Prime Mathematics 7

In each class different bar should be raised in same order.
Different colurs or shade are to be provided for different bars.

Example2: The data given below shows the number of boys and girls in different
classes of a school. Represent the data in multiple bar diagram.

Classes VI VII VIII IX X

No. of boys 20 18 15 18 26

No. of girls 24 22 25 18 12

Solution :

Exercise 13.2

1.a) Areas of oceans of the world in percentage are given below. Show the data in
simple bar graph.

Ocean Area (in %)

Pacific 46

Atlantic 27

Indian 19

Antarctic 5

Arctic 3

b) The table below shows the birth rate per thousand of different countries over a
certain period. Show the data in a simple bar graph.

Country USA France China India Pakistan

Birth rate/1000 18 20 12 36 45

c) The marks secured by a student in different subjects is given below. Represent the
data by a bar graph.

Subjects Maths English Nepali Social Studies Science

Marks 100 30 60 70 80

244 Statistics

d) The number of students enrolled in different classes is given below. Represent the
data by a bar graph.

Classes I II III IV V

No.of students 150 170 130 125 110

2. a) The table given below shows the number of students of a primary school.
Represent the data in a multiple bar chart.
Class I II III IV V
Boys 28 22 40 15 16
Girls 18 34 12 25 30

b) Number of patients who visited a doctor in a week are as follows. Show the data
in a multiple bar diagram.

Days Sun Mon Tue Wed Thur Fri Sat

Male Patients 10 14 10 8 20 8 12

Female Patients 15 6 12 8 8 4 20

c) The number of students passed in different years in S.L.C. exam are given below.
Represent the data by a multiple bar graph.

Years(B.S.) 2063 2064 2065 2066 2067 2068

No.of boys 45 60 70 65 75 80

No.of girls 50 25 40 45 30 50

d) The number of tourists who visited Nepal on different years from U.S.A. and India
is given below. The number of tourists is given in thousands. Represent the data
by a multiple bar graph

Years(A.D.) 2006 2007 2008 2009 2010 2011 2012

India 25 42 45 35 20 50 40

U.S.A. 32 35 50 65 80 60 75

3. a) Read the bar graph and answer the following questions.

Marks obtained by a student in
different subjects

i. What is the information given by the graph?
ii. In which subject, the students has scored highest mark and what is it?
iii. In which subject the student has scored the least mark and what is it?

Prime Mathematics Book - 7 245

b) Study the given bar graph and answer the following questions;
The number of students passed in S.L.C.
exam in different years of a school.

i. What is the number of students passed in S.L.C. exam in the year 2063 B.S.?
ii. In which year, the least number of students passed in S.L.C. exam ? What is

the number of students?
iii. What is the total number of students passed in S.L.C. exam from 2062 to

2066?
iv. In which year, maximun number of students passed in S.L.C. exam ? What is

the number of students?
c) Read the given multiple bar diagram and answer the following questions:

Number of boys and girls of
different classes of a school

No. of students.

i. What is the total number of students in grade VII ?
ii. What is the number of girls in grade VIII ?
iii. Which grade has the maximum number of students ? What is it ?
iv. What is the number of boys and girls in grade IX?
v. What is the total number of students from grade VII to grade IX ?

246 Statistics

d . Study the diagram given below and answer the following question.

Result of SLC examination of ABC school.

Distinction

i. What type of representation of data is it?
ii. What is the data about?
iii. Find the number of students passed in each year?
iv. In which year did the maximum number of students passed in distinction?
v. How many students were passed in distinction in the year 2068?

13.3 Analysis of data

After the data collected is classified and tabulated, the next step is to analyse it to
get further informations. Under analysis comes necessary calculations, like central
tendency, dispersion, skewness, kurtosis etc.

Central tendency: Literally central tendency means "characteristics of a group tend
(goes) towards the central value (average) i.e. average represents the whole data".
A number or a quantity which is typical or representative of a set of data is called
central tendency. Measures of central tendency are the averages (mean, median and
mode).

Mean (or arithmetic mean): The mean of a given set of data is obtained by dividing
the sum of all the observations by their number. It is denoted by x (read as x bar).
Let the given data is 8, 20,15,5 and 12.

The sum of the data = 8+20+15+5+12 = 60
Number of data = 5
sum of the observations 60
\ Mean (x) = number of observations = 5 = 12.

a) Mean of raw (individual) data: Let there be n numbers of observations: x1,x2,x3,..,xn

Then their mean x = x1+ x2+xn3+...+xn

Which is also written as x= 1 ∑ xn

n n=1 n

or simply written as x= ∑x
n

The symbol ∑ (Greek letter "sigma ") stands for “summation of”

Prime Mathematics Book - 7 247

Example 1 : Find the mean of the data.

27, 25, 37, 43, 55, 59, 67, 29, 14, 32

Solution:

Here ∑x = 27+25+37+43+55+59+67+29+14+32 = 388

And N = 10 Mean (x) = ∑x = 388 = 38.8
\ N 10

b) Mean of ungrouped data: Let x1, x2, x3, ……..,xn be N observations with frequencies
f1, f2, f3, …….,fn respectively. This means the observation x1 occurs f1 times, x2 occurs
f2 times and so on. Then the sum of all these observations = f1x1+f2x2+f3x3+…………+fnxn

∑= f xn or simply = ∑fx
i=1 i i

Also, the total number of observations = f1+f2+f3+……….+fn
= i∑=n1fi = N

\ The mean of the data = the sum of all the observations
total number of observations
i∑=n1fixi
or x= i∑=n1fi = ∑fx
N

For an ungrouped data

Prepare table with the columns of x, f and fx

Sum of f values gives N, sum of fx values gives ∑fx

Then x= ∑fx
N

Example 2 : Find the mean of the data:

x 10 20 30 40
f 9 15 14 12

Solution:

Here ∑fx = 10×9+20×15+30×14+40×12

= 90+300+420+480 = 1290

And N = ∑f = 9+15+14+12 = 50
\ Mean (x)
= ∑fx = 1290 = 25.8
N 50

248 Statistics

Alternate method :

The given data can be tabulated as

xf fx

10 9 90

20 15 300

30 14 420

40 12 480

N = 50 ∑fx = 1290
Here, we get
N = 50

∑fx = 1290 ∑fx 1290
N 50
\ Mean (x) = = = 25.8

Example 3: The marks recorded by 30 students in a unit test is as follows:
6, 7, 8, 9, 7, 6, 8, 9, 10, 6, 10, 4, 6, 7, 8, 6, 8, 7, 7, 7, 6, 8, 4, 5, 6, 7, 8,
6, 8, 10. Find the arithmetic mean.
Solution:
Here; No. of items (N) = 30
Sum of items(∑x) = 6+7+8+9+7+6+8+9+10+6+10+4+6+7+8+6+8+7+7+7+6+8+
4+5+6+7+8+6+8+10
= 214

\ Arithmetic mean (x) = Sum of items (∑x) = 214 = 7.1
No. of items (N) 30

This problem can also be solved by representing data in frequency distribution table.

Item(x) Tally marks Frequency (f) fx
4 2 8

5 1 5
6 8 48
7 7 49
8 7 56
9 2 18
10 3 30

N = ∑f = 30 ∑fx = 214

\ Arithmetic mean (x) = Sum of items (∑fx) = ∑fx
No. of items (∑f) N

= 214 = 7.13
30

Prime Mathematics Book - 7 249

Exercise 13.3
1. Find the arithmetic mean of the following data.

a) 28, 29, 30, 31, 32
b) 12, 10, 15, 9, 11, 6, 14
c) 15, 21, 91, 83, 68, 48, 58
d) 3, 6, 7, 8, 10, 13, 15, 16, 12
e) 11, 23, 34, 47, 67, 79, 81, 58
f) 110cm, 70cm, 120cm, 40cm, 210cm, 80cm
g) 35, 37, 29, 45, 29, 19, 15, 27, 37, 33, 27, 25, 19, 38, 36, 25, 13, 12, 22, 37

2. a) From the following data, find the mean height.

Heights(inches) 60 61 62 63 64

No. of children 23 8 7 5

b) Find the mean marks from the following data.

Marks(x) 15 16 17 18 19 20 21 22
93
No. of Students 3 9 12 24 30 10

c) From the following data, find the mean marks.

