Math 7

Variable class marks Tally marks Frequency

0–4 5

4–8 // 2

8 – 12 5

12 – 16 5

16 – 20 /// 3

The class (0 – 4) includes the marks
more than or equal to 0 but less than
4. Similarly, (4 – 8) includes the marks
more than or equal to 4 but less than 8

and so on.

Cumulative Frequency Distribution

The frequency distribution table in which frequencies are cumulated either from
top to bottom or from bottom to top is called cumulative frequency distribution. It
includes the sum of all frequencies upto the corresponding class interval as shown
below. Marks of 20 students are tabulated as below.

Variable (x) Tally Frequen- Cumulative • Write the frequency of first
marks cy (f) frequency class in corresponding c.f.
0–5 (c.f.)
5–10 // • Add c.f. of first class with f
10–15 //// 22 of second class, which is the
c.f. of the second class.
/// 4 2+4=6
• Continue this process till
8 6 + 8 = 14 the end.

15–20 5 14 + 5 = 19

20–25 /1 19 + 1 = 20

Note: variable = x, frequency = f, cumulative frequency = c.f.,
These are notations used in statistics



Oasis School Mathematics – 7 245

Exercise 20.1

1. (a) Prepare a discrete frequency table of the following marks obtained by the stu-
dents of class VII in an examination using tally mark.

14, 16, 8, 10, 15, 12, 14, 16, 15, 15, 16, 14, 14, 16, 10, 16,

15, 12, 16, 12, 14, 15, 16, 10, 12, 16, 15, 16, 12, 14.

(b) Followings are the weights (in Kg.) of children in a class of a school. Prepare a
discrete frequency table using tally marks.

18, 19, 20, 27, 25, 24, 23, 20, 18, 19, 20, 27, 20, 20, 25, 25, 19,

18, 20, 25, 20, 23, 20, 23, 23, 20, 24, 25, 25, 18, 20, 20, 20.

2. (a) In a village of 30 families, the number of family members are as follows. Group
them in different classes and construct the frequency table making class
interval (0-2), (2-4), (4-6), etc.

3, 4, 6, 10, 8, 7, 3, 9, 10, 10, 8, 5, 3, 4, 8, 7, 6, 7, 8, 7, 4, 5, 3, 4, 6, 7, 10, 8, 9, 4.

(b) Marks of 25 students are given below. Construct frequency distribution table
making the class interval of 5 as 0 – 5, 5 – 10, etc.

2, 4, 12, 13, 19, 8, 7, 10, 19, 13, 8, 3, 7, 6, 4, 19, 12, 10, 11, 13, 4, 7, 18, 14, 3

3. (a) Study the given table and answer the questions given below:

Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 –60
Frequency 8 4 6 9 10

(i) How many students get more than or equal to 10 and less than 20 marks?
(ii) How many students have marks between 50 – 60 ?
(iii) In which class does the marks 30 lie?
(iv) How many students are there altogether?
4. Construct the commutative frequency table from the given data:

(a) Wages 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

Number of workers 4 12 9 6 8

(b) Class 5 – 10 10 – 15 15 – 20 20 –25 25 – 30
Frequency 38964

(c) Marks 0 – 20 20 – 30 30 – 40 40 – 50 50 – 60
Frequency 47968

246 Oasis School Mathematics – 7

5. (a) Marks obtained by 30 students of class VII out of 50 full marks is given
below. Make the cumulative frequency table. Making the class interval 0 – 10,
10 – 20, etc.

15, 18, 25, 37, 40, 48, 39, 30, 25, 20, 19, 28, 29, 35, 19
27, 39, 42, 45, 30, 38, 5, 7, 9, 24, 23, 29, 28, 32, 45

(b) Daily wages of 50 workers of a factory are given below.
The minimum daily wage is Rs. 100. Prepare the cumulative frequency

distribution table from the given data. Making the class interval 100 – 120,
120 – 140, etc.
110, 115, 125, 165, 190, 195, 180, 173, 162, 161,
119, 117, 114, 125, 131, 134, 138, 149, 151, 162,
175, 174, 160, 155, 150, 140, 130, 138, 144, 180,
193, 178, 164, 154, 138, 142, 127, 118, 108, 105
124, 134, 136, 122, 102, 105, 144, 146, 148, 194



Answer
Consult your teacher.

Project Work

• Collect marks obtained by the students of your class in mathematics and prepare
the cumulative frequency table of the data.

• Collect the similar informations in other subjects also and prepare cumulative
frequency table on the chart paper and paste it in the wall of your classroom.

Oasis School Mathematics – 7 247

20.2 Bar Graph

Bar graph is also known as bar diagram. It is one of the ways of representation of the
collected data. By the help of bar graph, one can have a very clear picture about the
collected data. There are mainly three types of graph:

(i) Simple bar graph.

(ii) Multiple bar graph and

(iii) Sub – divided bar graph.
But in this class, we will discuss only first two types.

I. Simple bar graph

Let us consider the following figure of a bar graph which shows the different choic-
es of students of a school after their SEE exam.

From the above diagram students will reply the following:

No. of Students Y • What does the height of
45 bar represent?
Education X
40 • Are the widths of all
35 bars equal?
30
25 • Are the gaps between
20 any two bars the same?
15
10 • Which subject (stream)
is the choice of max-
5 imum and minimum
0 number of students?

O Arts Commerce Science Law
Subject choice

After observing the above data and replying all questions, we will be able to learn
the properties of bar diagrams.

(i) Suitable scales should be selected for the variable items along x – axis and respective
numbers along y – axis.

(ii) Each bar should be of equal width and the intervals between them should also be
maintained the same.

Bar graph is the simplest way to represent the different variables of items and their
respective numbers.

II. Multiple bar graphs:

Given bar graph shows the result of SEE examinations of three different schools in
different categories.

248 Oasis School Mathematics – 7

Result of SEE Exam 2075

45

40

Number of Students 35

30 Index

25 A+ Grade

20 A Grade
15 B+ Grade

10

5

0 B C
A
School


Students will observe the above diagram and will reply the following questions.
(i) What is the total number of students in school A, B and C ?
(ii) How many students were passed in A+ grade from each of A, B and C?
(iii) What is the total number of students who passed in A grade?

(iv) How many students of school C scored B grade marks?

tSh•tue dceoTnuotrssreecpaonrfeedsaeissncitluytswsstioounodrywmtihtoheredfrtiiahegnarndastmwanotoddrtaeetpaalcyohfethdreisftfauebrdoeevnnettvsbaywritiahllbeclehosem,lpme oaucfltrtioepsalscehtbheaerr. fOoln-
lowindgiaidgeraams. is used.

• It contains adjacent bars in order for each variable along x – axis and
respective height of bars shows the number along y – axis.

• Width of adjacent bars is maintained the same.

• Intervals between successive bars of different variables, are also kept the same.


Multiple bar graph helps us to have comparative study of different categories of
values for different variables.

Exercise 20.2

1. (a) Total number of students enrolled in a school in different year is given below.
Represent the given data is simple bar diagram.

Years 2069 2070 2071 2072 2073

No. of Students 250 260 280 300 350

Oasis School Mathematics – 7 249

(b) Average temperature of a town in different days are given. Represent these
informations in bar graph.

Day: Sunday Monday Tuesday Wednesday Thursday Friday Saturday

Temperature: 30° 32° 31° 28° 29° 31° 32°

(c) Marks obtained by Santosh in different subjects is give below. Represent this in-
formation in simple bar diagram.

Subjects: English Nepali Mathematics Social studies Science

Marks 75 45 80 60 75
obtained

(d) Students of class VII in a school are selected for different events as given below:
Represent them by the help of simple bar graph:

Games: Football Volleyball Cricket Basketball
12
No. of students 18 14 20

2. Given bar diagram shows the number of students in different sections of
class VII. Observe the given diagram and answer the questions given below.

Number of Students Y
50

40

30

20

10
0 A B C DX

School


(a) How many students are there in section A, B and C?

(b) Which section has greatest number of students and which has the least number?

(c) Find the difference between the number of students in section A and C.

3. Draw multiple bar graph from the following data:

(a) School A BC D
Boys 200 300 400 250
Girls 300 250 300 100

250 Oasis School Mathematics – 7

(b) Subjects English Nepali Maths Science
30 40
No. of passed students: 35 25 15 5

No. of failed students 5 10

(c) Village Kusunde Barpak Bhachchek Muchchok
Male 25 40 45 50
Female 35 30 45 60

(d) School A BC
No. of students who like milk 90 40 65
No. of students who like tea 50 50 75

4. (a) Study the following diagram and then answer the questions given below:

Number of Students 60
50

40 Gold
30 Silver
20 Bronze

10

0 A B

Team

(i) How many medals are won by team A?

(ii) How many medals are won by team B?
(iii) How many gold medals are won altogether?
(iv) Find the difference in number of silver medals won by team A and B.
(v) Who won the maximum number of medals?

(vi) Which medal is the least in number?
(b) Given bargraph shows the production of paddy (in metric tonnes) and wheat in

three different villages.
Study the given multiple bar diagram and answer the questions given below:

Production (in metric tonnes) 300

250 paddy

200 paddy wheat paddy
150 wheat

100 wheat

50

0 Rampur Krishnapur Mangalpur

Oasis School Mathematics – 7 251

(i) How much metric tonne paddy is produced in Rampur village?

(ii) How much metric tonne wheat is produced in Krishnapur village?
(iii) How much more paddy is produced than wheat in Mangalpur village?
(iv) Write all other remaining informations, which you can get from this multiple
bar diagram.

Answer
Consult your teacher.

20.3 Arithmetic Mean or Average

In statistics, generally we need to represent a certain value of data for the set of given
values. The average or the mean of given set of variables is known as Arithmetic

mean. It is denoted by X.

For example,
Let the marks of 10 students in a class be 20, 35, 40, 30, 45, 5, 25, 30, 40, 50 then the

average or mean marks is obtained as

20+35+40+30+45+5+25+30+40+50 = 320 = 32
10 10

Calculation for Mean: In this class we study about only

(i) Individual data and (ii) Discrete data.

I. Individual data
In order to find the mean from the given set of individual data, we follow the given steps.

(i) Arrange the given set of data.

(ii) Take the sum of all the data denoted by∑X.

(iii) Count the total number of terms (n).

