two ratios are proportional and the four quantities 4, 5, 16 and 20 are in
proportion.

If a:b and c:d are proportional, it can be written as a:b:: c:d and read as ‘a is to b as equal c is to d’. In
the equal ratios a:b::c:d the terms a and d are called extremes and the terms b and c are called means.

The ratios a:b::c:d also can be written as a = c . Then, by the cross–multiplication, it can be written as
b d

ad = bc. ad is the product of extremes and bc is the product of means. Thus, the product of extremes is

equal to the product of means in every proportion.

128 | Mathematics – 9 Algebra

Continued Proportion

Let’s consider three numbers 5,10 and 20. The ratio of 1st and 2nd numbers is 5:10 i.e. 1:2. The ratio of
2nd and 3rd numbers is 10:20 i.e. 1:2. So, 5:10 = 10:20, then we say that 5,10 and 20 are continued
proportion. 5:10 = 10:20 is called the continued proportional. If a:b = b:c, then ac = b2. i.e. b = ac .

Thus, three quantities are said to be in continued proportion if the mean proportional is equal to the
square root of the product of extremes.

Properties of Proportion

If a, b, c and d are in proportion, then the following properties of proportion exist.

(i) Invertendo: If a : b = c : d, then b : a = d : c

Proof: Here,

a:b=c:d

or, a = c
b d

Now, 1  a = 1  c [... Taking receproval of them]
b d

or, 1 × b = 1 × d
a c

or, b = d i.e. b : a = d : c
a c

(ii) Alternendo: If a : b = c : d, then a : c = b : d.

Proof: Here, a : b = c : d

or, a = c
b d

or, a × b = c × b [Multiplying both sides by b ]
b c d c c

or, a = b i.e. a : c = b : d
c d

(iii) Componendo: If a : b = c:d, then (a + b) : b = (c + d) : d

Proof: Here, a : b = c : d

or, a = c
b d

or, a + 1 = c + 1 [Adding 1 on both sides]
b d

or, a + b = c + d
b d

i.e. (a + b) : b = (c + d) : d

Algebra Mathematics – 9 |129

(iv) Dividendo: If a : b = c : d, then (a – b) : b = (c – d) : d.

Proof: Here, a : b = c : d

or, a = c
b d

or, a – 1 = c –1 [ Subtracting 1 from both sides]
b d

or, a – b = c – d
b d

i.e. (a – b) : b = (c – d) : d

(v) Componendo and dividendo: If a : b = c : d, then (a + b) : (a – b) = (c + d) : (c – d).

Proof: Here,

a : b = c : d,

By componendo, we have

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

By alternendo in equation (i), we have

a+b = b ... (ii)
c+d d

Again, by dividendo in a:b = c:d, we have

a–b = c – d ... (iii)
b d

By alternendo in equation (iii), we have

a–b = b ... (iv)
c–d d

From equation (ii) and equation (iv), we have

a+b = a – b
c+d c – d

By alternendo, we have

a+b = c + d
a–b c – d

(a + b) : (a – b) = (c + d) : (c – d)

(vi) Addendo: If a : b = c : d, then a : b = c : d = (a + c) : (b + d).

Proof: Here, a : b = c : d

or, a = c
b d

130 | Mathematics – 9 Algebra

By alternendo, we have

a = b
c d

By componendo, we have

a+c = b + d
c d

Again, by alternendo, we have

a+c = c
b+d d

 a = c = a+c
b d b+d

i.e. a : b = c : d = (a + c) : (b + d)

Similarly,

a = c = e = ................ = a + c + e + .....................
b d f b + d + f + .....................

Worked Out Examples

Example 1: If 1, a, 5 and 15 are proportional, find the value of a.
Solution:
Here,

1, a, 5 and 15 are proportional

1 = 5
a 15

or, 5a = 15

or, a = 15
5

 a = 3

Example 2: If 5, x and 20 are in continued proportion, find the value of x.
Solution:
Here,

5, x and 20 are in continued proportion.

 5 = x
x 20

or, x2 = 100

or, x = 100
 x = 10

Algebra Mathematics – 9 |131

Example 3: What constant number should be added to each of the terms 3, 5, 7 and
Solution: 10 so that they are in proportion?

Let the constant number to be added be x.