Marks 0 1 2 3 4 5 6 7 8 9 10

No.of students 1 1 3 6 7 9 6 4 7 4 2

d) The daily wages of 30 workers are given in the table. Find the mean wage.

Wages (in Rs.) 240 250 275 280 290 300

No. of workers 4 6 8 11 8 3

3. a) The number of members in 30 different families of a certain locality are given
below.

4, 3, 5, 2, 4, 6, 4, 5, 7, 4, 3,
7, 4, 5, 4, 6, 5, 3, 4, 7, 4, 5,
6, 3, 3, 4, 2, 6, 4, 5
Construct a frequency distribution table and calculate the mean.

b) Daily wages of 25 workers of a construction company (in Rs.) are given
below.
350, 300, 400, 500, 250, 350, 450, 300, 400, 350, 500,
350, 400, 300, 350, 450, 300, 400, 350, 300, 300, 450,
250, 400, 350
Construct a frequency distribution table (ungrouped) and also compute the mean
wage.

c) The number of accidents per day for a month of 30 days are as follows.
1, 2, 3, 1, 3, 1, 4, 5, 1, 2,
1, 0, 1, 2, 2, 3, 4, 1, 0, 1,
1, 2, 4, 1, 0, 1, 2, 3, 2, 3
Prepare a frequency distribution table and calculate the arithmetic mean.

250 Statistics

d) The number of students absent in class were recorded every day for 20 days.
0, 1, 2, 3, 4, 6, 7, 0, 1, 2,
3, 3, 2, 1, 0, 0, 1, 2, 3, 1

Prepare a frequency distribution table and calculate the arithmetic mean.
4. a) If ∑fx = 480 and ∑f = 20, find x .

b) If mean marks of 25 students is 16, find the total marks of all the
students.

c) If N = 70+a, x = 50 and ∑fx = 2000+55a, find the value of a.
d) The mean weight of some men is 56 kg and the sum of their weight is 392 kg,
Find the number of men.
e) In a series, the mean is 20, ∑fx = 400 + 20a and ∑f = 18+2a, Find the

value of a.
f) Marks obtained by 7 students in an examination are as follows 61, 67,

p, 77, 79, 81, 82. If the arithmetic mean is 74, Find the value of p.
g) If the average of 5, 7, 8, x, 9 and 10 is 7, Find the value of x.

5) a) If the mean age in the given data is 12.08 years, calculate the
missing frequency.

Age (in years) 10 11 12 13 14

No. of children 3 a 18 12 5

b) If the mean of the data given below is 9, find the value of a.
x 4 6 9 10 15
f 5 10 10 a 8

Unit Revision Test

1. Construct a frequency distribution table with tally marks from the date.
12, 15, 14, 18, 10, 14, 15, 10, 15, 16, 12, 16, 16, 15, 10, 15, 16, 14, 16, 15, 18,
15, 12, 18, 12, 15, 14, 14, 16

2. Pocket money (in Rs) of 50 students of class VIII of a school are as follows.
22, 55, 49, 27, 30, 27, 30, 25, 42, 40, 13, 24, 38, 10, 24, 30, 33, 27, 29, 40, 15,
32, 36, 32, 27, 35, 17, 36, 18, 41, 20, 41, 51, 29, 27, 44, 43, 15, 10, 32, 50, 29,
18, 54, 34, 45, 13, 14.
Construct a frequency distribution table with one of the class 10-20 exclusive.

3. Construct a cumulative frequency distribution table of the following data with
group of 10.
40, 71, 99, 61, 51, 90, 59, 60, 59, 90, 85, 89, 50, 49, 65, 67, 79, 49, 75, 65, 70,
45, 40, 90, 79, 97, 98, 70, 81, 70

4. Represent the data given below in a bar diagram.

Years (B.S) 2066 2067 2068 2069 2070

No of Students 25 40 34 48 55

Prime Mathematics Book - 7 251

5. Represent the given data in a multiple bar diagram.

Subjects Maths Science English Nepali

No. of boys 40 20 30 20

No. of girls 30 50 70 60

6. Read the given multiple bar diagram and answer the following questions:
i. What is the total number of students in grade VII ?
Number of boys and girls of
different classes.
ii. What is the number of girls in grade VIII ?
iii. Which grade has the maximum number of
students ? What is it ?
iv. What is the number of boys and girls in class IX?
v. What is the total number of students from
grade 7 to grade 9 ?

7. Find the arithmetic mean of the data:
11, 23, 34, 47, 67, 79, 81, 58.

8. Find the mean marks from the following data.

Marks(x) 15 16 17 18 19 20 21 22

No. of Students 3 9 12 24 30 10 9 3

9. If N = 70+a, x = 50 and ∑fx = 2000+55a, find the value of a.

10. The mean of the data given below is 32. Determine the value of b.
x 10 20 30 40 50 60
f5 8 b 9 7 1

Answers

Show to your teacher. Unit:13 Statistics - Exercise: 13.1

1. Show to your teacher. Exercise: 13.2
2. Show to your teacher.
3. a) i) Marks obtained by a student in different subjects.
ii) Mathematics, 100 iii) Nepali, 60 iv) Respectively 90, 70, 85 sum=245
b) i) 80 ii) B.S. 2064 BS, 60 iii) 440 iv) B.S. 2066, 110
c) i)150 ii) 80 iii) viii,180 iv) boys 80, girls 90 v) 500
d) i) multiple bar diagram ii) Result of S.L.C. exam of ABC school
iii) 2067,42; 2068,36; 2069,36 iv) 2069 v) 5

1. a) 30 b) 11 Exercise: 13.3 d) 10
e) 50 f) 105 c) 54.86
b) 18.5 g) 28 d) 274 b) 7
2. a) 62.4 b) 364 c) 5.48 d) 2.1
3. a) 4.47 b) 400 c) 1.9 d) 7
4. a) 24 f) 71 c) 300 5. a) 12
g) 3
e) 2

252 Statistics

14UNIT ALGEBRA

Estimated periods 28 Objectives:

At the end of this unit students will be able to
● know about the types of algebraic expressions according to number of terms

and degree of variable.

● know about the polynomials and find the values of polynomials by substituting
the values of variables.

● Perform the four fundamental mathematical operations on algebraic expressions
● know about the geometrical concept of (a±b)2 and its application.
● use of law of indices.
● solve linear equation in one variable.
● translate the word problem into linear equation in one variable and solve.
● represent inequality in a number line.
● represent relation between two quantities in a function machine and a graph.
● draw the graph of liner equation of two variables.

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

Teaching Materials:
Chart of algebraic expressions, tiles, local materials, graph paper, function machine etc.

Activities:
It is better to

● discuss the way of writing the algebraic expression of simple variable
problems.

● discuss the activities of four fundamental mathematical operation on algebraic
expression by using the tiles.

● drill the problems of algebraic expression
● discuss about the rule of indices and its applications
● discuss the way of solving the equation of one variable
● discuss the way of representing an inequality of one variable in a number line
● discuss the relation between two quantities in a function machine
● discus about the graph of liner equation of two variables.

14.1 Algebraic Expression Estimated Period - 12

When 3 is added to x, the sum is x + 3. When 4 is subtracted from z, the

difference is z - 4. When x is multiplied by 5, the product is 5x. When x is

divided by 7, 7xthearequtohteienatlgeibs ra7xic. Here, the mathematical statements x + 3,
z - 4, 5x and expressions. So, the mathematical operations

(+, -, ×, ÷ ) are used between the variable (like x, z) and constant (like 3, 4 etc). The

mathematical statements formed by using the mathematical operations (+, -, × ÷)

between the variables and constants are called algebraic expressions. Some more
a2 + 2ab + b2
examples of algebraic expressions are 7, 5y + z, 4xy, 2x3 + 3x - 7, x2- 1 +7, x ,
x
y2 – 2 y +7

14.1.1 Classificaton of algebraic expression:

Algebric expression are classified by two ways which are
(i) according to terms (ii) according to degree

(i) According to terms : There are following algebraic expression.
(a) Monomial expression: An expression which contains only one term is called

monomial expression. For example: 5, 2x, 4x2, -3xyz
(b) Binomial expression: An expression which contains two terms is called binomial

expression. For example: a+b, 3x-y, 2xy+7, 4x3-3xy
(c) Trionomial expression: An expression which contains three terms is called trinomial

expression. For example: x+y-z, 2x2-3xy+y2, 4x3-7xy+8.
(d) Multinomial expression: An expression which contains more than three terms is

called multinomial expression. For examples: x + 2y – z + 7, a3+4x2-7x +5y -8.
(ii) According to degree, there are following algebraic expressions.
a) First degree expression: An expression in which the heighest power of the variable in

any term is 1 is called first degree expression. For example: 3x, x+2, x+y, 4x-y+7 etc.
(b) Second degree expression: An expression in which the highest power of variable

in any term is 2 is called second degree expression. For example: 5y2, 3x2+7x,
z2-2z+7, x-xy, x2+2xy-6 etc.
(c) Third degree expression : An expression in which the highest power of variable
in any term is 3 is called third degree expression. For example: 7x3, xyz, x3+4x2-7,
6x2y – 7xy + 5 etc.