(iv) Finally, in order to calculate mean (X), use the formula.

x = ∑x
n

Example:

Find the mean of: 2, 3, 5, 6, 1, 4, 7, 8, 10, 5

Solution:

Here, arranging into order,
1, 2, 3, 4, 5, 5, 6, 7, 8, 10

Now, ∑x = 1 + 2 + 3 + 4 + 5 + 5 + 6 + 7 + 8 + 10 = 51

And n = 10

We have,

∴ x = ∑x = 51 = 5.1
n 10

252 Oasis School Mathematics – 7

II. Discrete data

If the variables occur repeatedly and the individual sum of them is huge, we need to
reduce them for the easy and convenient calculation:

In order to find the mean of discrete data:

• We need to construct frequency table with the columns of variables, frequencies
and their product.

• The sum of frequencies will be noted and denoted by ∑f.

• The sum of the product of variables and respective frequencies is noted and is
denoted by ∑fx.

• The ratio of step 3 to step 2 is obtained which is the value of mean given by

X = ∑fx
∑f

Example:

Find the mean marks of the following data:

Marks 1 23 4 5
Frequency 4 55 4 3

Solution:

Marks (x) Frequency (f) fx
1 4 4
2 5 10
3 5 15
4 4 16
5 3 15
∑fx = 60
∑f = 21

Here,

Mean (X) = ∑X = 60 = 2.86
∑f 21

Worked Out Examples

Example : 1

Daily wages of 30 workers are given below. Prepare the frequency distribution
table and hence find mean.

60, 65, 60, 50, 55, 70, 75, 50, 65, 60, 75, 55, 60, 65, 70
55, 50, 60, 75, 70, 70, 75, 65, 60, 65, 50, 55, 60, 65, 70

Oasis School Mathematics – 7 253

Solution: Tally makrs Frequency (f) fx.
Wages (x) IIII 4 200
50 IIII 4 220
55 7 420
60 IIII II 6 390
65 IIII I 5 350
70 IIII 4 300
75 IIII
N =∑f = 30 ∑fx= 1880

We have,

∴ Mean ( x ) = ∑fx = 1880 = 62.67
N 30

Exercise 20.4

1. Find the arithmetic mean of the following numbers.
(a) 1, 3, 5, 4, 8 and 10.
(b) 12, 15, 18, 20 and 10.
(c) 100, 125, 150, 200, 250, 375.
2. The marks obtained by a student in unit test are given below, find the mean mark:
1, 8, 15, 19, 5, 16, 20.
3. Calculate the mean of the following data.

Marks 12345

No. of students 5 4 6 7 8

4. Find the daily mean income of 40 people form the following data.

Income (In Rs.) 150 175 200 225 250

No. of people 5 6 4 16 9

5. Find the arithmetic mean of the following data.

Age (In years) 12 13 14 15 16
No. of students 15 12 10 20 3

6. Marks obtained by the students in class VII are given below. Prepare frequency
distribution table and hence find mean.

254 Oasis School Mathematics – 7

(a) Section 'A'

60, 65, 60, 80, 85, 90, 75, 70, 75, 65, 75, 80, 85, 90, 90, 85, 75, 70, 65, 65, 60,
65, 70, 75, 75, 80, 85, 90, 65, 60

(b) Section 'B'

85, 80, 65, 60, 65, 60, 70, 75, 85, 85, 80, 55, 60, 55, 55, 60, 50, 50, 55, 50, 65,
60, 55, 60, 75, 70, 80, 85, 80, 85

(b) 15  2. 12
Answer 4. 211.25 6. (a) 74.33 (b) 67.17
(c) 200
1. (a) 5.167 5. 13.73
3. 3.3

Assessment Test Paper

Attempt all the questions Full Marks: 17

Group 'A' [3 × 1 = 3]

1. (a) Find the arithmetic mean of 5, 7, 9 and 11.

(b) Which number does the tally marks IIII IIII represent?
(c) If ∑x = 45, n = 5, find x .

Group 'B' [3 × 2 = 6]

2. (a) If the mean of 12, 18, 20, 24 and x is 20, find the value of x.

(b) From the given table, construct the cummulative frequency distribution table.

Class 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 3 4 6 7 98

(c) Construct the frequency distribution table of

3, 8, 7, 9, 4, 3, 8, 7, 9, 8, 7, 9

4, 3, 3, 4, 9, 7, 4, 7, 9, 8, 7, 9

Group 'C' [ 2 × 4 = 8]

3. Marks obtained by 40 students of a class is given below. Prepare the cumulative
frequency table from the given data. Make the class of 0-10, 10-20, etc.

25, 24, 27, 31, 44, 48, 45, 35, 37, 46, 9, 12, 17, 23, 19, 26, 32, 37, 41, 12

27, 28, 29, 34, 35, 32, 44, 45, 7, 18, 24, 28, 29, 32, 37, 38, 39, 36, 12, 28.

4. Draw the multiple bar diagram of the number of boys and girls in different sections
of class VII in a school.

Oasis School Mathematics – 7 255

Section A B C D
Boys Girls Boys Girls Boys Girls
Boys Girls 25 10 20 15 30 5

Number of 15 20
students

Objective Questions

Choose the correct alternatives.

1. Tally marks IIII IIII II represents

(i) 10 (ii) 12 (iii) 7

2. Arranging numerical data in ascending or descending order is called

(i) array (ii) collection of data (iii) tabulation of data.

3. Which one of the following is not a feature of bar diagram?

(i) Every bar should have equal height

(ii) Every bar should have equal width

(iii) Gap between every bar should be the same.

4. Sum of all the frequency in the discrete data is denoted by

(i) ∑fx (ii) x (iii) N

5. The mean of first 5 natural numbers is

(i) 5 (ii) 3 (iii) 2.

6. Which of the following relation is not true?

(i) N = ∑f (ii) x= ∑X (iii) ∑f = ∑x× N
n

Project Work

• Collect data of number of passed and failed boys and girls separately. Show the
result in Bar Graph.

• Collect the informations of literacy rate of male and female of any five district.
Represent it in Bar diagram.


256 Oasis School Mathematics – 7

Algebra

28Estimated Teaching Hours

Contents

• Addition, Subtraction, Multiplication and Division of
Algebraic Expressions

• Indices
• Factorisation of Algebraic Expression
• Simplification of Rational Expressions
• Equations and Inequalities

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following
competencies:

• Add and subtract the given algebraic expressions
• Multiply two algebraic expressions containing three terms
• Divide the algebraic expression containing three terms by

the algebraic expression containing one term
• Solve the simple problems of indices using its laws
• Factorise the algebraic expression by taking common, using

the formula of a2–b2, a3+b3 and a3–b3
• Simplify the given rational expressions
• Solve the simple algebraic equations containing one variable
• Solve the simple verbal problems of one variable
• Solve the simple inequality and to express the solution on

the number line

Teaching Materials

• Chart paper, A4 size paper, scissor, glue, etc.

Oasis School Mathematics – 7 257

Unit Algebraic
Expressions
21

21.1 Algebraic Expression

Algebra is a branch of mathematics in which letters and symbols are used to repre-
sent quantities.

In grade six, we already discuss about variable, terms (like and unlike terms), ex-
pressions (monomial, binomial, trinomial, etc ............), coefficient, index, etc.

Numerical values: When the numerical values of literal coefficients of an expression
are given, then we can find the numerical values of that expression.

For example,

If x = 2, y = 1 and z = –1 then,

(i) x² + y² – z² = (2)² + (1)² – (–1)² = 4 + 1 – 1 = 4

(ii) 2xyz – z² + y² = 2.2.1.(–1) – (–1)² + (1)² = –4 –1 + 1 = –4

Types of algebraic expressions

Monomial : Algebraic expression containing one term is monomial.
e.g. 2x, 3x2, 4x3, etc.
Binomial : Algebraic expression containing two terms is binomial.
e.g. 2x+5, 3x+5, etc.
Trinomial : Algebraic expression containing three terms is trinomial.
e.g. x2–3x+2, ax2+bx+c, x2+xy+y2, etc.

Exercise 21.1

1. If x = 1, y = 2, z = 3, find the values of following:

(a) 2xyz (b) x² + y² + z² (c) x – y + z

(d) xy (–z)² (e) xy – yz + zx

2. If a = 5, b = 3, c = 1, find the value of

(a) 2a + 3b – 2bc (b) ab + bc – ca (c) a²b – b²c + c²a
a+b+c

(d) (a + b)² - (b + c)² (e) abc – 2bc + ca (f) ab² + b²c – ca²
a+b+c

258 Oasis School Mathematics – 7

3. The area of triangle (A) = 1 × b × h. Find the area of triangle, if
2

(a) base (b) = 5cm, height (h) = 10 cm (b) base (b) = 4 cm, height (h) = 5cm

(c) base (b) = 8 cm, height (h) = 10 cm (d) base (b) = 12 cm, height (h) = 8 cm

4. The area of a rectangle (A) = length × breadth. Find the area of rectangles, if

(a) length = 8 cm, breadth = 4cm

(b) length = 6 cm, breadth = 4cm

(c) length = 15 cm, breadth (b) = 8cm

(d) length = 10 cm, breadth (b) = 5 cm

5. The perimeter of rectangle is p = 2(l + b). Find the perimeters of the following
rectangles.

(a) l = 8cm, b = 4 cm (b) l = 6 cm, b = 4 cm

(c) l = 7 cm, b = 4 cm (d) l = 9 cm, b = 7 cm

6. Identify whether the given expressions are monomial, binomial or trinomial.

(a) 2x (b) 3x+5 (c) 2x2–x+5 (d) 3x–2

(e) ax2+3x+5 (f) 4–2x (g) x–5 (h) 3x2+x+2

Answer 
1. (a) 12
(b) 14 (c) 2 (d) 18 (e) - 1
2. (a) 13
(b) 13 (c) 71 (d) 48 (e) 14 (f) 29
3. (a) 25 cm² 9 9
4. (a) 32 cm²
5. (a) 24 cm (b) 10 cm² (c) 40 cm² (d) 48 cm²

(b) 24 cm² (c) 120 cm² (d) 50 cm²

(b) 20 cm (c) 22 cm (d) 32 cm 6. Consult your teacher

21.2 Addition and Subtraction of Algebraic Expressions

Addition: 2x, 3x and 5x are like terms because these terms have the same base x and
each base has equal index (power) 1.

While adding these algebraic terms, we should add the numerical coefficients of like
terms. i.e. the sum of 2x, 3x and 5x is 2x + 3x + 5x = 10x.