Then, by the question,

3+x = 7+x
5+x 10 + x

or, (3 + x)(10 + x) = (7 + x) (5 + x)
or, 30 + 3x + 10x + x2 = 35 + 7x + 5x + x2
or, 13x + x2 – 12x – x2 = 35 – 30

 x=5

Hence, the required number to be added is 5.

Example 4: If x = 2 , find the value of 6x + y .
Solution: y 3 3x + 2y

Here,

x = 2
y 3

or, x = y [ by altrnendo]
2 3

or, x = y = k(suppose)
2 3

 x = 2k and y = 3k

Now, 6x + y = 3 6 × 2k + 3k = 12k + 3k = 15k = 5
3x + 2y × 2k + 2 × 3k 6k + 6k 12k 4

Example 5: If a = b prove that a = a – b
Solution: b c a+b a – c

Here, a = b
b c

or, b = c by invertendo
a b

or, a + b = b + c by componendo
a b

 a a b = b b c ... (i) by invertendo
+ +

Again, a a b = b b c
+ +

or, a = a + b by alternendo
b b + c

or, a–b = a + b – b – c by dividendo
b b + c

or, a – b = a – c
b b + c

132 | Mathematics – 9 Algebra

or, a – b = b b c ... (ii)
a – c +

From (i) and (ii), we get

 a a b = a – b Proved
+ a – c

Example 6: If a = bc, prove that ba + b2 = a2 + b2
Solution: b + c b2 + c2

Here a = b
b c

 a + b = b by addendo
b + c c

 ba + bc2 = b2 ... (i) [... Squaring on both sides]
+ c2

Again a = b
b c

 a2 = b2
b2 c2

 a2 + b2 = b2 + c2 by componendo
b2 c2

 a2 + b2 = b2 ... (ii) by alternendo
b2 + c2 c2

 ab + b2 = a2 + b2 from (i) and (ii)
+ c b2 + c2

Proved.

Example 7: If a = b , prove that (a – b)2 = (b – c)2
Solution: b c a c

Here, a = b
b c

or, b = c by invertendo
a b

or, a –b = b – c by dividendo
a b

or, (a – b)2 = a2 [...squaring on both sides]
(b – c)2 b2

or, (a – b)2 = a2 [... b2 = ac]
(b – c)2 ac

or, (a – b)2 = a
(b – c)2 c

 (a – b)2 = (b – c)2 Proved.
a c

Algebra Mathematics – 9 |133

Example 8: If a = b , prove that ab = c2 cd
Solution: b c a2 – b2 – d2

Here, a = c
b d

or, a2 = c2
b2 d2

or, a2 – b2 = c2 – d2 by dividendo
b2 d2

or, a2 b2 = c2 d2 by invertendo
– b2 – d2

or, a2 b2 × a = c2 d2 × c ... a = dc
– b2 b – d2 d b

 ab = cd Proved
a2 – b2 c2 – d2

Example 9: If a:b = c:d, prove that ac = a2 + c2 .
Solution: bd b2 + d2

Here,

a:b = c:d

Let a = c =k
b d

a = bk and c = dk

Now, L.H.S = ac = bk.dk = bdk2 = k2
bd bd bd

R.H.S = a2 + c2
b2 + d2

= (bk)2 + (dk)2
b2 + d2

= b2k2 + d2k2 = k2(b2 + d2) = k2.
b2 + d2 b2 + d2

L.H.S = R.H.S proved.

Example 10: If a:b = c:d = e:f, prove that a2 + c2 + e2 = ce .
b2 + d2 + f2 df

Solution: Here,

a:b = c:d = e:f

or, a = c = e = k(suppose)
b d f

 a = bk, c = dk and e = fk

Now, L.H.S = a2 + c2 + e2
b2 + d2 + f2

= (bk)2 + (dk)2 + (fk)2 = b2k2 + d2k2 + f2k2 = k2(b2 + d2 + f2) = k2.
b2 + d2 + f2 b2 + d2 + f2 b2 + d2 + f2

134 | Mathematics – 9 Algebra

R.H.S = ce = dk.fk = dfk2 = k2
df df df

 L.H.S = R.H.S proved

Example 11: If a, b, c are in continued proportion, prove that:

Solution: a3 + b3: b3 + c3 = a(a – b):c(b – c).
Here,

a, b, c are in continued proportion.

 a = b = k (say)
b c

 b = ck

a = bk = ck.k = ck2

Now, L.H.S = a3 + b3 = (ck2)3 + (ck)3 = c3k6 + c3k3 = c3k3(k3 + 1) = k3
b3 + c3 (ck)3 + c3 c3k3 + c3 c3(k3 + 1)

R.H.S = a(a – b) = ck2(ck2 – ck) = c2k3(k – 1) = k3
c(b – c) c(ck – c) c2(k – 1)

 L.H.S = R.H.S proved.