Note:

The degree of an expression having single variable is the heighest power of the
variable of any term contained in that expression
The degree of an expression having two or more variables is the heighest sum of

254 Algebra

the power of variables of any terms contained in that expression
An expression containing constant term only Is called zero degree expression
because constant term has a variable with zero exponent.

Polynomial: An algebraic expression having one or more than one terms where as the

power of the variables of the terms is whole number is called polynomials. Therfore,
a polynomial is a special type of expression with the variable having whole number as
the power. For examples: 4x, 5, 7x2+2, xy + 7y-9, 3x3+2x2+4x-9 etc.

Example 1. Identify the following algebraic expressions are which expression according to terms.
a) 3x b) x2-y c) 4x2y d) x+y-2+z e) 3x2+2xy-y2

Solution
a) 3x is a monomial expression as it contains only one term.
b) x2-y is a binomial expression as it contains two terms.
c) 4x2y is a monomial expression as it contains only one term.
d) x+y-2+z is a multinomial expression as it contains more than three terms.
e) 3x2+2xy-y2 is a trinomial expression as it contains three terms.

Example 2. Write down the degree of the following polynomials:
a) 5x b) 3xy2 c) 2x2-5xy3 d) x3y3+y3-x4+x2y3

Solution:
a) 5x is first degree polynomial because the power of x is 1.
b) 3xy2 is third degree polynomial because the sum of the powers of x and y2 is

1+2=3.
c) 2x2-5xy3 is fourth degree polynomial because the sum of the powers of x and y in

2nd term is 1+3=4.
d) x3y3 + y3 – x4 +x2y3 is sixth degree polynomial because the sum of the powers of x

and y in 1st term is 3+3=6 which is highest among other terms.

Example 3. Find the degree of the polynomial 5x3-2x5+x.
Solution:
Given polynomial is 5x3-2x5+x
The power of variable x in 5x3 is 3, the power of x in 2x5 is 5 and the power of x in x is 1.
So, the highest power of x is 5. Hence the given polynomial is fifth degree polynomial.

Example 4. Find the degree of the polynomial x5-6x2y3 + 4x4y3 – 8x2y3z.
Solution:
Given polynomial is x5-6x2y3 + 4x4y3 – 8x2y3z
The sum of exponents of variable in x5 = 5
The sum of exponents of variables in 6x2y3 = 2+3 = 5
The sum of exponents of variables in 4x4y3 = 4+3 = 7
The sum of exponents of variables in 8x2y3z = 2+3+1 = 6
The highest sum of the exponents of variables is 7.

Prime Mathematics Book - 7 255

Hence the given polynomial is seventh degree polynomial.

Example 5. Find the value of the polynomial 2x2+3xy3-4xy when x = 2 and y = 3.

Solution :
Here,

2x2+3xy3-4xy

= 2(2)2+3×2×(3)3 - 4×2×3 [x = 2 and y = 3]

= 2×4+6×27-24

= 8+162-24

= 170-24

= 146

Exercise 14.1.1

1. Identify the following algebraic expressions according to terms.
a) 3x + z b) 3x3 + 4xy + 7 c) 5x d) 7x2y e) x2 - 6x

f) 2abc g) 4x ÷ 2y h) x - 2y - 3z + 4xy i) 5x - 4y + 3z

j) x7 - 4x2y + 3x3y2 - 5xyz + 8 k) 9x2 - 5
y

2. Classify the following algebraic expressions according to their degrees.
a) 3x2 b) 4x2y c) 3x - 2y2 d) x4 + 2x2y - x
e) x3 - 2y2x + 4x2y3 f) 7x3yz2 g) 4x5 - 7x2y3 + 8xyz6 - 8 h) 3x2y - 2z5

3. Identify which of the following expressions are polynomials? Give reason also.
2
a) 3x2 b) 7x2 - 3 c) 3 x + 4y2 + 2xy d)5
f) 5x - 9
e) 2x - 3 +4 g) x2 - 2x + x + 5
x2

h)9x3 -2x2y + 5xy2 + 9 i) x3 - 2x2y-1 + 5y2

4. Write down the degree of each of the following polynomials:
a) 5 + 2x2 b) 7x2y3 c) 5x2 - 2x + 6x4 - 3 d) 5y2 - 6y3 + 8
e) 5x2 + 6x2y + 7y2 f) a2y4 + 3a5y + y7 g) x4 - 6xy5 + 4x4y2 + y4

5. Find the value of the following algebraic expressions is x = 2, y = -1 and z = 3.
a) 2x2 - 3y b) 3x + 4y2 c) x3 + 2y2 - 2z d) x2 - 3y3 + 2xy - z

e) x2 - 2y2 + 3zy f) 2x + 3y g) 3x2 + 4y - z2
z 2y2z

6. What is the value of 5 - 6a when a = -4, -2 and 0?
4

7. Find the perimeter of the given figures when x=3, y=2, z=5 and a=4.

256 Algebra

14.1.2 Addition and subtraction of Algebric Expressions

When carrying out addition and subtraction of algebraic expressions, we first connect
the given algebric expressions with the mathematical operations (+ and -) and then we
select the like terms and finally add or subtract the coefficient of like terms.
Study the following examples.
Example 1. Add 3x - 4y + 3 and 2x + 6y - 1
Solution
Here,
Given expressions are 3x - 4y + 3 and 2x + 6y - 1.

(3x - 4y + 3) +(2x + 6y - 1)
= 3x - 4y + 3 + 2x + 6y - 1
= 3x + 2x - 4y + 6y + 3 - 1
= 5x + 2y + 2
Example-2 . Add: 2x3 + 3y3 - 10z3, 5x3 - 7y3 + 12z3 and -6x3 + 3y3 + 2z3
Solution
Addition by horizontally method.

(2x3 + 3y3 - 10z3) + (5x3 - 7y3 + 12z3) + (-6x3 + 3y3 + 2z3)
= 2x3 + 3y3 - 10z3 + 5x3 - 7y3 + 12z3 - 6x3 + 3y3 + 2z3
= 2x3 + 5x3 - 6x3 + 3y3 - 7y3 + 3y3 - 10z3 + 12z3 + 2z3
= x3 - y3 + 4z3
Addition by vertical method

2x3 + 3y3 - 10z3
5x3 - 7y3 + 12z3
-6x3 + 3y3 + 2z3
(+)

x3 - y3 + 4z3

Example 3: Subtract 8x2 - 5x - 2 from 1 + 3x2 - 6x.
Solution

Subtractly horizontally, we get

(1 + 3x2 - 6x) - (8x2 - 5x - 2) Subtracting Vertically
3x2 - 6x + 1
= 1 + 3x2 - 6x - 8x2 + 5x + 2 (Like terms are arranged 8x2 - 5y3 - 2
= 3x2 - 8x2 - 6x + 5x + 1 + 2 in the same column and (-) (+) (+)
= -5x2 - x + 3 then add or subtract the -5x2 - x + 3
coefficients)

Prime Mathematics Book - 7 257

Example 4: What should be added to 2a + 3b - c so that the result will be 4a - b + 3c?
Solution:

Let the required expression be A.
Then, (2a+3b-c) + A = 4a – b + 3c
or, A = (4a - b + 3c) - (2a + 3b - c)

= 4a - b + 3c - 2a - 3b +c
= 4a - 2a - b - 3b + 3c + c
= 2a - 4b + 4c
\ (2a - 4b - 4c) should be added to (2a + 3b - c), so that the result will be 4a - b + 3c.