Similarly,

2x² + 3x² = 5x²

2xy + 3xy + 7xy = 12 xy

Subtraction: Similarly while subtracting the algebraic terms, we should subtract the
numerical coefficients of like terms. For example:

4x – 3x = x

7xy – 4xy = 3xy etc.

Oasis School Mathematics – 7 259

We change the sign of terms of the expression to be subtracted and we add the like
terms according to the new signs.

Worked Out Examples

Example: 1

Add the following expressions

(i) 12xy + 13xy + 7xy
Solution:
Here, 12xy + 13xy + 7xy
= 32xy.

(ii) 3x²y + 5x²y and – 2x²y
Solution:
Here, 3x²y + 5x²y + (–2x²y)
= 8x²y – 2x²y = 6xy².

Example: 2

Add the following expressions

2x³ – 3x² + 5x + 7, 2x³ – 3x² – 4x + 2 and 4x³ + 5x² – 10 x + 5

Solution:

Here, 2x³ - 3x² + 5x +7
2x³ - 3x² - 4x +2


+ 4x³ + 5x² - 10x + 5

8x³ - x² - 9x + 14


Example: 3

Simplify:

(2a – b + 3c) + (5a + 7b – 2c) – (a – b)
Solution:
Here, (2a – b + 3c) + (5a + 7b – 2c) – (a – b)
= 2a – b + 3c + 5a + 7b – 2c – a + b
= 2a + 5a – a – b + 7b + b + 3c – 2c
= 6a + 7b + c

260 Oasis School Mathematics – 7

Example: 4 Alternative method
(x2–3xy+3y2) – (2x2–8xy+2y2)
Subtract: = x2–3xy+3y2–2x2+8xy–2y2
= –x2+5xy+y2
2x² – 8xy + 2y² from x² – 3xy + 3y²
Solution:
Here,

x² – 3xy + 3y²

2x² – 8xy + 2y²

– + –

– x² + 5xy + y²

Example: 5

From what expression 2x² – 5x + 6 must be subtracted to get x² + 2x + 7?

Solution: Alternative Method

Let, 2x² – 5x + 6 be subtracted x² + 2x + 7

from P to get x² + 2x + 7, + 2x² – 5x + 6

Now, P – (2x² – 5x + 6) = x² + 2x + 7 3x² – 3x + 13

P = (x² + 2x + 7) + (2x² – 5x + 6)

= x² + 2x + 7 + 2x² – 5x + 6

= 3x² – 3x + 13

Note: While subtracting the signs are changed '+' to '–' and '–' to '+' of the subtracted
expression.

Exercise 21.2

1. Add the following:
(a) a, 2a, 5a and 5a.
(b) 3xy, 4xy, 7xy, and 2xy.
(c) –2a²b, 4a²b, –7a²b and 5a²b.
(d) 6x² – 3x + 5 and x² – 2x + 3.
(e) 4p² – 3pq + 4q² and 3p² + 2pq – 2q².
(f) 4xy – 6yz + 2zx, and 3xy – 5yz + zx and 2xy + 2yz + 3zx.
(g) x² + xy – y², 3x² – 7xy + 3y² and – 3x² + 2xy – 5y².

Oasis School Mathematics – 7 261

2. Subtract:
(a) 4x from 5x.
(b) – 3x² from 7x².
(c) 2a + 3b – 4c from 4a – 2b + 3c.
(d) 2ab + 4bc – 3c from 3ab – 2bc + 5c.
(e) 2x³y³ – 3x²y² + 4xy + 1 from – 6x³y³ – 2x²y² – 8xy – 4.
3. Simplify:
(a) 5xy + 2x – 3xy + 4x + y.
(b) 7a – 4b – 2a + 9b.
(c) 10a + 4b – (2a + 3b).
(d) 43a² –(33b² – 14c²) + a² – (18a² – 25b² + 27c²).
(e) 12x – (5x + 4y) – (2x + 3y) + 15x.
4. (a) What should be added to 2x + 3y – 4z to get 3x – 2y + 5z?
(b) What should be added to x² – y² to get 2x² + y² – z²?
(c) What should be subtracted from 10x – 3y + 6 to get 3x + 5y – 6?
(d) From what expression 12x² – 15b² + 10 must be subtracted to get 5x² + 10b² + 7?
(e) From what expression a³ + 3a²b + 4b² be subtracted to get 2a³ – a²b – b³ ?
5. (a) If a side of an equilateral triangle is 3a + 2b + c, find the perimeter of the triangle.
(b) If three sides of a triangle are 2a + 3b – c, 4a – 2b + c and a + b – 2c, find its

perimeter.
(c) The perimeter of a triangle is 14x² + 10x + 7 and its two sides are 10x² + 3x + 1 and

2x² + 7x – 3. Find its third side.
(d) If the perimeter and any two sides of a triangle are 8a + 3b + 5c, a + b + c and

2a – b + c respectively. Find its third side.

6. If x = 1, y = 2 and z = 3, find the perimeter of the following figures:

(3x + 2y)cm (2x + 3y + z)cm (x+y+z)cm (2x
(–2x+3y+ z)cm
(b)


(a) (2x + y)cm (c) + y + z)cm
(5x + y + 2z)cm (x+y+2z)cm
(x + 4y)

262 Oasis School Mathematics – 7

Answer

1. (a) 13a (b 16 xy (c) 0 (d) 7x² - 5x + 8

(e) 7p² - pq + 2q² (f) 9xy - 9yz + 6zx (g) x² - 4xy - 3y²

2. (a) x (b) 10x² (c) 2a – 5b + 7c (d) ab – 6bc + 8c

(e) –8x³y³ + x²y² – 12xy – 5 3. (a) 2xy + 6x + y (b) 5a + 5b

(c) 8a + b (d) 26a² - 8b² - 13c² (e) 20x - 7y 4. (a) x - 5y + 9z

(b) x² + 2y² - z² (c) 7x – 8y + 12 (d) 17x² - 5b² + 17 (e) 3a³ + 2a²b – b³ + 4b²

5. (a) 9a + 6b + 3c (b) 7a + 2b – 2c (c) 2x² + 9 (d) 5a + 3b + 3c

6. (a) 20 cm (b) 48 cm (c) 29 cm

21.3 Indices

Let us consider an algebraic term 2xa. Here, '2' is coefficient, 'x' is the base and 'a' is

the index (or power) of the base. An index of a base means how many times the base

is multiplied. The plural form of index is called indices. There are different laws of

indices, some of them are as follows: 2 xa Index
i. a² × a³ = a × a × a × a × a = a5 = a²+³
Base

a4 × a2 = a × a × a × a × a × a = a6 = a4+2 Co-efficient

i.e. am × an = am+n it is product law of indices.

ii. a5 ÷ a² = a×a×a×a
a×a

= a3 = a5–2

a6 ÷ a2 = a×a×a×a×a×a
a×a

= a4 = a6–2

am ÷ an = am = am–n
i.e., an

It is quotient law of indices.

iii. (a²)³ = a² × a² × a² = a6 = a2×3
(a³)² = a³ × a³ = a6 = a3×2

i.e. (am)n = am×n . It is power law of indices.

Some more illustrations for power law.

(a) (a×b)m = am × bm

( ) a m am
b bm
(b) =



Oasis School Mathematics – 7 263

iv. 20 = 2a–a = 2a =1
2a

30 = 3a–a = 3a =1 i.e. a0 = 1 . It is law of zero index.
3a

Worked Out Examples

Example: 1

Simplify.
(a) a² × a³ × a5 (b) (ab)². (a²b)³ . (ab²)² (c) (ab3)3 ÷ (ab²)²
Solution:

(a) a² × a³ × a5 (b) (ab)². (a²b)³ . (ab²)² (c) (ab³)³ ÷ (ab²)²

= a2+3+5 = a²b². a6b³. a²b4 = a³b9 ÷ a²b4
= a10
[∵ (ab)m = ambm] = a3b9
Example: 2 = a2+6+2. b2+3+4 a²b4

Evaluate: = a3–2.b9–4

= a10.b9 = ab5

1

(a) 53.43 (b) 2² × 24 (c) (64) 3

Solution: 26 1
(a) 5³.4³
= (5.4)³ 2² × 24 (64)3
= (20)³ 26
= 8000 1

= (26)3

= 2² × 24 = (2)6× 1
26 3

= 26 = 2²
26 =4

=1

Example: 3 =1

Simplify: (b) xa+b. xb+c. xc+a
(xa.xb.xc)²
(a) xb–c . xc–a. xa–b
Solution: (b) xa+b. xb+c. xc+a
(a) xb–c. xc–a. xa–b (xa.xb.xc)²
= xb–c+c–a+a–b
= x0 xa+b+b+c+c+a
=1 = x²(a+b+c)
= x2a+2b+2c

x²a+2b+2c

= x2a+2b+2c–2a–2b–2c
= x0 = 1

264 Oasis School Mathematics – 7

Exercise 21.3

1. Write the product into their exponent form

(a) a × a × a × a (b) 1 × 1 × 1 (c) (–2) × (–2) × (–2) × (–2) × (–2)
x x x

(d) b2 × 2 × 2 × 2 × 2 × 2 (e) 5x × 5x × 5x × 5x
b b b b b

2. Simplify and write the result in exponent form 1 3. 1 4. 1
2 2 2
(a) x³ × x4 × x5 (b) 25 × 24 × 26 ( ( () ) )(c)
(d) (ab)4. (ab)5. (ab)3
(e) (– 1 )4, (– 1 )2, (– 1 )5
3 3 3

( ) ( ) ( ) (f) 2 6.2. 2 3 (g) (2x)3(2x)5.(2x)2 (h) 54 × 53 × 52
3 3 3


3. Write the quotient into their exponent form

(a) a5 ÷ a3 (b) x13 ÷ x10 (c) (2x)6 ÷ (2x)4

5 2 2xy 4 2xy
z z
÷ ÷
(d) (–2x)10 ÷ (–2x)5 ( ) ( )(e) 2 2 ( ) ( )(f)
3 3

4. Evaluate and write the answer in exponent form

(a) (a5)2 (b) [(2x)3]6 (c) [(3a)4]³

(d) (x–2)6 (e) [(xy)2]–3

5. Simplify and write the answer in exponent form

(a) 4–1 × 83 (b) 93 × 32 × 272 (c) 23 × 16 × 2–5 (d) 25 ÷ 16–2

(e) 164 ÷ 42 ( ) ( )(f) 84 2 5 (g) 34 ÷ 273 (h) (8)4 ÷ (2)6
6. Evaluate: 27 3
(a) (27)1/3 ÷