Example 12: If a, b & c are in continued proportion, show that (a2+b2) (b2+c2) = b2(a+c)2

Solution: Since a, b and c are in continued proportion, a = b
b c

 b2 = ac
L.H.S. = (a2 + b2) (b2 + c2)

= (a2 + ac)(ac + c2)

= a(a + c).c(a + c)
= ac(a + c)2
= b2(a + c)2 = RHS

Proved.

Example 13: If a, b, c and d are in continued proportion, show that (a+d) (b+c) –(a+c)
(b+d) = (b–c)2

Solution: Since a, b, c and d are in continued proportion, a = b = c
b c d

 b2 = ac, c2 = bd and ad = bc

L.H.S. = (a+d) (b+c) –(a+c) (b+d)

= (ab+bd+ac +cd)–(ab+bc+ad+cd)

= ab + bd +ac +cd –ab –bc –ad –cd

= c2 +b2 –bc –bc [... bc = ad]

= b2–2bc+c2

= (b–c)2 = R.H.S. Proved.

Algebra Mathematics – 9 |135

Example 14: If a, b, c and d are in continued proportion, prove that a3 – b3 – c3 = a .
b3 – c3 – d3 d

Solution: Here,

a, b, c and d are in continued proportion,

 a = b = c = k (say)
b c d

 c = dk

b = ck = dk.k = dk2

a = bk = dk2.k = dk3

Now, L.H.S = a3 – b3 – c3 = (dk3)3 – (dk2)3 – (dk)3 = d3k9 – d3k6 – d3k3
b3 – c3 – d3 (dk2)3 – (dk)3 – d3 d3k6 – d3k3 – d3

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

R.H.S = a = dk3 = k3.
d d

L.H.S = R.H.S proved.

Example 15: If x:a = y:b = z:c, prove that x3 + y3 + z3 = xyz .
a3 + b3 + c3 abc

Solution: Here,

x = y = z = k (suppose)
a b c

 x = ak, y = bk and z = ck

Now, L.H.S = x3 + y3 + z3
a3 + b3 + c3

= (ak)3 + (bk)3 + (ck)3 = a3k3 + b3k3 + c3k3 = k3(a3 + b3 + c3) = k3
a3 + b3 + c3 a3 + b3 + c3 a3 + b3 + c3

R.H.S. = xyz = ak.bk.ck = abck3 = k3
abc abc abc

L.H.S = R.H.S proved

Example 16: If x = y = z , prove that xy – yz + zt = (x – y + z) (y – z + t)
y z t

Solution: Here,

x = y = z = k (suppose)
y z t

z = tk
y = zk = tk.k = tk2
x = yk = tk2.k = tk3

Now, Algebra
L.H.S = xy – yz + zt
136 | Mathematics – 9

= tk3.tk2 – tk2.tk + tk.t
= t2k5 – t2k3 + t2k
= tk2 k – tk k + t k
= t k (k2 – k + 1)

R.H.S = (x – y + z)(y – z + t)
= (tk3 – tk2 + tk)(tk2 – tk + t)
= tk(k2 – k + 1).t(k2 – k + 1)
= t2k(k2 – k + 1)2
= t k (k2 – k + 1)

 L.H.S = R.H.S proved.

Exercise 4.4.2

1. (a) If 3, 4, 6 and a are proportional, find the value of a.

(b) If 4, x, 10 and 15 are proportional, find the value of x.

(c) If 4, x and 9 are in continued proportion, find the value of x.

(d) If 3, x and 27 are in continued proportion, find the positive value of x.

(e) If 3, 12, a and 192 are in proportion, find the value of a.

2. (a) What is the value of x if 3, 4, x and 28 are in proportion.

(b) Find the fourth proportion of 3, 4, 9.

(c) What is the value of x when x, 6 and 4 are in continued proportion?

(d) Find the third proportion of 14, 10, 25.

3. (a) Find the value of x in the following proportions:

(i) 4:12: :x:6 (ii) 4:x: :8:12 (iii) 5:10: :20:x

(b) What constant number should be added to each of the terms 8, 13, 10 and 16 so that
they are in proportion?