Example 5. What should be subtracted from 2x2 - 5xy - 7, so that the result will be 2xy?
Solution:
Let the required expression be A.
Then, (2x2 - 5xy - 7) - A = 2xy
or, 2x2 - 5xy - 7 - 2xy = A
or 2x2 - 7xy - 7 = A.
\ 2x2 - 7xy - 7 is subtracted from 2x2 - 5xy - 7, so that the result will be 2xy.

Exercise 14.1.2

1. Add the following expressions.

a) 2a + b and 3a + 4b - c b) 2a + 3b, 2b - 4a and 7a - 2b - 6

c) 2x + 5y - 3 and 3x – 2y -2 d) 3a + 5, 3b + 2c and 3a - b - 2

e) 2x3 + 5y3 - 10z2, 5x3 - 7y3 + 11z2 and -3x3 + 4y3 + 2z2 + 5

f) 4x2y + 7xy2 – 7, -2xy2 - x2y + 6 and 5 + 2x2y – 2xy2

2. Subtract.

a) 2a + 3b from 4a - b + 7 b) 3x - 3y2 from 5x - y2 + 2
c) 4a2 + 4ab - 2b2 from 7a2 - 2ab + b2 d) 2a + b -3c from 4a + 3b -2c
e) 2m3 - 4m2 + 7m - 5 from m3 - 3m2 + 6m + 7
f) x2y - 4xy2 + 5 from 2 - 2xy2 + 3x2y

3. Simplify.
a) (3a2 + 2a + 5) + (4a2 - 3a – 2)
b) (9a2 – 4ab + 7b2) + (4a2 – 3ab + 2b2)
c) (7x2 – 3xy – 2y2) – (5x2 – 5xy + y2)
d) (2m2 – 4mn + 3n2) – (5m2 + 2mn – 5n2)
e) (4a + 15b + 2c) + (3a + 2b – 5c) – (5a + 6b – 2c)
f) (2p2 + 10p + 6) – ( 4p2 + 2p – 3) + (5 + 6p + 6p2)

4. a) What should be added to 3x + 5y - 3z, so that the result will be 2x + 5z - y ?
b) What should be added to 5xy + 4x2 + 7y2, so that the result will be
2x2 – 20xy + 16y2 ?
c) What should be subtracted from m+n-3, so that the result will be 5n-2m-4 ?
d) What should be subtracted from 3a2-2ab+y2, so that the result will be ab?

258 Algebra

14.1.3 Multiplication of Algebraic Expressions

Review

Mulitplicaton of monomial expressions
When a monomial is multiplied by another monomial, we should first multiply the
coefficient of the expressins and then multiply the variables of the expressions.
Study the following examples.
Example 1. Multiply 2ab2 by 3a2
Solution :
Here, 2ab2 × 3a2 = 2 × 3 × a × a2 × b2 = 6a3b2
Example 2. Multiply: 4x2y × -3xz × 5z2y
Solution:
Here, 4x2y × (-3)xz × 5z2y

= 4 × (-3) × 5 × x2y × xz × z2y
= -60x2+1 y 1+1 z1+2
= -60x3y2z3

Multiplying a binomial or Trinomial by a monomial.

When a binomial or Trinomial expression is multiplied by a monomial expression we
should multiply each term of the binomial or trinomial expression by the monomial
expression.
Study the following examples:
Example 3. Multiply (2x+3xy) by 4x2
Solution:
Here, (2x+3xy) × 4x2

= 2x × 4x2 + 3xy × 4x2
= 2 × 4 × x × x2 + 3× 4 × x × x2 × y
= 8 x1+2 + 12 x 1+2 y
= 8x3+12x3y
Example 4. Mulitply (3x2y2 + 2xy - 4y2) by 2xy.
Solution:
Here, (3x2y2 + 2xy - 4y2) × 2xy
= 3x2y2 × 2xy + 2xy × 2xy - 4y2 × 2xy
= 3 × 2 × x2 × x × y2 × y + 2 × 2 × x × x × y × y – 4 × 2 × y2 × y × x
= 6x2+1y2+1 + 4x1+1y1+1-8y2+1x

= 6x3y3 + 4x2y2 - 8y3x.

Prime Mathematics Book - 7 259

Exercise 14.1.3

1. Multiply: b) 2ab2 × -3bc c) 2m2ny3 × -5mn2
a) 3xy × 2x2

d) -2b × 6ab2c e) 2a× 4bc × 5ax2 f) -3x2 × 4xy2 × -5xz2

g) (-4ab) × 3b2c × (-7c3a) h) 3p2y × 7q2r × 2pqr3 ( )i)1 x2y x 4xy
2
( ) ( )j) ( )k) (16x2y) x
2 a2b x 9 ab2 3 4x2y2
3 4 8

2. Calculate the area of each of the following rectangles.

a) b) c)

3. Calculate the volume of the following solid cuboids.
a) b) c)

5c cm 2 cm

3b cm 1.3y cm

7a cm 3.5x cm

4. Multiply. b) (3x - 2y) by 5xy2
a) (4a + 2b) by 3ab d) (5m2 - n2) by (- 2mn)
c) (4x2y + 3xy2) by 2xy f) (2p2 – pq–4) by 4p2q
e) (x2 - xy2 +3y3) by 2xy2

5. Calculate the area of each of the following rectangle.
a) b) n c)

n

260 Algebra

14.1.4 Multiply a binomial expression by another binomial expression

What is the area of the adjoining rectangle of
length (x+y) cm and breadth (a+b) cm ?

In the given adjoing figure, the rectangle is divided
into two parts x cm and y cm along the length.
Similarly it is divided into two parts a cm and b cm
along the breadth. So, that the whole rectangle is
divided into four parts.

From the figure, area of the whole rectangle =
sum of the area of four parts.

We know that, = (ax + ay + bx + by) cm2
Area of rectangle = length × breadth

= (x +y) cm × (a+b)cm

= (x+y) (a+b) cm2

Now, (x+y)(a+b)cm2 and (ax+ay+bx+by)cm2 are the area of the same rectangle

\ (x+y) (a+b) = ax+ ay + bx + by

Here, (x+y) is multiplied by (a+b) means the first (x+y) is multiplied by a and then again
(x+y) is multiplied by +b

The process of multiplication is shown below.

Hence, when a binomial expression is multiplied by another binomial expressions we
first multiply each term of the second expression by each term of the first exrpessin
seperetely and then the products are simplified.

Example 1. Multiply : (2x - 3y) × (3x + 4y)
Solution:
Here, (2x -3y) × (3x + 4y)

= 2x(3x + 4y) – 3y(3x + 4y)
= 2x × 3x + 2x × 4y - 3y × 3x - 3y × 4y
= 6x2 + 8xy – 9xy – 12y2
= 6x2 – xy – 12y2

Prime Mathematics Book - 7 261

Example 2. Find the product of (3x + y) and (4x - 3y). If x = 2 and y = 3, what is the
value of the prouduct?

Solution:
Here,
The product of (3x + y) and (4x - 3y)

= (3x + y) × (4x – 3y)
= 3x(4x – 3y) + y(4x – 3y)
= 12x2 - 9xy + 4xy - 3y2 = 12x2 -
5xy - 3y2
The value of the product
= 12(2)2 - 5 × (2) × (3) - 3(3)2
= 12 × 4 - 30 - 3× 9
= 48 - 30 - 27
= 48 - 57 = -9

Multiplying a Trinomial expression by a Binomial Expression

What is the area of the adjoining rectangle of length (a+b+c) units and breadth (x+y)
units ?
In the adjoining figure, the whole rectangle is divided into three parts a units, b units
and c units along the length. Similarly, it is divided into two parts x units and y units
along the breadth.
So, that the whole rectangle is divided into six parts.
From the figure, area of the whole rectangle = sum of the are of six parts.

= (ax + bx + cx + ay + by + cy) sq. units.