(b) (125)1/3 (c) (16)3/4 ( )(d) 8 1/3
27

( ) 125 1/3 ( )(f) 125 -1/3 ( )(g) 32 1/5
(e) 216 27 243


7. Simplify: a²b5c³ 3
(a) (a2b5)3 × (ab2)4 ab4c²
(b) (35.23) × (322–1)2 ( )
(c)

(d) 25² × 5³ × 125² (e) 34 × 3² × 3
54 × 25 9² × 27

8. Simplify:

(a) 100a4 (b) 324a² × 64b² (c) xa–b . xb–c . xc–a

(d) xa+b. xb+c . xc+a (e) (xa)b–c. (xb)c–a. (xc)a–b
x²a . x2a . x2c x²a+2b. x2b+2c

(f) (x²)a–b. (x²)b–c.(x²)c–a (g) (xa . x2b . xc)²

Oasis School Mathematics – 7 265



Answer

( )1. (a) a4 (b) 1 3 ( )(c) 2 6
x (–2)5 (d) b (e) (5x)4

2. (a) x12 ( )(b) 215 (c) 1 8 ( ) ( )(c) 1 11 2 10
2 (d) (ab)12 – 3 (f) 3

(g) (2x)10 (h) 59 3. (a) a2 (b) x3 (c) (2x)2

( )(d) (–2x)5 (e) 2 3 ( )(f)
3 2xy 3 4. (a) a10 (b) (2x)18
z 5. (a) 27

(c) (3a)12 (d) x–12 (e) (xy)-6 (b) 314 (c) 22

(d) 213 ( )(e) 212 (f) 2 7 (g) 3-5 (h) 26
3
2 5 3
6. (a) 3 (b) 5 (c) 8 (d) 3 (e) 6 (f) 5

(g) 2 7. (a) a10b23 (b) 2.39 (c) a3b3c3 (d) 57 (e) 1
3 (b) 144ab (c)1
8. (a) 10a2 (d) 1 (e) 1 (f)1 (g) 1

21.4 Multiplication of Algebraic Expressions

In case of multiplication of algebraic expressions the co–efficients of the terms are
multiplied and the power of the same bases are added.

Multiplication of a binomial by a monomial.

(i) To multiply a binomial by a monomial we use an identity;

a × (b + c) = a × b + a × c

or, (a + b) × c = a × c + b × c

For example:

(2xy – 3yz) (–2xy) = 2xy (–2xy) – 3yz (–2xy)

= –4x²y² + 6xy²z. Multiply each term of
binomial by monomial.
Multiplication of a binomial by another binomial.

To multiply a binomial by another binomial we use an identity:

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

For example:

(2a – 3b) (2a + 4b) = 2a (2a + 4b) – 3b(2a + 4b)

= 4a² + 8ab – 6ab – 12b²

= 4a² + 2ab – 12b²

Here, we see that when multiplying an algebraic expression by another expression,
each term of the first expression is multiplied by each term of the second expression
and the result is then simplified by summing like terms.

Geometrical interpretation of multiplication of binomial and bionomial.

266 Oasis School Mathematics – 7

Lets take any two binomials (a+b) and (x+y). Aa Gb B

Draw a rectangle having length (a+b) and breadth

(x+y) as shown in the figure. x ax bx
ay
Product of (a+b) and (x+y) gives the area of given

rectangle. yE I F
In the given figure, D C
by

H

Area of rectangle AEIG (A1) = ax

Area of rectangle GBFI (A2) = bx

Area of rectangle EDHI (A3) = ay

Area of rectangle HCFI (A4) = by

Now, area of rectangle ABCD = A1 + A2 + A3 + A4
∴ (a+b) (x+y) = ax + bx + ay + by.

Worked Out Examples

Example: 1 Alternative Method

Multiply: x² – 2x + 3 by x + 2 x² – 2x + 3
Solution: × x + 2
(x + 2) (x² – 2x + 3) x³ – 2x² + 3x → Multiplying by x
= x(x² – 2x + 3) + 2(x² – 2x + 3) + 2x² – 4x + 6 → Multiplying by 2
= x³ – 2x² + 3x + 2x² – 4x + 6 x³ – x + 6 → Adding
= x³ – x + 6

Example: 2

Simplify: (a – b) (a² + ab + b²) + (a + b) (a² – ab + b²)
Solution:
(a – b) (a² + ab + b²) + (a + b) (a² – ab + b²)
= a(a² + ab + b²) – b(a² + ab + b²) + a(a² – ab + b²) + b(a² – ab + b²)
= a³ + a²b + ab² – a²b – ab² – b³ + a³ – a²b + ab² + a²b – ab² + b³
= a³ – b³ + a³ + b³
= 2a³

Oasis School Mathematics – 7 267

Exercise 21.4

1. Multiply:

(a) 3ab × (4a²b2) (b) 3x2 × (–4y2) (c) 4x5y2 × 10x6y5

(d) 3x7y9 × (–4x9y5) (e) 4x2y3 × (–8x3y2)

2. Multiply:

(a) (a² – b²) by ab ( )(b) (ab + bc + ca) by abc (c) 2 x²y – 3 xy² by xy
3 5

(d) (4b + 3c) by (– 3bc) (e) 9x²y³ by (3xy – 2xy²)

(f) (27a2+ 3ab + 3b2 ) by (– 4ab)
7 7

3. Evaluate:

(a) ab(2a – 3b) (b) (x³ – 3x² + 2x – 1) (x – 1)

(c) (a + b + c) (a – b) (d) (2a + 3b) (2a – 3b)

(e) (4a + 2b + c) (b + c) (f) (2x²y + 3xy³). (x² – y²)

4. Simplify:

(a) 2(x – 3) + 4(x – 4) + 2(x – 2)

(b) z² (x² – y²) + x² (y² – z²)

(c) 2x (a – b) + 3x (b – c) + 4x (c– a)

(d) (2 – x) (4 + 2x + x²) + (2 + x) (4 – 2x + x²)

5. (a) If a = x – 3 and b = x + 3, show that 2ab = 2x² – 18.

(b) If x = a – b, y = b – a, find the value of 2xy – x².

6. Area of rectangle = l × b, using this relation, find the area of rectangle having,

(a) l = 3x + 2y and b= 2x + y (b) l = x + y + 2z and b = x + y

7. Find the area of rectangle counting the area of rectangles on it.

(a) 3x 2y (b) 4a 2b

b

2x 3a

y 5x 2y
3x
(c) p q (d) y

2x
y



268 Oasis School Mathematics – 7

Answer

1. (a) 12a³ b³ (b) –12x²y² (c) 40 x11y7 (d) – 12 x16. y14

(e) – 32 x5y5 2. (a) a3b - ab3 (b) a²b²c + ab²c² + a²bc²

(c) 2 x³y² – 3 x²y³ (d) - 12b²c - 9bc² (e) 27x³y4 - 18x³y5 (f) -8a3b – 12a2b2 – 12ab3
3 5 (b) x4 - 4x3 + 5x² - 3x + 1 (c) a² - b2 +7ac - bc 7 7
3. (a) 2a²b - 3ab²

(d) 4a2 - 9b² (e) 4ab + 4ac + 2b² + 3bc + c² (f) 2x4y – 2x²y³ + 3x³y³ – 3xy5

4. (a) 8x - 26 (b) x²y2 - z2y2 (c) –2ax + bx + cx (d) 16

5. (b) –3a² + 6ab – 3b² 6. (a) 6x² + 7xy + 2y² (b) x² + y² + 2xy + 2yz + 2zx

7. (a) 6x2+7xy+2y2 (b) 12a2 + 10ab + 2b2 (c) 2px+2qx+py+qy (d) 15x2+11xy+2y2

21.5 Factorisation

Resolving certain expression into two or more than two parts is called factorisation
and the parts are called the factors of that expression.

Let us see the following factorisations: x2 = x × x
(i) x² = x × x

(ii) 5x³ = 5 × x × x × x factors
(iii) a²x + b²x = x(a² + b²)

(iv) ax – ay + ab = a(x – y + b) etc. are some examples of factorisation to get the
factors of given expression.

Here, factors are the terms which are multiplied to give the required expression.

In arithmetic, factors of given number are obtained by simple division. The least
value of factor of a number is considered to be one. Here in order to get the factors,
we need to resolve into its prime factors so that their product will give us the same
number.

e.g. The factors of 20 are given by 20 = 2 × 2 × 5

Here, 2 and 5 are called prime factors of 20.

Factorisation of Algebraic expression having common in each term:

Worked Out Examples Factorise 12 and
15 into their prime
Example: 1 factors and then
take common from
Resolve into factor: 12x² – 15y² each term of the
Solution: expression.
Here, 3 is the common prime factor in each term. So,
Here, 12x² – 15y²
=2×2×3×x×x–3×5×y×y
= 3(4x² – 5y²)

Oasis School Mathematics – 7 269

Example: 2

Factorise: 2x²y – 4xy + 6y
Solution :
Here 2y is the common factor to each term of given expression so, we have,
2x²y – 4xy + 6y
= 2y (x² – 2x + 3)

Example: 3

Factorise: a²(x – y) + b²(x – y) + c²(x – y)

Solution:

We have (x – y) as the common factor in each of the terms in above expression. Hence
taking common,

a²(x – y) + b²(x – y) + c²(x – y)

= (x – y)(a² + b² + c²) [Taking (x – y) common from each term.]

Example: 4

Simplify by using factors.

650 × 150 – 500 × 150

Solution:
650 × 150 – 500 × 150
= 150 (650–500)
= 150 × 150 = 22500

Example: 5

Factorise: p²(x – y) + q²(x – y) – r²(y – x)
Solution:
Here, taking – ve sign common to change (y – x) into (x – y) in order to get common factor
in each term of the expression, we have

p²(x – y) + q²(x – y) – r²(y – x)

= p²(x – y) + q²(x – y) +r2 (x – y)

= (x – y) (p² + q² + r²)

Example: 6

Factorise: a² + ab + ca + cb
Solution:
Here, combining first two and last two terms in a separate group.