(c) What number should be added to each of the terms 2,4,7 so that the sums will be in
continued proportion?

(d) What number should be subtracted to each of the terms 4, 5, 7 and 9 so that the
differences will be in proportion?

(e) What number must be deducted from each of the terms 6, 10, 18 so that the difference
will be in continued proportion?

4. (a) If a:b = 4:3, find the value of 3a – 2b . (b) If x:y = 5:6, find the value of 5x + 2y .
3a + 2b 5x – 2y

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

Algebra Mathematics – 9 |137

5. If a = c by using properties of propertion, prove that:
b d

(a) a a b = c c d (b) a–b = c – d (c) 4a + 7b = 4a – 7b
+ + a+b c + d 4c + 7d 4c – 7d

(d) c2 = a2 + c2 (e) ac + bd2 = a2 + b2 (f) a2 + c2 = c
d2 b2 + d2 + c2 + d2 b2 + d2 d

(g) ba – cd2 = a2 (h) ab + cd = a2 + c2
– b2 ab – cd a2 – c2

(i) ma + nc = a2 + c2 (j) a(c – a)2 = b(a2 + c2)
mb + nd b2 + d2 c(b – d)2 d(b2 + d2)

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

(a) a – b = c – d (b) a2 – b2 = b2
a + b c + d c2 – d2 d2

(c) ac = a2 – c2 (d) a2 + b2 = (a + b)2
bd b2 + d2 c2 + d2 (c + d)2

(e) a2 – c2 = c (f) 3a + 4c = a2 + c2
b2 – d2 d 3b +4d b2 + d2

(g) ad + bc = 2bd (h) a2 + ab + b2 = a2 – ab – b2
a2 + c2 ab + cd c2 + cd + d2 c2 – cd – d2

7. If a:b = c:d = e:f, prove that:

(a) (a2 + c2 + e2) (b2 + d2 + f2) = (ab + cd + ef)2 (b) ba + 2c – 33ef2 = c(a + e)
+ 2d – d(b + f)

(c) ac + ce + ae = a2 (d) a2 + c2 – e2 = ab + cd – ef
bd + df + bf b2 ab + cd – ef b2 + d2 – f2

(e) pa + qc + re = 3 ace (f) a3 + c3 + e3 = ace
pb + qd + rf bdf b3 + d3 + f3 bdf

(g) a + c + e = a (h) (a – 2c + e)3 = a3 + c3 + e3
b + d + f b (b – 2d + f)3 b3 + d3 + f3

8. If a = b by using properties, prove that:
b c

(a) a = a – b (b) (a + b)2 = (b + c)2 (c) (a – b)2 = (b – c)2
+ a – c b2 c2 a c
a b

(d) a3 + b3 = a(a – b) (e) a2 – b2 = a (f) a – 2b + c = (b– c)2:c
b3 + c3 c(b – c) b2 – c2 c

(g) a2 + ab = b2 + bc (h) (a + b)2 = a2 + b2 (i) a2 + ab + b2 = a
b2 c2 (b + c)2 b2 + c2 b2 + bc + c2 c

9. If a, b and c are in continued proportion, prove that:

(a) abc (a + b + c)3 = (ab + bc + ca)3 (b) a2 – ab + b2 = a2 + b2
b2 – bc + c2 b2 + c2

(c) (a + b + c)2 = a + b + c (d) (a + b + c)(a – b + c) = a2 + b2 + c2
a2 + b2 + c2 a – b + c

(e) b2c2 + c2a2 + a2b2 = a3 + b3 + c3 (f) a3 + b3 + c3 = 1 (a3b3 + b3c3 + c3a3)
a b c abc

138 | Mathematics – 9 Algebra

10. (a) If p, q and r are in continued proportion, show that: p3 + q3 + r3 = 1 + 1 + r13.
p2q2r2 p3 q3

(b) If x : y : : y : z, prove that x2y2z2x13 + 1 + 1  = x3 + y3 + z3.
y3 z3 

(c) If a : b = b : c, prove that 1 (ab + bc + ca)3 = (a + b + c)3.
abc

(d) If x : y : : y : z, prove that y2 + yz + z2 : x2 + xy + y2 = z : x.

11. (a) If a, b, c and d are in continued proportion, prove that: (a3 + b3 + abc) : (b3 + c3 + bcd) = a:d.