Area of the rectangle = length × breadth
= (a + b + c) × (x + y)
= a(x + y) + b(x + y) + c(x + y)
= ax + ay + bx + by + cx + cy
= (ax + bx + cx + ay + by + cy) sq.units.

Now, (a + b + c) (x + y) sq. units and (ax + bx + cx + ay + by + cy) sq.units are the area
of the same rectangle.

\ (a + b + c) (x + y) = ax + bx + cx + cy + by + cy

Here, (a + b + c) is multiplied by (x + y) means the first (a + b + c) is multiplied by x and

then again (a + b + c) is multiplied by +y.

The precess of the multiplication is shown below.

262 Algebra

Hence, when a trinomial expression is multiplied by another binomial expression, we
first multiply each term of the second expression by each term of the first expression

separately and then simplify the terms.

Example 3. Multiply : (2x + 3y - z) × (a + 2b)
Solution :
Here, (2x + 3y -z) × (a + 2b)

= 2x(a + 2b) + 3y (a + 2b) - z(a + 2b)
= 2ax + 4bx +3ay + 6by - az - 2bz

Example -4. Find the product of (3a - 2b) and (6a + 7b - 8c). Also, find the value of the
product when a = -2, b = 1 and c = 3.

Solution :
Here,
Product of (3a - 2b) and (6a + 7b - 8c) = (3a - 2b) × (6a + 7b - 8c)

= 3a (6a + 7b - 8c) -2b (6a + 7b - 8c)
= 18a2 + 21ab - 24ac - 12ab - 14b2 + 16 bc
= 18a2 + 21ab - 12 ab - 24ac + 16bc - 14b2
= 18a2 + 9ab - 24ac + 16bc - 14b2
The value of the product = 18 (-2)2 + 9 × -2 × 1 - 24 × -2 × 3 + 16 × 1 × 3 - 14(1)2
= 18 × 4 - 18 + 144 + 48 - 14 × 1
= 72 - 18 + 144 + 48 -14
= 264 - 32
= 232
Example 5. The length and breadth of a rectangular garden are (12a - 3b) units and
(6a - 2b - 2c) units respectively. Find the area of the garden . Also find the
actual area of the garden when a = 5, b = 2 and c = -1.
Solution:
Here,
Length (l) = (12a - 3b) and breadth (b) = (6a - 2b - 2c)
Area of the garden (A) = l × b
= (12a - 3b) × (6a - 2b - 2c)
= 12a(6a - 2b - 2c) - 3b(6a - 2b - 2c)
= 72a2 - 24ab - 24ac - 18ab + 6b2 + 6bc
= (72a2 - 42ab - 24ac + 6bc + 6b2)sq.units.
Actual area of the garden
= 72 × (5)2 - 42 × 5 × 2 - 24 × 5 × (-1) + 6 × 2 × (-1) + 6× (2)2
= 72 × 25 - 420 + 120 - 12 + 6 × 4
= 1800 - 420 +120 - 12 +24
= 1800 + 120 +24 - 420 - 12
= 1944 - 432
= 1512 sq units

Prime Mathematics Book - 7 263

Exercise 14.1.4

1. Multiply
a) (x + y) × (x - 2y) b) (a + 2b) × (2a + b) c) (m - n) × (m - n)
d) (2x + z) × (2x - z) e) (x + 2a) × (x - 5a) f) (5p - 6q) × (6p - 5q)
g) (x2 + y2) × (2x -3y) h) (3x - 5y) × (3x + 5y) i) (a2 - b) × (a2 + b)

( )( ) ( )( )j)5x+3 5x - y k) x+ 1 x - 1
y 5x y 5x 2y 2y

2x( )( )l) + 1 2x + 1
2x 2x

2. Find the product of (2m +n) and (3m - 2n). Also find the value of the product when
m = 3 and n = 2.
3. If the length and breadth of a rectangle are (5a + 2b) and (4a - b) respectively,
what will be the area ? Also find the area in the numeral value when a = 3m and
b = 2m.
4. Find the product of (5m + 3) and (3m - 2). Also find the value of the product when
m = 2.
5. Multiply:
a) (x + 2) (x2 - 3x + 3) b) (2m + 3) (m2 + 2m + 3) c) (x + 2) (x2 + 2x - 3)
d) (x - y) (x2+ xy + y2) e) (a2 - b2) (2a + 3b - 4c) f) (m + n)(m2- mn + n2)

6. Find the product of (p-2) and (p2 - 3p + 4). If p=2, what is the actual value of the
product ?

7. If the length and breadth of a rectangle are (2x + y + 3) cm and (x + 2y) cm
respectively, find its area, If x = 2 and y = -1, Find its actual area.

8. Calculate the area of each of the following rectangles.
a) b) c)

d) e)

264 Algebra

14.1.5 Division of Algebraic Expression

Divide Binomial or Trinomial Expression by a Monomial Expression
When the Binomial or Trinomial polynomial is divided by a monominal, we divide each
term of the polynomial by the monominal.
Study the following example,
Example 1: Divide 2x2 - 6xy2 by 2x
Solution

Here, (2x2 - 6xy2) + 2x

2x2 - 6xy2
= 2x
2x2 6xy2
= 2x - 2x (Separating the terms)

= x - 3y2
Next method.

2x) 2x2 - 6xy2 (x-3y2
-2x2
0 - 6xy2
+- 6xy2
0

\ quotient = x - 3y2

Example : 2 Divide 4m2n - 12mn2 + 16m2n3 by 4mn
Solution :
Here, (4m2n - 12mn2 + 16m2n3) + 4mn

= 4m2n - 12mn2 + 16m2n3
4mn

4m2n 12mn2 16m2n3
= 4mn - 4mn + 4mn

= m-3n + 4mn2
Next Method
4mn) 4m2n - 12mn2 16m2n3(m-3n+4mn2

- 4m2n
- 12mn2
+- 12mn2
16m2n3
- 16m2n3
0

\ Quotient = m-3n + 4mn2

Prime Mathematics Book - 7 265

Dividing Binomial or Trinomial Polynomial by a Binomial

When the binomial or trinomial polynomial is divided by a binomial, we divide by the
long division method as used above in the examples. Study the following examples.

Example 3 Divide (4x2 - y2) by (2x + y) [ ]\ 4x2 + 2x = 4x2 = 2x
Solution: 2x
Here
(4x2 - y2) + (2x + y)

2x+y) 4x2 - y2 (2x-y (2x + y) ×2x = 4x2 + 2xy
- 4x2 +- 2xy
- 2xy - y2 [ ]\ -2xy + 2x = -2xy = -y
-+ 2xy -+y2 2x
0
(2x + y)×-y = -2xy - y2
\ Quotient = 2x - y
Example 4. Divide (a2 + 7a + 12) by a + 3 [ ]\ a2 + a = a2 =a
a

a+3) a2 + 7a + 12 (a+4 (a+3)×a = a2 +3a
- a2 +- 3a
4a + 12 [ ]4a
- 4a +-12 \ 4a + a = a = 4
0 (a+3)×4 = 4a +12

Example 5. If the length and area of a rectangular garden are (3x + 7) units and
(3x2 - 5x - 28) sq.units respectively, find

a) the breadth of the garden.
b) the actual length, breadth and area of the garden when x = 10m.
Solution
Here,
a) length (l) = (3x + 7) units, area (A) = (3x2 - 5x - 28) sq units.
breadth (b) = ?
We know that
Area (A) = length(l) × breadth(b)

\ Breadth (b) = Area (A) = 3x2 - 5x - 28
Now, length(l) 3x +7

3x+7) 3x2 - 5x - 28 (x-4
-3x2 +-7x
-12x - 28
+-12x -+28
0 \ Breadth of the garden = (x - 4) units.