(a² + ab) + (ca + cb)

270 Oasis School Mathematics – 7

= a(a + b) + c(a + b)
= (a + b) (a + c)
= (a + b) (a + c)

Example: 7

Factorise: x² – xy – x + y [Combining the first two and last two in separate groups.]
= x(x – y) – 1(x – y)
= (x – y)(x – 1)


Exercise 21.5

1. Resolve into factors:

(a) 2a + 4 (b) 5a + 15 (c) 6a + 24b (d) 3x – 9

(e) 7p – 21q (f) x² – 8x (g) ax² – bx² (h) p²q – pq²

(i) abx + acy (j) 16x² – 48x + 24 (k) 3p² – 4pq – 5p²q²

(l) a²x + bx + c²xy (m) 11p²q – 33pq² + 88q³ (n) a²bc + ab²c + abc²

(o) 8x²y³ – 12xy4 – 24y6 (p) 10a³b – 20a²b² – 35ab³

(q) 17xy – 34xz + 51xyz
2. Factorise:
(a) p²(a – b) + q²(a – b) + r²(a – b)

(b) 20(a – b) – 25(a – b)

(c) 4(2a – 5) + 8(2a – 5) – 7(2a – 5)

(d) 3x(p – q) + 4y (p – q) + 5z (p – q)

(e) x (a - b) –y2 (b – a) + z2(a – b)

3. Simplify the following by using factors:

(a) 12 × 22 + 12 × 18 (b) 17 × 30 + 17 × 20 (c) 31 × 45 + 31 × 55
(d) 42 × 49 + 42 × 51 (e) 62 × 60 – 42 × 60 (f) 82 × 12 – 12 × 32

(g) 250 × 450 – 150 × 450 (h) 75 × 70 – 70 × 25

4. Resolve into factors:

(a) x² + xz + xy + yz (b) y² – yw + yz – wz
(c) a² + ab – ac – bc (d) pq – pr – q² + qr
(e) 2a² – 6ab + 3ab² – 9b³ (f) 2x³ – y³ – x²y + 2xy²

(g) 4c³ – 5cd² – 4c² + 5d² (h) 5x²y – 2xy – 5x + 2

(i) 2x³ – y³ + 2xy² – x²y (j) ap – bq + bp – aq

Oasis School Mathematics – 7 271

Answer b. 5(a + 3) c. 6 (a + 4b) d. 3 (x – 3)

1. a. 2(a + 2) f. x (x – 8) g. x²(a – b) h. pq( p – q)
e. 7 (p – 3q)
i. a(bx + cy) j. 8 (2x² – 6x + 3) k. p (3p – 5pq² – 4q)
l. x (a² + b + c²y)
o. 4 y³ (2x² – 3xy – 6y³) m. 11q(p² – 3pq + 8q²) n. abc (a + b + c)
2. a. (a – b) ( (p² + q² + r²)
d. (p-q) (3x+4y+5z) p. 5ab (2a² – 4ab –7b²) q. 17x (y – 2z + 3yz)
3. a. 480 b. 850 c. 3100
4. a. (x + y) (x + z) b. 5 (b – a) c. 5(2a– 5)
e. (a–3b) (2a+3b²)
h. (5x – 2) (xy – 1) e. (a-b) (x2+y2+z2)

d. 4200 e. 1200 f. 600 g. 45000 h. 3500

b. (y – w) (y + z) c. (a + b) (a - c) d. (p – q) (q – r)

f. (2x – y) (x² + y²) g. (c – 1) (4c² – 5d²)

i. (2x – y) (x² + y²) j. (a + b) (p – q)

21.6 Factors of Algebraic Expression in the form of Difference of Two Squares

We know,
(a + b) (a – b) = a(a – b) + b(a – b) = a² – ab + ab – b² = a² – b²

i.e. a² – b² = (a + b) (a – b)

Here the factors of a² – b² are (a + b) and (a – b). So we can use the above form is in
standard form to use as formula.

Thus the difference of squares of two numbers is Remember !
equal to the product of their sum and the difference. a² – b² = (a + b) (a – b)

Geometrical interpretation of a² – b²

• Take a square sheet of paper having a side 'a' units.

Its area is a² a

• a

From the square sheet take 'b' units from each side as shown in the figure.

b a

b
a
•
Area of that part = b²

Area of the shaded part = a² – b² .............. (i)

a

Cut the shaded part.

b a

b
a–b

272 Oasis School Mathematics – 7

• Again, cut it along the dotted line.

a

a–b

a
bb

a–b

• Join these two parts as shown in the figure.

ab

a–b a–b

ba

Then the new sheet is a rectangle having two sides (a + b) and (a – b)
ab

a–b a–b

ba

Area of shaded part

∴ a² – b² = (a + b) (a – b)

Worked Out Examples

Example: 1 Example: 2

a² – 9 3x² – 12y²
=(a)² – (3)² = 3(x² – 4y²)
= (a + 3) (a – 3) = 3{(x)² – (2y)²}
= 3(x + 2y)(x – 2y)

Example: 3 Example: 4

(a + b)² – (w + x)² a4 – 16b4
= {(a + b) + (w + x)} {(a + b) – (w + x)} = (a²)² – (4b²)²
= (a + b + w + x) (a + b – w – x) = (a² + 4b²)(a² – 4b²)
= (a² + 4b²){(a)² – (2b)²}

Example: 5 = (a² + 4b²)(a + 2b)(a – 2b)

Find the values of following by the method of factors.

(a) 65² – 55² (b) 49 × 51

Oasis School Mathematics – 7 273

Solution: (b) 49 × 51
(a) 65² – 55² = (50 – 1) (50 + 1)
= (65 + 55)(65 – 55) = (50)² – (1)²
= 120 × 10 = 2500 – 1
= 1200 = 2499


Exercise 21.6

1. Resolve into factors: (b) x² – 4y² (c) 9x² – 16y²
(a) x² – y² (e) 100xy² – 169xz² (f) 50a²b² – 18b4
(d) 25a² – 49b²

(g) a² – 16 (h) 4p² – 81r² (i) a² – x4
b² 9q² b4 y²

(j) 400a²b – b3 (k) x³y³ – 64xy (l) 8x³ – 18xy²

(m) a4 – b4 (n) 16x4 – 81y4 (o) 81c²d4 – 256a4c²

2. Evaluate by using formula:

(a) 35² – 25² (b) 41² – 31² (c) 62² – 52² (d) 120² – 110²
(e) 21 × 19 (f) 32 × 28 (g) 82 × 78 (h) 73 × 67



Answer

1. a. (x + y) (x – y) b. (x + 2y) (x – 2y) c. (3x + 4y) (3x – 4y)
d. (5a + 7b) (5a – 7b)
e. x(10y + 13z) (10y - 13z) f. 2b² (5a + 3b) (5a - 3b)

( ) ( ) g.a+ 4 a – 4 ( ) ( )h.2p+ 9r 2p – 9r ( ) ( )i.a+x²a – x²
b b 3q 3q b² y b² y
j .
b(20a + b) (20a - b) k. xy(xy + 8) (xy – 8) l. 2x(2x + 3y) (2x – 3y)

m. (a² + b²) (a + b) ( a – b) n. (4x² + 9y²) (2x + 3y) (2x – 3y)

o. c² (9d² + 16a²) (3d + 4a) (3d – 4a)

2. a. 600 b. 720 c. 1140 d. 2300

e. 399 f. 896 g. 6396 h. 4891

21.7 Some Special Products and Formulae

An inference expressed with the help of some symbols is known as a formula. Here
are some formulae for the multiplication of two algebraic expressions containing
two or more terms each.

274 Oasis School Mathematics – 7

(i) Square of the sum of two terms

We have,

(a + b)² = (a + b) (a + b)

= a (a + b) + b(a + b)

= a² + ab + ab + b²

= a² + 2ab + b².

∴ (a + b)² = a² + 2ab + b²

(ii) Square of the difference of two terms

We have, ( a – b)² = (a – b) (a – b)

= a(a – b) – b(a – b)

= a² – ab – ab + b²

= a² – 2ab + b².

∴ (a – b) = a² – 2ab + b²

Geometrical interpretation of the identity (a+b)² = a² + 2ab + b²

Make a square AEGH of length (a + b) units Aa B bH

where, a a
AB = AD = DC = BC = a units a² ab

BH = CI = FG = IG = CF = DE = b units D CI
From the figure,
b ab b² b

It is clear that (a + b)² is the area of square AEGH. E a F bG

Area of square AEGH = Area of square ABCD + Area of rectangle BCIH + Area of
rectangle DCFE + Area of square CFGI

∴ (a + b)² = a² + ab + ab + b²

= a² + 2ab + b²

Geometrical interpretation of the identity (a–b)² = a² – 2ab + b²

Make a square ABHE with the sides AB, AE, EH and BH equal to 'a' units.

Take points, G,C,F and D on AB, BH, EH and AE respectively, aG B

such that GB = CH = FH = DE = b units. A

Then AD = AG = (a – b) units. a–b
a–b
Now, a–b
Area of square ADIG = (a – b)² a (a–b)² b(a–b)
Area of rectangle DEFI = b (a – b)
Area of rectangle GBCI = b (a – b) D IC

b b(a–b) b² b

E a–b F b H

Oasis School Mathematics – 7 275

Area of square ICHF = b²
From the figure,
Area of square ADIG = Area of square ABHE – Area of rectangle GBCI – Area of

rectangle DEFI – Area of square ICHF
(a – b)² = a² – b(a – b) – b (a – b) – b²
= a² – ab + b² – ab + b² – b²
= a² – 2ab + b²

∴ (a–b)2 = a2–2ab+b2

Note: a2 + b2 = (a + b)2 – 2ab
a2 + b2 = (a – b)2 + 2ab


Worked Out Examples Example: 2

Example: 1 Find the square of 2x – 3y.
Solution:
Simplify: (2a + b)² Suppose 2x = a, 3y = b.
Solution: (a – b)² = a² – 2ab + b²
(2a + b)² = (2a)² + 2.2a.b + (b)² (2x – 3y)² = (2x)² – 2.2x.3y + (3y)²
= 4a² + 4ab + b². = 4x² – 12xy + 9y²

Example: 3 25x² + 10xy + y²
= (5x)² + 2 (5x) y + y²
Express 25x² + 10xy + y² as a perfect square.
25x² + 10xy + y² a² 2 a b b²
Solution:

25x² + 10xy + y² = (5x)² + 2.5x.y + (y)²

= (5x + y)²

Example: 4

If a – 1 = 3, find the value of a² + 1 .
a a²

Solution:

( )We a² 2.a.1a 1 1 ²
have, – + a² = a – a

( )or, a² + 1 = a – 1 ²+2
a² a

= (3)² + 2 = 9+2 = 11

276 Oasis School Mathematics – 7

Example: 3

Simplify: (a – b)² – (a + b)²
Solution: (a – b)² – (a + b)² = (a² – 2ab + b²) – (a² + 2ab + b²)

= a² – 2ab + b² – a² – 2ab – b²

= – 4ab.