(b) If a, b, c and d are in continued proportion, prove that: a + b : c + d = a2 + b2 + c2: b2 + c2 + d2.

(c) If a, b, c and d are in continued proportion, prove that: (a + d)(b + c) – (a + c)(b + d) = (b – c)2.

(d) If a, b, c and d are in continued proportion, prove that: a2 – b2 : b2 – c2 : :b2 – c2 :c2 – d2.

12. (a) If p = q = r , prove that: (p + q + r)(q + r + s) = pq + qr + rs .
q r s

(b) If p = q = r , prove that: (p2 + q2 + r2)(q2 + r2 + s2) = (pq + qr + rs)2.
q r s

(c) If a = b = c , prove that: (b – c)2 + (c– a)2 + (d – b)2 = (a – d)2.
b c d

(d) If a = b = c , prove that: (a – c)2 + (b – d)2 = (a – d)2 – (b – c)2.
b c d

13. (a) If p = q = r , prove that: pq – qr + rs = (p – q + r)(q – r + s) .
q r s

(b) If a, b, c, d, e are in continued proportion, prove that: a : e = a4 : b4.

(c) If x = y = z , prove that: x3 – y3 – z3 = (x – y – z)3 .
a b c a2 b2 c2 (a – b – c)2

(d) If a, b, c, and d are in continued proportion, prove that:

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

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

4.5 Simultaneous Equations

4.5.1 Linear Equation Y

Let's consider an equation 2x + 3y = 5 in x and y. The variables y = 3x - 2
quantities x and y of the equation 2x + 3y = 5 have the power 1. Such (4,10)
equation is called the first degree equation. The first degree equation
is known as the linear equation. (3,7)

Every linear equation can be expressed in the form of y = mx + c, X’ (1,1) X
where m is the slope of the straight line represented by the equation O (0,-2)
and c is the y–intercept made by the line in the y–axis.
(-1,-5)

Y’

Algebra Mathematics – 9 |139

Graph of Linear Equation

Let's consider an equation y = 3x – 2. In the equation, the value of y depends on the value of x.
Substituting the different values of x, the corresponding values of y can be calculated. When we put
the values of x as –1, 0, 1, 3, 4 in the equation y = 3x – 2 successively, then the corresponding values
of y are obtained which are –5, –2, 1, 7, 10 respectively.

We present these values in a table given below. 1 34
1 7 10
x –1 0
y –5 –2

From the above table the order pairs (–1, –5), (0, –2), (1, 1), (3, 7) and (4, 10) are obtained. We plot
these order pairs on a graph paper. By joining these points, a straight line is obtained which is shown
in the above diagram. From the graph we can see that the plotted points of y = 3x – 2 lie on the
straight line. So it is concluded that the graph of any linear equation represents a straight line.

Simultaneous Equations

Let's take an equation x + y = 7. The equation x + y = 7 also can be written as y = 7 – x. The equation
has as many pairs of solutions as substituting the values of x. A few pairs of solutions of the equation
are shown on the following given table.

x 0 2 –1 5 –3 8
y 7 5 6 2 10 –1

From the above table, (0, 7), (2, 5), (–1, 6), (5, 2), (–3, 10), (8, –1) are a few pairs of solutions that
satisfy the equation x + y = 7.

Let's take another equation x – y = 3. This equation has also as many pairs of solutions as substituting
the values of x. A few pairs of solutions of this equation are (0, –3), (1, –2), (–1, –4), (3, 0), (–3, –6),
(5, 2), (6, 3). These pairs of solutions satisfy the equation x – y = 3. But, a pair solution i.e. (5, 2)
satisfies both the equations. Such pair of equations which are satisfied by only one pair of solution is
called the simultaneous equations. Thus, simultaneous equations are a pair of equations. Both the
equations have one unique solution which satisfies on the both equations at the same time.

We can solve the simultaneous equations in two variables by the different methods. But, according to
the curriculum of class IX, we have discussed three methods for solving the simultaneous equations.
The three methods are substitution method, elimination method and graphical method.

(i) Substitution Method

Let’s consider the equations

x + 3y = 7 and 3x – y = 11

Here,

x + 3y = 7 ... (i)

3x – y = 11 ... (ii)

From equation (i)

x = 7 – 3y ... (iii)

140 | Mathematics – 9 Algebra

Substituting the value of x in equation (ii) from equation (iii), we have the equation in one variable
which is
3(7 – 3y) – y = 11
or, 21 – 9y – y = 11
or, –10y = 11 – 21
or, –10y = –10

y = 1
Now, again substituting the value of y in equation (iii), we get
x=7–3×1=7–3=4

 Solution of the equations (i) and (ii) is x = 4 and y = 1.