266 Algebra

b) actual length of the garden = (3x + 7) = 3 × 10 + 7 = 37m
actual breadth of the garden = x - 4 = 10 - 4 = 6m
actual area of the garden = (3x2 - 5x - 28)

= 3 × 102 - 5 × 10 - 28
= 300 - 50 - 28
= 222m2

Exercise 14.1.5

1. Divide b) 24m3n2 + (-8m2n) c)48x5y3z + (-1.2x2y2)
a) 15a2b + 5a e) (5m4 + 3m2 - 4mn) + m2 f) (-2x4+8x3) + 2x2
d) (4a2b + 6ab2) + 2ab h) (4a3 + 6a2b + 8ab2) + 2ab
g) (x3y - x2y3 + x4y2) + x2y
b) (4m2 - 9n2) + (2m+ 3n)
2. Divide:
a) (4a-4b) + (a-b)

c) (4x2 + 12x) + (2x6) d) (a2 + 5a + 6) + (a+2)
e) (x2 - 7x + 12) + (x-4) f) (4b2 + 13b + 21) + (b+8)
g) (16x2 + 24xy + 9 y2) + (4x + 3y) h) x2 - 9 by x + 3

3. Find the quotient and remainder when (4x2-13x-21) is divided by (x-8) check your answer
by the rule of divisor algorithm. Also find the value of 4x2 - 13x - 21 when x = 2.

4. Area of a rectangle is (6x2 - 25x + 2) sq. units.
a) If the breadth of the rectangle is 6x - 7 units, what is its length?
b) Write the expression which represents the perimeter of the rectangle.
c) If x = 10m, find the actual length, breadth and area of the rectangle.

5. If Rs. (x2 + 6x - 16) is divided equally among (x - 2) people, how much amount will
get by each people ? Find it. If x = 15, find the total amount, the number of people
and the amount by each people.

Prime Mathematics Book - 7 267

14.1.6 Geometrical concept and Application of (a ± b)2.

1. Geometrical concept of (a+b)2.

First we draw a square ABCD of a side ‘a’ unit. The
sides AB and AD of the square ABCD are increased
by the length ‘b’ unit upto the points E and H
respectively. Now, through the point E and H, two
straight lines EG and HG of the length (a + b) each
are drawn perpendicularly at G. Then a greater
square AE GH of the length (a + b) unit is formed.

In the figure alongside, square ABCD, square CFGI,
rectangle DHIC and rectangle BCFE are formed.

From the above figure,
Area of the square AEGH = Area of square ABCD +
area of square CFGI + area of rectangle DHIC + area of rectangle BCFE

or, (a + b)2 = a 2 + b 2 + a × b + a × b [ Area of a square = (side)2 and Area

of a rectangle = length × breadth]

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

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

In the above figure, the length of the square AEGH = AE = AB + BE = a + b

Now, area of the square AEGH = (AE)2
= (a + b)2
= (a + b) × (a + b)
= a(a + b) + b(a + b)
= a2 + ab + b2 + ab
= a2 + 2ab + b2

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

2. Geometric concept of (a-b)2:

First we draw a square ABCD of the length of a side ‘a’ unit.
\ AB = AD = DC = BC = ‘a’ unit.
All the four sides AB, AD, DC and BC of the
square ABCD are decreased by the length ‘b’ unit
respectively.
\DR = BP = CQ = CT = ‘b’ unit
R and T, P and Q are joined.
Then, AP = AR = RS = PS = DQ = BT = (a - b) unit.
In the figure alongside, square APSR, square CTSQ,
rectangle BPST and rectangle RSQD are formed.

268 Algebra

Now,
Area of the square ABCD = area of square APSR + area of square CTSQ +

area of rectangle BPST + area of rectangle RSQD

or, (AB)2 = (AP)2 + (CQ)2 + PS × PB + DQ × DR
or, (a)2 = (a - b)2 + b2 + (a - b) × b + (a - b) × b
or, a2 = (a - b)2 + b2 + ab - b2 + ab -b2
or, a2 = (a - b)2 + 2ab - b2
or, a2 -2ab + b2 = (a - b)2
\(a - b)2 = a2 - 2ab + b2

To Check,
(a - b)2 = (a - b) × (a - b)

= a(a - b) - b(a - b)
= a2 - ab - ab + b2
\(a-b)2 = a2 - 2ab + b2
From the above Geometrical concept,

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

From the above result, = a2 - 2ab + b2 + 4ab = (a - b)2 + 4ab
(a + b)2 = a2 + 2ab + b2 = a2 + 2ab + b2 - 4ab = (a + b)2 - 4ab
(a - b)2 = a2 - 2ab + b2 = (a - b)2 + 4ab

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

Example 1. Find the square of (x + 4) by using the formula and without using the
formula. Also find the value by using geometrical concept figure.

Solution:

The square of (x + 4) by using the formula,
square of (x + 4) = (x + 4)2

= (x)2 + 2.x.4 + (4)2 [(a + b)2 = a2 + 2ab + b2]
= x2 + 8x + 16

The square of (x + 4) without using the formula, 269
Square of (x + 4) = (x + 4)2

=(x + 4) × (x + 4)
= x(x + 4) + 4(x + 4)
= x2 + 4x + 4x + 16
= x2 + 8x + 16

Prime Mathematics Book - 7

The square of (x + 4) by using geometrical concept figure,
square of x + 4 = (x + 4)2
= x2 + x × 1 + x × 1 + x × 1+ x × 1 + x × 1 + x × 1 + x × 1+ x

× 1+ 1 × 1 + 1 × 1 +1 x 1 +1 × 1 +1 × 1 +1 × 1 +1 × 1 +1 × 1
+1 × 1 +1 × 1 +1 × 1 +1 × 1 +1 × 1 +1 × 1 +1 × 1 +1 × 1

= x2 + x + x + x + x + x + x + x + x +1+1+1+1+1+1+1+1+1+1+1

+1+1+1+1+1

= x2 + 8x + 16

Example 2. Find the square of (3a + 2b) and (2a - 3b2).

Solution: = (3a + 2b)2
Here, = (3a)2 + 2x3a × 2b + (2b)2
the square of (3a + 2b) = 9a2 + 12ab + 4b2
the square of (2a - 3b2) = (2a - 3b2)2

= (2a)2 - 2 × 2a × 3b2+(3b2)2
= 4a2 - 12ab2 + 9b4

Example 3. Find the square of (x + b + c)by

a) using the formula

Solution : b) without using the formula
Here,
supposing x + b = a and c = b

a) the square of (x + b + c) = (x + b + c)2 (a + b)2 = a2 + 2ab + b2

= {(x +b) + c}2

= (x + b)2 + 2(x + b).c + c2

= x2 + 2.x.b + b2 + 2cx + 2bc + c2

= x2 + 2bx + b2 + 2cx + 2bc + c2

= x2 + b2 + c2 + 2bx + 2cx + 2bc

b) the square of (x+b+c) = (x+b+c)2

= (x+b+c) × (x+b+c)

= x(x+b+c) + b(x+b+c) + c(x+b+c)
= x2 +bx + cx + bx + b2 + bc + cx + bc + c2
= x2 + b2 + c2 + 2bx + 2cx + 2bc

270 Algebra

( ) ( )Example 4 : Find the square of x3 + 1 and x+ 1 .

Solution : x x

Here, 1 ( )= 1 2

the square of x3+ x x3+ x

( )= (x3)2 + 2.x3. 1 + 12

x x

= x6 + 2x2 + 1
( ) ( ) ( )the square of 1 2
1 x2
x- = x- x2
x2
1 12
= (x)2 - +
2.x.x2 x2

= x2 - 2 + 1

x x4

Example 5 : If x + 1 = 6, Find the value of x2 + 1 .
Solution :
x x2 Alternative Method

Here, 1
1 x +
x+ x =6 x =6
2
( ) ( )Square of 1 1 x2 + 1 = ?
x + = x + = (6)2 = 36 x2
Now, x x
12
( )x2 + 1 ( )Now, 1 2
= (x)2 + x 1
x2 x + x - 2.x.
x
( )= (x)2 + 2. x. 1 1 2

x + x -2 = (6)2 - 2

( )= + 1 2

x x -2 = 36 - 2

= 36 - 2

= 34 = 34

( ) ( )Example 6. If m - 1 1 2 m2 + 1
m m m2
= 8 , Find the values of m+ and .

Solution:

Here, m- 1 =8
m
( )= 2
1 1
square of m - m m - m = (8)2 = 64

Prime Mathematics Book - 7 271

Now, ( )m +

1 2 ( )= 1 2 1 [ (a+b)2 = (a-b)2 + 4ab]
m m m
m- + 4m.