Exercise 21.7

1. Find the square of:

(a) a + 5 (b) 2b + 3 (c) a + 1 (d) 2a – 3b (e) 2x – 1
a 2x

2. Express the following as perfect squares:

(a) 4a² + 4a + 1 (b) 9a² + 24ab + 16 b² (c) x4 + 2 + 1
(e) 25p² – 40pq + 16q² x4
(d) x² – 4xy + 4y²
(f) x² + 1 – 2
(g) 4y² – 4 + y1² x²
3. Find the value of:
(h) (5x + y)² + 2(5x + y) (2x – y) + (2x – y)²

(a) 98 × 98 + 2 × 98 × 2 + 2 × 2. (b) 1.4 × 1.4 – 2×1.4 × 1.1 + 1.1 × 1.1

(c) (1.8)² + 2 × 1.8 × 1.2 + (1.2)² (d) (2.1)² – 2 × 2.1 × 1.1 + (1.1)²

4. Simplify:

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

(c) (51)² – (50)² (d) (a + 3)² – 3(2a + 3).

5. Find the square of following number by using the formula (a + b)² or (a –b)²

(a) 12 (b) 102 (c) 999 (d) 990

6. (a) If a + b = 5, ab = 6, find the value of a² + b².

(b) If 2a – 3b = 3 and ab = 12, find the value of 4a² + 9b².

7 (a) If a + a1 = 9, find the value of a² + 1 .
a²

(b) If p – p1 = 8, find the value of p² + p1².

(c) If x + 1 = 2, find the value of x2 + 1 .
x x²

(d) If m – m1 = 4, find the value of m2 + m1².



Oasis School Mathematics – 7 277

Answer

1. (a) a² + 10a + 5 (b) 4b² + 12b + 9 (c) a² + 2 + 1 (d) 4a² - 12ab + 9b²
a²
1 ( )(c) 2
(e) 4x² - 2 + 4x² 2. (a) (2a + 1)² (b) (3a + 4b)² x² + 1
x²

(d) (x - 2y)² (e) (5p - 4q)² ( )(f) 1 2 ( )(g) 1 2
x– x 2y – y


(h) 49x² 3. (a) 10000 (b) 0.09 (c) 9 (d)1
4. (a) (x + y)² (b) - 4ab (c) 101 (d) a²
5. (a) 144 (b) 10404
6. (a) 13 (b) 33 (c) 998001 (d) 980100

7. (a) 79 (b) 66 (c) 2 (d) 18



(iii) Product of sum and difference of two terms. Ab a–b

We have,

(a + b) (a – b) = a(a – b) + b(a – b) b b² b a

= a² – ab + ab – b²

= a² – b² a–b

Shaded part of the given figure represents (a² – b²). a

Worked Out Examples

Example: 1

Find the product of:

(i) (a + 2b) (a – 2b) (ii) (a – b) (a + b) (a² + b²)
Solution:

(i) (a + 2b) (a – 2b) (ii) (a – b) (a + b) (a² + b²)

= (a)² – (2b)² [∵ (a+b) (a–b) = a² – b²] = {(a)² – (b)²} (a² + b²)

= a² – 4b² = (a² – b²) (a² + b²)

= (a²)² – (b²)²

Example: 2 = a4 – b4

Using the formula (a + b) (a – b) = a² – b², find the product of : 51 × 49

Solution:

51 × 49 = (50 + 1) (50 – 1)
= (50)² – (1)²
= 2500 – 1 = 2499

278 Oasis School Mathematics – 7

Example: 3

Simplify: 1.2 × 1.2 – 1.1 × 1.1
1.2 + 1.1
Solution:

= 1.2 × 1.2 – 1.1 × 1.1 = (1.2)² – (1.1)² = (1.2 + 1.1)(1.2 – 1.1)
(1.2 + 1.1) (1.2 + 1.1) (1.2 + 1.1)

= 1.2 – 1.1 = 0.1

Exercise 21.8

1. Find the products of the following expressions:

(a) (a + 1) ( a – 1) (b) (2x + y) (2x – y)

(c) (x² – y²) (x² + y²) (d) (a + b + c) ( a + b – c)

(e) (p – q + r) (p – q – r)

2. Simplify:

(a) (x – y) (x + y) (x² + y²) (b) (a – 2) (a + 2) ( a² + 4)

(c) (a – 2b) (a + 2b) (a² + 4b²) (d) (2x – y) (2x + y) (4x² + y²)

(e) (3p – 1) (3p + 1) (9p² + 1)

3. Find the products of the following numbers by using formula (a + b) (a – b) = (a² – b²)

(a) 105 × 95 (b) 82 × 78

4. Simplify:

(a) (102)² – (98)² (b) (58)² – (42)² (c) 21² – 19²

(d) 1001² – 999² (e) 3.6 × 3.6 – 1.5 × 1.5 () 0.5 × 0.5 - 0.2 × 0.2
3.6 – 1.5 (0.5 + 0.2)

Answer 

1. (a) a² - 1 (b) 4x² - y² (c) x4 - y4 (d) a² + 2ab + b² - c²
(e) p² - 2pq + q² - r² 2. (a) x4 - y4 (b) a4 - 16 (c) a4 - 16b4
(d) 16x4 – y4 (e) 81p4 - 1 3. (a) 9975 (b) 6396
4. (a) 800 (b) 1600 (c) 80 (d) 4000 (e) 5.1 (f) 0.3

21.8 Cubes of the Sum of Two Quantities

We have, Remember !
(a+b)3 = a3 + 3a2b + 3ab2 + b3
(a + b)³ = (a + b) (a + b)² = a3 + b3 + 3ab (a + b)
a3 + b3 = (a + b)3 – 3ab (a + b)
= (a + b) (a² + 2ab + b²)

= a(a² + 2ab + b²) + b(a² + 2ab + b²)

= a³ + 2a²b + ab² + a²b + 2ab² + b³

Oasis School Mathematics – 7 279

= a³ + 3a²b + 3ab² + b³

Again, a³ + 3a²b + 3ab² + b³ = (a + b)³

(a + b)³ = a³ + b³ + 3ab(a + b)

Cubes of the Difference of the Two Quantities

We have, Remember !
(a – b)³ = (a – b) (a – b)2 (a – b)3 = a3 – 3a2b + 3ab2 – b3

= (a – b) (a² – 2ab + b²) or, (a – b)³ = a³ – b³ – 3ab ( a – b)

= a(a² – 2ab + b²) – b(a² – 2ab + b²) a³ – b³ = (a – b)³ + 3ab (a – b)

= a³ – 2a²b + ab² – a²b + 2ab² – b³

= a³ – 3a²b + 3ab² – b³

And, a³ – 3a²b + 3ab² – b³ = (a – b)³

Worked Out Examples

Example: 1

Find the cubes of: (b) x – 2y
(a) (2a + b)
Solution:

(a) Cube of 2a + b is (2a + b)³
(2a + b)³ = (2a)³ + 3. (2a)².b + 3.(2a).b² + (b)³
= 8a³ + 12a²b + 6ab² + b³.

(b) Cube of x – 2y is (x – 2y)³
(x – 2y)³ = (x)³ – 3.(x)². (2y) + 3.x.(2y)² – (2y)3
= x³ – 6x²y + 12xy² – 8y³

(a + b)3 = a3 + 3a2b + 3ab2 + b3
∴ (2a+b)3 = (2a)3 + 3(2a)2.b + 3.2a(b)2 + (b)3.

Example: 2

Express a³ + 6a²b + 12ab² + 8b³ as a cube of a binomial.
Solution:

a³ + 6a²b + 12 ab² + 8b³
= (a)³ + 3.a².2b + 3.a. (2b)² + (2b)³ = (a + 2b)³

280 Oasis School Mathematics – 7

Example: 3

Simplify: (a +b)³ + 3(a + b)² . (a – b) + 3(a + b) (a – b)² + (a – b)³
Solution:

Here, (a + b)³ + 3(a + b)². (a – b) + 3(a + b)(a–b)² + (a – b)³
= [(a + b) + (a – b)]³
= (a + b + a – b)3
= (2a)³ = 8a³

Example: 4

Find the value of:

(i) x³ + 1 , when x + 1 =2 (ii) Find the value of
x³ x a³ + b³ if a + b = 5 and ab = 2.

We have,
a³ + b³ = (a)³ + (b)³
We have, = (a + b)³ – 3.a.b (a + b)
= (5)³ – 3.2.5
( ) ( ) x³ + 1 = x+ 1 3 3x 1 x+ 1 = 125 – 30
x³ x x x
–

= (2)³ – 3.2

= 8 – 6 = 2

= 95

Exercise 21.9

1. Find the cubes of:

(a) a + 2 (b) 2a + b (c) ax + by (d) 3x – 2y

(e) 2a – 1 (f) p – 1
2a p

2. Simplify:

(a) 8a³ + 12a²b + 6ab² + b³

(b) x³ + 3x²y + 3xy² + y³

(c) (x + y)³ + 3(x + y)² (x – y) + 3(x + y) (x – y)² + (x – y)³

(d) (2x + 3y)³ – 3(2x + 3y)² (2x – 3y) + 3 (2x + 3y) (2x – 3y)² – (2x – 3y)³

( ) ( ) ( ) ( ) ( ) ( ) (e) a+13–3a+ 1 2 a – 1 +3 a+ 1 a – 1 2 – a – 1 3
a a a a a a

3. Find the value of :

(a) a³ + b³ when a + b = 3, ab = 4

(b) x³ + 8y³ when x + 2y = 6 and xy = 4.

Oasis School Mathematics – 7 281

(c) a³ + 1 when a + 1 = 5
a³ a
x1³, 1
(d) x³ – when x – x = 6

(e) 8a³ – 1 , when 2a – 1 = 10.
8a³ 2a

4. (a) If 2a + b = 3, find the value of 8a³ + b³ + 18ab.

(b) If a – b = 4, find the value of a3 – b3 – 12ab.