Check: Substituting the values of x = 4 and y = 1 in equation (i) and (ii), we have

4+3×1=7 and 3 × 4 – 1 = 11

or, 7 = 7 or, 11 = 11

Which are true.

From the solution of above example, it is concluded that the following steps are applied to
solve the simultaneous equations by the substitution method.

 Express the value of x in terms of y or the value of y in terms of x by taking any one
equation from the given two equations.

 Substitute the value of x(or y) in the next equation and make the equation in the single
variable.

 Solve the equation and find the value of x(or y).

 Again, substitute the value of x(or y) in any one equation and solve to find the value of
y(or x).

(ii) Elimination Method

The method of eliminating one of the two variables in a pair of simultaneous equation is to
make the same coefficient of the variables to be eliminated. This can be done by multiplying
the equations by such quantities so that the coefficient of x or y should be the same in two
resulting equations. Then we add or subtract the resulting two equations to eliminate one of the
two variables, we get an equation in a single variable. Then, we solve the equation with single
variable and find the value of the variable. After getting the value of the variable, it is
substituted in one of the equation to get the value of the remaining variable. Let’s learn the
process from the following example.

Consider the two equations

5x + 3y = 5 and 3x – 4y = 32

Here, 5x + 3y = 5 ... (i)

3x – 4y = 32 ... (ii)

Here, we are eliminating the variable y. So, to make the same coefficient of y. Multiplying
equations (i) by 4 and equation (ii) by 3. Then adding the resulting equations.

Algebra Mathematics – 9 |141

20x + 12y = 20

9x – 12y = 96

29x = 116

or, x = 116
29

x=4
Now, substituting the value of x in equation (i), we get

5 × 4 + 3y = 5

or, 20 + 3y = 5

or, 3y = 5 – 20

or, y = – 15
3

y = – 5
Hence, x = 4 and y = –5.

From the solution of the above example, it is concluded that the following steps are applied to
solve the simultaneous equations by the elimination method.

 Make the same coefficients of the variables x or y in the both equations by multiplying
both sides of the equations with the suitable number.

 Eliminate one of the variables x or y by adding or subtracting the resulting equations
according to these equal quantities that have opposite or same sign.

 Solve the equation with the single variable and find the value of the variable.

 Substitute the value of the variable in any one of the equations and find the value of the
remaining variable.

Worked Out Examples

Example 1: Solve by the substitution method: 2x – 3y = 4 and 3x + y = –5
Solution:
Here,

2x – 3y = 4 ... (i)

3x + y = –5 ... (ii)

From equation (i),

2x = 4 + 3y

or, x = 4 + 3y ... (iii)
2

Substituting the value of x in equation (ii) from equation (iii), we get

34 +23y + y = – 5

or, 12 + 9y +y=–5
2

142 | Mathematics – 9 Algebra

or, 12 + 9y + 2y =–5
2

or, 12 + 11y = –10

or, 11y = –10 – 12

or, y = – 22
11

 y = –2

Now, substituting the value of y in equation (iii), we get

x = 4 + 3(–2) = 4 – 6 = – 2 = –1
2 2 2

Hence, x = –1 and y = –2.

Example 2: Solve the following equations by elimination method: 7x– 4y = –9 and
3x–y = 4

Solution: Here,

7x – 4y = – 9 ... (i)

3x – y = 4 ... (ii)

Multiplying equation (ii) by 4 and subtracting equation (i),

12x - 4y = 16
-7x -+4y = -9
5x = 16

or, x = 25
5

x=5

Now, multiplying equation (i) by 3 and equation (ii) by 7 and substituting, we get

21x - 12y = -27
-21x +- 7y = -28

-5y = -55

 y = 11

Hence, x = 5 and y = 11

Note: It is better to use elimination method in the first step and substitution method in the next steps which
we cal mixed method.

Example 3: Solve the equations: x – 2 = 1 and x + 3 =3
Solution: 3 y 4 y

Here,

x – 2 =1 ... (i)
3 y

x + 3 =3 ... (ii)
4 y

Algebra Mathematics – 9 |143


Click to View FlipBook Version