= 64 + 4

1 = 68
Again, m - m =8

squaring on the both sides,
( )m -
1 2 = (8)2
m

( )or, 1 12
(m)2-2.m. m+ m = 64

or, (m)2 - 2 + 1 = 64
m2

or, m2 + 1 = 64 + 2
m2

\ m2 + 1 = 66
m2

Example 7. If a + 1 = 11, Prove that:
a
( )(i) 1 2
1 a
a2 + a2 = 119 (ii) a - = 117

Solution:

Here, a+ 1 = 11
a
1
(i) To prove: a2 + a2 = 119

Now, a+ 1 = 11
a

Squaring on the both sides,

( )a + 1 2 = (11)2
a

( )or, 1 12
(a)2 + 2. a. a + a = 121
= 121
or, a2 +2+ 1
a2

272 Algebra

1
or, a2 + a2 = 121 - 2

1
or, a2 + a2 = 119 proved.

( )(ii) To prove:1 2
a
a- = 117

Now,

( ) ( )a -1 2 1 21
a a - 4.a. a
= a + [\ (a-b)2 = (a+b)2-4ab]

= (11)2 - 4 = 121 - 4 = 117

( ) 1 2
a
\ a- = 117 proved

Example 8. Simplify: (m + 2n)2 + 2(m + 2n)(m - 2n) + (m - 2n)2

Solution:
Here, (m + 2n)2 + 2(m + 2n) (m - 2n) + (m - 2n)2

= (m)2 + 2.m.2n + (2n)2 + 2(m2 - 2mn + 2mn - 4n2) + (m)2 - 2.m.2n + (2n)2
= m2 + 4mn + 4n2 + 2m2 - 8n2 + m2 - 4mn + 4n2
= 4m2

Exercise 14.1.6

1. Find the square of the following by using the formula and without using the
formula.

a) x+2 b) m-3 c) a-5 d) 2a+3b
e) x2+2y f) 4m-3n2 h) 3a2-2y3
g) x2y-2y2

2. Find the square of the following by using the geometrical concept figure.

a) a+3 b) x-1 c) x+2 d) 2p+3q

3. Find the square of c) x2y+y2z
f) 3x-2y+5p
a) m2- 3n b) xy+ab
d) a+2b+c e) x-a+b

4. Find the square of 1 1
1 1
a) x2 + b) a2 + a2 c) a3 + a d) 3p3 +
x2 6p3

Prime Mathematics Book - 7 273

( )5. 1 1 1 2
a of and a 2+ a+ a
a) If a + = 5, find the value a2 and

( )b) 1 p 2+ 1 1 2
p p
If p + = 4, find the value of and p2 and p -

( )6. 1 1 2 m2 1
m m
a) If m - = 10, find the value of and m + and + m2

( )b) If n +1 13, find n2 + 1 1 2
n = the value of and and n
n2 n -

( )7. a) If x -
1 = 5, prove that: (i) x2 + 1 = 27 (ii) 12
x
x2 x + x = 29

( )b) If b +
1 = 14, prove that : (i) b2 + 1 = 194 (ii) 12
b
b2 b - b =192

c) If a2 - 6a + 1 = 0 then find the value of a2 + 1

a2

8. Simplify:

a) (x + y)2 + (x - y)2

b) (c - b)2 - (a + b)2

c) (3c + 2d)2 + (5c - 6d)2

d) 2(x - 3)2 - 2(x - 5)(x + 6)

274 Algebra

14.2 Indices Estimated Period - 6

Introduction

Study the following pattern of the product of the same factor.

Factors Short forms Way of reading

3 × 3 = 9 (product of two 3’s) 32 = 9 Index 2 of 3

3 × 3 × 3 = 27 (product of three 3’s) 33 = 27 Index 3 of 3

3 × 3 × 3 × 3 = 81(product of four 3’s) 34 = 81 Index 4 of 3

3 × 3 × 3 × 3 × 3 = 243(product of five 3’s) 35 = 243 Index 5 of 3
3×3×........ n times(product of n 3’s) 3n = 3n Index n of 3

x×x×x×........ n times(product of n x’s) xn Index n of x

In the above table, in 34, 3 is the base and 4 is the index. Similarly, in xn, x is the base

and n is the index.
The Students can understand easily as

34 Index(exponent)
Base

The process of multiplication of the same factor continuously is represented by using
exponent of the factor.
Therefore, an is said to be in an exponential form of the factor a which multiply itself
n times.
In an, a is the base and n is the index. an is called the power. The base ‘a’ is either
positive or fraction.

Laws of Indices
Law 1 : Multiplication of power having the same Base:
We obtain a law for writing the product of two or more than two quantities in an
exponential form with the same base and a single exponent. For example,

22 × 23 = (2 × 2) × (2 × 2 × 2) = 2 × 2 × 2 × 2 × 2 = 25 = 22+3
product of
product of three 2’s 2+3=5
two 2’s product of five 2’s

Similarly,

53 × 54

= (5 × 5 × 5) × (5 × 5 × 5 × 5) = 5 × 5 × 5 × 5 × 5 × 5 × 5 = 57 = 53+4

product of product of 3+4=7
three 5’s four 5’s product of seven 5’s

Again, xm×xn = (x × x × x...... m times) × (x × x × x...... n times)
= x × x × x........ m times × x × x × x........ n times
= x × x × x........ (m+n) times
= xm+n

\ xm×xn = xm+n, where x g 0 and m,n are integers.

Prime Mathematics Book - 7 275

Thus, the multiplication of two powers having the same base is the base raised to the
sum of exponents.

Law 2: Division of powers having the same Base:

We obtain a rule for writing the division of two quantities in an exponential form with
the same base and a single exponent.
For example,

34+32 = 34 = 3×3×3×3 = 3 × 3 = 32 = 34-2
32 3×3

Similarly, 55+52 = 55 = 5×5×5×5×5 = 5 × 5 × 5 =53 = 55-2
52 5×5

Thus, 2m+2n = 2m = 2m-n
2n
In general, xm + xn = xm-n, where x ≠ 0 and m, n are integers. Thus, the division
of two powers having the same base is the base raised to the difference of
exponents.

Law 3. Meaning of zero exponent:

Study the following examples,

32+32 = 32 = 3 × 3 =1
32 3 × 3

By using the law - 2 of indices,

32+32 = 32 - 2 = 30 = 1

We know that, 2 - 2 = 0

Then, 32-2 = 30
32
or, 32 = 30

or, 3 × 3 = 30
3 × 3
or, 1 = 30

\ 30 = 1

From the above example, we can say that,

50 = 1, (100)0 = 1, (2000)0=1, x0 = 1 etc
Thus, any base raised to power zero is equal to 1, so, x0 = 1, where x ≠ 0. This is called

the law of zero index.

276 Algebra

Some other laws.

(i) (xm)n = xmn

(ii) x-m = 1

xm
(iii) xm×ym = (xy)m

(iv) (x/y)n = xn .
yn

Example 1. Express into the exponential form:

(a) 7 × 7 × 7 × 7 (b) (-5)×(-5)×(-5)×(-5)×(-5)
(c) (3x) × (3x) × (3x) × (3x)

Solution
(a) 7 × 7 × 7 × 7 = 7 1+1+1+1 = 74
(b) (-5)×(-5)×(-5)×(-5)×(-5) = (-5)5

(c) (3x) × (3x) × (3x) × (3x) = (3x)4

Example 2. Find the product of

(a) 23 × 24 (b) (3a)2 × (4b)3 (c) 32 × 3-3 × 34

Solution:
(a) 23 × 24 = 23+4 = 27 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128
(b) (3a)2 × (4b)3 = (3a × 3a) × (4b × 4b × 4b)

= 9a2 × 64b3 = 576a2b3
(c) 32 × 3-3× 34 = 32-3+4 = 33 = 3 × 3 × 3 = 27

Example 3. Express the following as the exponent form. 2 648
2 324
(a) 400 (b) 648 2 162
3 81
Solution : 3 27
(a) 400 = 4 × 100 = 2 × 2 × 10 × 10 39

=2×2×2×5×2×5 3
= 24 × 52
(b) 648 = 2 × 2 × 2 × 3 × 3 × 3× 3 = 23 × 34

Exmple 4. Express 60000 as the power of 10.
Solution
Here,
60000 = 6 × 10000 = 6 × 10 × 10 × 10 × 10

= 6 × 104

Prime Mathematics Book - 7 277

Express 5. Express into the simplest form.