(c) If a – 1 = 2, find the value of a³ – 1 + 6.
a a³

Answer 

1. (a) a³ + 6a² + 12 a + 8 (b) 8a³ + 12a²b + 6ab² + b³ (c) a³x³ + 3a²b x²y + 3ab² xy² + b³y³

(d) 27x³ - 54x²y + 36 xy² - 8y³ (e) 8a³ - 6a + 3 – 1 (f) p³ – 3p + 3 - 1
2a 8a³ p p³
8
2. (a) (2a + b)³ (b) (x + y)³ (c) 8x³ (d) 216 y³ (e a³

3. (a) - 9 (b) 72 (c) 110 (d) 234 (e) 1030

4. (a) 27 (b) 64 (c) 20

21.9 Division of Algebraic Expressions

When the dividend and divisor have the same sign, the sign of the quotient will be
positive. If they are of opposite signs, then the sign of the quotient will be negative.

Remember !

+ ve = + ve, – ve = + ve, + ve = – ve, – ve = – ve.
+ ve – ve – ve + ve


Worked Out Examples

Example: 1 In case of division of a
polynomial by a monomial,
8x³y² ÷ 2xy each term of the polynomial is
divided by the monomial and
Solution: then they are simplified.

8x³y² ÷ 2xy = 2 × 2 × 2 × x × x × x × y × y In case of division of polynomial
2×x×y by another polynomials firstly
we should arrange the terms
= 4x²y of divisor and dividend in
descending or ascending order
Example: 2 of power of common bases.
We should start the division
Divide: (3a²b – 2ab + 3ab²) by ab. dividing the term of dividend
with the highest power.
Solution:

3a²b - 2ab + 3ab² = 3a²b – 2ab + 3ab²
ab ab ab ab


= 3a – 2 + 3b

282 Oasis School Mathematics – 7

Example: 3 • Divide x² by x
x² ÷ x = x (quotient)
Divide: x² + x – 6 by x– 2 • Multiply (x – 2) by x
Solution: (x – 2) × x = x² – 2x.
• Subtract x² – 2x from x² + x – 6
x – 2 x² + x – 6 x + 3 = 3x – 6, Repeat the same process.

x² – 2x

(–) (+)

3x – 6

3x – 6 ∴ quotient = x + 3

(–) (+)

(×)

Exercise 19.10

1. Divide. (b) 4x³y³ ÷ 2x²y² (c) – 29 ab²c ÷ 5abc
(a) 16abc ÷ 2ab (e) 26x3y4 ÷ (–6xy)

(d) – 24 x³y7 ÷ –4x2y4

2. Divide:

(a) (4x² + 6x) by 2x (b) (3x²y – 2xy²) by xy

(c) (7a³b² – 14a²b³ – 21b²) by 7b². (d) (8a³b² + 4a²b³ – 2ab²) by (–2ab²)

(e) 24x²y² – 6x³y + 12 xy³ by 6xy (f) (18a²b²c² + 12a3bc2) by 3 abc

3. Divide:

(a) (x² + 2x + 1) by (x + 1) (b) (x² – 4) by (x + 2)
(d) (6x² – 11x – 10) by (2x – 5)
(c) (x² + 7x + 12) by (x + 3) (f) (x² – 2x – 35) by (x + 5)
(h) (x3 – 6x2 + 11x – 6) by (x – 3)
(e) (x³ + 3x²y + 3xy² + y³) by (x + y)

(g) (a4 – b4­) by (a – b)

(i) (8x3 – y3­) by (2x – y)

4. (a) The product of two algebraic expressions is 2x² + 11x + 12 and one of the expres-
sions is (x + 4). Find the other.

(b) The product of two algebraic expressions is x2 – 5x + 6. If one of them is (x – 2),
find the other.

5. (a) If the area of rectangle is (x² + 10x + 9) square units and the breadth of the
rectangle is (x + 9) units, find the length.

(b) If the area of the rectangular field is (x2 + 3x + 2) square units and the length is (x
+ 1) units, find the breadth.



Oasis School Mathematics – 7 283

Answer

1. (a) 8c (b) 2xy (c) –29 b (d) 6xy³ (e) – 13 x²y³
5 3
2. (a) 2x + 3 (b) 3x - 2y (c) a³ - 2a²b - 3 (d) - 4a² - 2ab + 1
(e) 4xy – x² + 2y² (f) 6abc + 4a²c 3. (a) x + 1
(b) x - 2 (c) x + 4 (d) 3x + 2
(f) x - 7 (g) a³ + a²b + ab² + b³
(e) x² + 2xy + y² (i) 4x² + 2xy + y² 4. (a) (2x + 3)
(h) x² – 3x + 2 5. (a) (x + 1) (b) (x + 2)
(b) (x – 3)

21.10 Simplification of Rational Expressions

We have,

1 + 1 = 3+2 = 5 (in arithmetic)
2 3 6 6

Similarly, in algebra,

1 + 1 = c+a [Where the L.C.M of ab and bc is abc).
ab bc abc

For, multiplication and division of fractions, we have certain rules in arithmetic.

Same rule is used in algebra too.

Worked Out Examples

Example: 1 Example: 2

Multiply: 4a³b² × 10 a² b³ Divide: 5x²y ÷ 25 x³y³
5ab² 12 a4 b² 10xy² 35xy

= 2a5 b5 = 5x²y × 35xy
3a5b4 10xy² 25x³y³

= 2 b. = 7x³y² = 7
3 10x4y5 10xy³

Example: 3 Example : 4

Simplify: 1 + 1 Simplify : 2 + x
2a 4a x2–4 x2–4
Solution: Solution:

1 + 1 = 2+1 = 3 = 2+x
2a 4a 4a 4a x2 – 4

= 2 + x = (x+2) = 1
(x)2 – (2)2 (x+2) (x-2) (x–2)

284 Oasis School Mathematics – 7

Exercise 21.11

1. Multiply:

(a) 2x³ × 9a² (b) 4y³z³ × 12yz (c) 5a²bc³ × 24b³c²
3a³ 8x² 8y²z4 16y²z3 16bc² 25abc

2. Divide:

(a) yx³³ ÷ x (b) 4x³y³ ÷ 10xy (c) 6y³ ÷ 18y³
y 5ay² 8ay² 5x4 10x6

3. Simplify:

(a) 52yx²³ × 10yx
20xy³

(b) 3x²y × 2x³ × 16ayz
4y³a 8z³ 6xz

(c) xa³y³ ² × c³x ÷ a²c²
ab² b²xyz

(d) x² – 9 ÷ (x – 3)
x+3

( ) (e) x² – 1 ÷ (x + 1) ×6
3 (x + 1)²

4. Simplify:

(a) x + 2 (b) 5a – 4a – 1 (c) 3a + 2a
x+2 x+2 a+1 a+1 4 4

(d) x 4 y – 1 (e) x² x y² + x² y y² (f) a + x a 2
+ x+y – – x+2 –

(g) x 1 y – x 1 y (h) 4a – 12
+ – a2–9 a² – 9



Answer

1. (a) 3x (b) 3 (c) 3 ab²c² 2. (a) x² (b) 16 x²y² (c) 2 x²
4a 8z³ 10 y² 25 3
x4
3. (a) x² (b) 2yz³ (c) cz (d) 1 (e) 2(x² - 1) (x + 1)
5y5 xy

4. (a) 1 (b) 1 (c) 5a (d) 3 (e) 1
(f) x 2²a–x4 4 x+y x–y

(g) –2y (h) 4
x² – y² a+3

Oasis School Mathematics – 7 285

Unit Equations and
22 Inequality

22.1 Equation

Let us consider some statements such as:
(a) 3 added to a number 'x' makes 8, i.e. x + 3 = 8
(b) 5 subtracted from a number 'x' results 15, i.e. x – 5 = 15
(c) 2 times a number 'x' added with 6 gives 12, i.e. 2x + 6 = 12 etc.
All above statements can be expressed in terms of relation between two expressions

on the either side of equal to sign. Such an expression is called an equation. Hence,
an equation is a statement which states that the two terms or expressions are equal.
On solving above equations, we have,
(a) x + 3 = 8 or x = 8 – 3 = 5
i.e. the number is 5.

(b) x – 5 = 15 or x = 15 + 5 = 20
i.e. the number is 20 and

(c) 2x + 6 = 12 or 2x = 12 – 6 or 2x = 6 ∴ x = 3
i.e. the number is 3.

Solution of an Equation

As shown in above three cases, we consider the number as any unknown value x.
Then we may follow the following rules.

Rule – 1
We may add the same number on either side of equation.

Example: x – 10 = 6

or, x – 10 + 10 = 6 + 10 (adding 10 on both sides)

or, x = 10 + 6

∴ x = 16

286 Oasis School Mathematics – 7

Rule – 2

We may subtract equal number from both sides of the equation.

Example: x + 8 = 20

or, x + 8 – 8 = 20 – 8 (subtracting 8 from both sides)

or, x = 20 – 8

∴ x = 12
Rule – 3

We may multiply both sides of equation by the same constant.

Example: x = 6
or, 5 = 6 × 5 (multiplying both sides by 5)
x
5 × 5

or, x = 5 × 6

∴ x = 30
Rule – 4

We may divide both sides of equation by the same constant.

Example: 4x = 20
or,
or, 4x = 20 (dividing both sides by 4)
4 4

x = 20
4

∴ x = 5

Note: The inverse operation of addition is subtraction and that of multiplication is division.
While solving equation, the inverse operation helps us to solve the equation by
leaving the variable on one side and constant on another side to get its value

Worked Out Examples

Example: 1 Alternative Method

Solve for x: 12x – 8 = 10x + 16 12x – 8 = 10x + 16
Solution: or, 12x – 10x = 16 + 8

Here, 12x – 8 = 10x + 16 or, 2x = 24
or, 12x –8 – 10x = 10x + 16 – 10x or,
or, or, x = 24
or 2x – 8 = 16 2
2x –8 + 8 = 16 + 8
x = 12

or 2x = 24

∴ x = 12 (on dividing both sides by 2)

Oasis School Mathematics – 7 287

Example: 2

Solve: x (x + 5) = x² – 3x + 32 Alternative Method

Solution: x(x + 5) = x² – 3x + 32

Here, x (x + 5) = x² – 3x + 32 or, x² + 5x = x² – 3x + 32

or, x² + 5x = x2 – 3x + 32 or, x² + 5x – x² + 3x = 32

or, x² + 5x – x² = – 3x + 32 or, 8x = 32
or,
or, 5x = – 3x + 32 ∴ x = 32
8
or, 5x + 3x = – 3x + 3x + 32
x = 4
or, 8x = 32

or, x = 32 (dividing both sides by 8)
8

∴ x = 4

Example: 3

Solve: 3x + 4x = 57
4
Solution:

Here, 3x + 4x = 57
or 4

3x + 16x = 57
4

or 19x = 57 × 4 (multiplying both sides by 4)

or x = 57 × 4 (dividing both sides by 19)
19

∴ x = 12

Example: 4

Solve: x + 5 + x – 2 =x
3 2

Solution: x+5 + x–2 = x
Here, 3 2

or, 2(x + 5) + 3(x – 2) = x
or, 6

2x + 10 + 3x – 6 = 6x (Multiplying both sides by 6)

or, 5x + 4 = 6x

or, 4 = 6x – 5x

or, 4 = x

∴ x = 4

288 Oasis School Mathematics – 7

Example: 5

Solve for x: (x – 3) (x – 5) = (x + 2) (x + 7)
Solution:

Here, (x – 3) (x – 5) = (x + 2) (x + 7)

or, x² – 3x – 5x + 15 = x² + 2x + 7x + 14

or, x² – x² – 8x + 15 – 9x – 14 = 0

or, – 17x + 1 = 0

or, – 17x = – 1

∴ x = 1 [Dividing both sides by 17]
17

Exercise 22.1



1. Solve:

(a) x + 5 = 8 (b) 19 + 2x = 20

(c) 6x + 5 = 35 (d) 8x + 10 = 2x – 14

(e) 4x – 15 = 10 – 6x (f) 5(x + 2) = 4(x + 8) + 24

(g) 2 ( 5x + 10) = 3x – 1 (h) 4(4x + 5) = 5 (3x + 5)

(i) 5 (x + 4) = 2(x + 7) (j) 8 (2x – 5) = 40

(k) 7 (x – 5) + 6x = 30 (l) 3 (2x + 15) = 36

(m) 4x – (x + 4) = 8(x – 18) (n) 3 (2x – 5) – 2x = 3(x + 7)

(o) 8(x –6) – 4x = 3(x + 5) (p) 15 (2x – 10) – 20x = 8(x + 6)

2. Solve:

(a) (x + 5) (x – 6) = x² – 40 (b) (x – 6) (x – 7) = x(x – 15) + 38

(c) 2(x + 1) (x – 1) = 2x² – x + 18 (d) (x + 1)² = (x – 3)² + 8

(e) 3(x – 5) (7 + x) = 3x² – (7x + 14)

(f) (x – 5) (3x – 8) = (3x – 4) (x – 6) + 13

3. Solve:

(a) x + 3x = 20 (b) x + x = 5 (c) x + x = 10
3 4 3 2

(d) x + x = 50 (e) 73x – 4x = 37
3 2 7

(f) 3x – 2 – x – 3 = 7 (g) x – 12 + x = 5
5 4 6 12 2

(h) x+1 + x+2 + x+3 =4 (i) 51x6 – 3x = 2x – 11
2 3 6 4 8


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Answer (b) 1 (c) 5 (d) –4 (e) 5 (f) 46 (g) – 3 (h) 5
1. (a) 3 2 2
(k) 5
(i) –2 (j) 5 (c) 20 (l) –3 (m) 28 (n) 36 (o) 63 (p) 99
2. (a) 10 (e) 21 2 (b) 4
(c) 12 (b) –2 (d) 2 (e) 7 (f) 3 3. (a) 6

(d) 60 (f) 19 (g) 18 (h) 37 (i) 2

22.2 Problems of Linear Equation of One Variable

Word problems of one variable are generally discussed in order to develop the un-
derstanding capacity of the students for the formation of equations and steps of
their functions.

Following are a few important steps to form the equations from the given word
functions.

• Suppose the "unknown" value by any variable say 'x', 'y', 'z' etc.

• According to the condition given by the question, form the linear equation
according to the condition given in the question.

• Solve the given equation and find the value of the variable.

• Verify the answer with the condition given by the question.

Worked Out Examples

Example: 1

The difference of two numbers is 20. If the smaller number is 32, find the greater one.
Solution:

Let the greater number = x,

Now, by the condition,

x – 32 = 20

or, x = 32 + 20

= 52

∴ The required number = 52

Example: 2

A number is twice the other, if their difference is 11, find the numbers.
Solution:

Let the required numbers be x and 2x.

From the given condition,

290 Oasis School Mathematics – 7

2x – x = 11
or, x = 11

∴ Required numbers are, x = 11

2x = 2 × 11 = 22

Example: 3

Divide Rs. 1400 into the ratio of 1:2:4.
Solution:

Let First part be = x

Second part be = 2x

Third part be = 4x

By the question, x +2x + 4x = Rs. 1400

or, 7x = 1400
or,
x = 1400
7

x = 200

∴ First part = Rs. 200

Second part = 2 × 200 = Rs. 400

Third part = 4 × 200 = Rs. 800

Example: 4

Length of a room is three times its breadth, if its perimeter is 80m, find its length and
breadth.

Solution:

Let, the breadth of the room = x,

then, the length of the room = 3x

Perimeter = 80m.

We have, P = 2(l + b)

or, 80 = 2(3x + x)

or, 80 = 2 × 4x

or, 80 = 8x
or,
or, x = 80
8

x = 10m.

∴ Breadth of the room = x = 10m

Length of the room = 3x = 3 × 10m = 30m.

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Example: 5

Sum of two consecutive odd numbers is 32, find the numbers.
Solution:

Here,

L et the required number be x and (x + 2) Alternative Method

Now, x + (x + 2) = 32 Two consecutive natural numbers
are x and (x+1)
or, 2x + 2 = 32
Two consecutive even numbers are x
or, 2x = 32 – 2 and x+2

or, 2x = 30 Two consecutive odd numbers are x
or, and x+2
or, 30
x = 2

x = 15

∴ Required numbers be x = 15
and x + 2 = 15 + 2 = 17

Example: 6

The sides of ∆ABC are (x + 1) cm, (x + 2) cm and (x + 3) cm. If the perimeter is 30 cm, find
the value of x and each side of the triangle.

Solution:

Here,

(x + 1) + (x + 2) + (x + 3) = 30 cm

or, (3x + 6) cm = 30 cm

or, 3x = 30 cm – 6 cm A
B x+2 C
or, = 24 cm

x = 24cm x+1 x+3
3

= 8 cm

∴ x + 1 = (8 + 1) cm = 9 cm

x + 2 = (8 + 2) cm = 10 cm

x + 3 = (8 + 3) cm = 11 cm

292 Oasis School Mathematics – 7

Exercise 22.2

1. (a) The sum of two numbers is 25. If one of them is 12, find the other.

(b) The sum of two numbers is 31. If the smaller is 8, find the greater one.

(c) The sum of two numbers is 94. If one of them is 1 , find the other.
3

(d) The difference of two numbers is 11. If the greater number is 34, find the smaller

number.

(e) The difference of two numbers is 5. If the smaller one is 13, find the greater one. ]

(f) The difference of two numbers be 31 , if the greater number be 1 , find the
smaller number. 2

2. (a) The difference between the age of father and that of his son is 26 years. If the
present age of father is 49 years, find the present age of son.

(b) Age of father and his son are (x + 30) years and x years respectively If the sum
of their ages is 50 years, find their ages.

3. (a) Sum of two consecutive numbers is 43, find the number.

(b) Sum of two consecutive odd numbers is 24, find the number.

(c) Sum of two consecutive even numbers is 34, find the number.

(d) The sum of three consecutive even numbers is 90. Find them.

4. (a) A sum of Rs. 1000 is divided among three persons in the ratio of 2: 3: 5, find
their parts.

(b) Three angles of a triangle are in the ratio 2: 3: 4. Find the size of each angle.

(c) The angles of a quadrilateral are in the ratio. of 1: 2: 3: 4. Find the value of each
angle.

5. (a) A rectangular field is 30 m long. If the perimeter is 110 m, find its breadth.

(b) A rectangular field whose length is 10 m longer than breadth has its perimeter
180 m, find its length and breadth.

(c) Length of a room is twice its breadth. If its perimeter is 36m, find its length and
breadth.

(d) A square has its perimeter 440 m, find its length.

(e) The perimeter of a triangle is 60 cm. If the ratio of its sides is 3: 4: 5, find the
measure of each side.

(f) The perimeter of a rectangle is 56 cm. If the ratio of its breadth and length is 2:5.
Find its length and breadth.

6. Find the value of x, if the perimeter of the given figure is given as below. 293

Oasis School Mathematics – 7

(a) x + 3 (b) (c) x+3
x
2x + 2

3x + 5

P = 35 cm
x–5
x –10
2x + 1
x+2

x+2 2x + 10

P = 30 cm P = 120 cm

Answer (b) 23 (c) 1 (d) 23 (e) 18 (f) 1
1. (a) 13 9 6
2. (a) 23 years. (b) 10 years, 40 years 3. (a) 21, 22 (b) 11, 13 (c)16, 18

(d) 28, 30, 32 4. (a) Rs. 200, Rs. 300, Rs. 500 (b) 40°, 60°, 80°

(c) 36°, 72° 108°, 144° 5. (a) 25m (b) 50 m, 40 m (c) 12m, 6m (d) 110 m

(e) 15 cm, 20 cm, 25 cm (f) l = 20m, b = 8 m

6. (a) x = 10 cm (b) x = 20 cm (c) x = 2.2 cm

22.3 Graph of Linear Equations

Look at the given equations, x +y = 7, 2x – y = 5, 5x + 4y = 16, etc.

These all equations are in the form of ax+by+c = 0.

In the first equation, a = 1, b = 1, c = -7

In the second equation, a = 2, b = -1, c = -5

In the third equation a = 5, b = 4, c = -16

Every linear equations are in the form of ax + by + c= 0.

In this part, we will discuss how to represent the linear equation by graph.

Lets take the first equation

x+y=7

Let's find some points which lie on this equation.

when x = 0, y = 7 Y

when x = 1, y = 6 (0, 7)
(1, 6)
when x = 2, y = 5 (2, 5)
(3, 4)
when x = 3, y = 4

∴ Points (0, 7), (1, 6), (2, 5), (3, 4).... etc. lie on the

given equation. Let's plot there points on the graph

paper and join these points. X' X

The line obtained by joining these points is the graph- O
ical representation of given line. Y'



294 Oasis School Mathematics – 7