(a) b4 × b5 × b2 (b) (2a)2 × (3a)3 × (4a)2

Solution:
(a) b4 × b5 × b2 = b × b × b × b × b × b × b × b × b × b × b = b11
(b) (2a)2 × (3a)3 × (4a)2 = (2a × 2a) × (3a × 3a × 3a) × (4a × 4a)

= 4a2 × 27a3 × 16a2
= 4 × 27 × 16 × a2 × a3 × a2
= 1728 a2+3+2 = 1728 a7

Example 6 : Simplify the following by using the law of Indices

(a) 7x5 × 2x3 × 3x-2 b) a4 × a5 c) 4y6 × 7y5
a3 14y11

Solution :

Here,

(a) 7x5 × 2x3 × 3x-2
= 7 × 2 × 3 × x5+3-2 [ \xm × xn = xm+n]
= 42x6

(b) a4 × a5 = a4+5 = a9 + a3 = a9-3 = a6
a3 a3

(c) 4y6 × 7y5 = 4 × 7 × y6+5
14y11 14y11

= 28 2 (y11 + y11)
14

= 2y11-11 = 2y0 = 2 × 1 = 2

Example 7: If a = 5 and b = 7, find the value of 4a2- 4ab+b2 by using the law of
2a-b
indices.

Soultion:

Here, 4a2- 4ab+b2
2a-b

= (2a)2-2 × 2 × a × b+(b)2
2a-b

= (2a-b)2 = (2a-b)2-1 = 2a — b = 2 × 5 — 7 = 10 — 7 = 3
2a-b

278 Algebra

Example 8 : Simplify: x( )a+b a-b × x( )b+c b-c × x( )c-a c+a

Soultion:

Here,

x x x( )a+b a-b × ( )b+c b-c × ( )c-a c+a

x x x x x= ×(a+b)(a-b) ×(b+c)(b-c) [(c-a)(c+a) . .. ( m)n= mn]

b2-c2 c2-a2
x x x= × ×a2-b2

x x x x= [ . ..a2-b2+b2-c2+c2-a2 m × n = m+n]

= x0 =1

Exercise 14.2
1. Express the following in the index form.

a) 2 × 2 × 2 b) (-4) × (-4) × (-4) × (-4) c) (2a) × (2a) × (2a) × (2a) (2a)

d) 24 × 23 × 2-1 e) (3x)3 × (3x)5 × (3x)7 ( ) ( ) ( ) ( )(f)-13×-124× -126× - 1 5
2 2

2. Express the following as the power of 10.

a) 200 b) 5000 c)160000 d) 560000 e) 17000000

3. Find the value of:

a) 3 × 102 b) 6 × 103 c) 23 × 104 d) 12 × (-10)5 e) 7 × 20

4. Express the following numbers as the power form. 64
125
a) 800 b) 32 c) 432 d) 27000 e) 1728 f)4500 g)

5. Expresse the following into single base.

a) 34 × 32 b) x6 × x2 × x-4 c) 56 ÷ 52 d) (11)9 + (11)4

4a 7 4a 4 4a 3 f)(5b)7 + (3b)5

( ) ( ) ( )e) 5b × 5b × 5b

g) xb × xc h) (3b)7× (3b)5
xd (3b)2

Prime Mathematics Book - 7 279

6. What is the value of

a) yo b)4o c) (3q)o ( )2 × xo ( )2 o

d) 3 e) - 3

( )f)(105)o - 4xc o 3x6 × 2x4 7 × a4 × a9 × a-13
3y x10 9
g) h) i)

7. Simplify.

a) 2a5 × 7a3 × 3a-4 b) (2x)2 ×(3x)5 × (4x)2 c) 3a7 × 2a4 d) x6 × x5
5a3 x11

e) (3m)3 × (4m)2 f) 24x2y3z2 g) (7xy)2 × x3y2
18m 8x2yz2 1
3 x2y2 × (2xy)2

8. Find the value of:

a) (x + 3)a-2, where a=5 and x = 2

b) a2+2ab+b2 , where a=7 and b=-2
a+b

c) x2-4xy+4y2, where x = 3 and y = -3
x - 2y

d) m4 × m3 ×m5 , where m=3
m9

9. Simplify:

a) xp-q × xq-r × xr-p b) (ax)y-z × (ay)z-x ×(az)x-y

c) p2x-y ×p2y-x x x xd) ( )a-b c × ( )b-c a × ( )c-a b
px-y

e) (am+n)m-n × (an+p)n-p × (ap+m)p-m

f) (x-a)b x (x-b)c x (x-c)a ( ) ( ) ( )g)xb3× xc 3× xa 3
x x xx x-ab -bc -ca xc xa xb

( ) ( ) ( )(h)xaa+b×xb b+c× xc c+a
xb xc xa

280 Algebra

14.3 Equation, Inequality and Line - Graph Estimated Period - 10
14.3.1 Linear equation in one variable.

Let us consider an open mathematical statement containing equal sign ‘=’ be x+7 = 13.
In x+7=13, x is a variable and the exponent of x is 1. x+7 =13 is an equation. So x+7 = 13

is called a linear equation in one variable.
From the above example, Linear equation in one variable is defined as an open
mathematical statement containing equal sign (=) having one variable with exponent
1 only.

The equation x+7=13 is true only for x=6. Therefore, x=6 is called the solution of
the equation x+7=13. The solution of the equation means finding the value of the

variable which contain in the equation. The process of finding the value of the variable
containing in an equation is called solving the equation.

Solution of Linear Equations of one variable.

For the solution of Linear equation of one variable, we isolate variables by using the
rules of trichotomy (class VI reference).

Consider an equation, 2x+5=11.

Then, 2x+5=11

or, 2x+5 -5 =11-5 [ if a = b, then a - c = b - c]

or 2x=6 3 b
c
or, 2x = 6 [ if a = b, then a = ]
2 c
2
\ x = 3. which is the required solution for the given equation.

Example 1: Solve and check the answer.

(a) 5x - 3 = 22 (b) 3x + 8 = x + 12

Solution

(a) Here,

5x-3 = 22 [add 3 in both side] To check the answer
or, 5x- 3 + 3 = 22+3
5x - 3 = 22
or, 5x = 25
or, 5 × 5 - 3 = 22
or, 5x = 255 [divide both side by 5] or, 25 - 3 = 22
5 5 or, 22 = 22, which is true.

\ x = 5

Prime Mathematics Book - 7 281

(b) Here, To check the answer

3x+8 = x+12 3x + 8 = x + 12
or, 3x+8-8 = x+12-8[subtract 8 in both sides]
or, 3x = x+4 or, 3 × 2 + 8 = 2 + 12
or, 3x-x = x+4- x[subtract x in both side] or, 6 + 8 = 14
or, 2x = 4 or, 14 = 14, which is true.

or, 2x= 4 2 [ divide both sides by 2]
2 2

\ x = 2

Example 2: Solve and check the answer 7y- 1 = y + 3
3 2 4

Solution:

Here,

7y - 1 = y + 3 To check the answer
3 4
2

or, 7y - 1 + 1 = y + 3 + 1 [add 1 in both sides] 7y - 1 = y + 3
3 3 2 4 3 3 3 2 4

or, 7y = y + 9+4 or, 7 × 1 - 1 = 1 × 1 + 3
12 6 3 6 2 4
2

or, 7y - y = y + 13 - y [Subtract y in both sideso]r, 7 - 1 = 1 + 3
2 12 6 3 12 4
1 2 2 2
7 - 2 1+9
14y-y 13 or, 6 = 5 12
or, = 12 or, 5
2 6 10 6
= 12
or, 13y = 13
12 5 5
12 6 6

or, = , which is true.

or, 13y × 2 = 13 2
13 ×
12 12 6 13

\y= 1
6

282 Algebra

Exercise 14.3.1

1. Identify weather the following statements are true or false.

a) In x+3=7, x is a variable.
b) In y-4=6, 6 and -4 are variables.
c) 2x+3=9 is an equation with two variables.

d) -6+z=4 is a mathematical open statement in one variable.

2. Write the answer of the following questions.
a) Define an equation in one variable with an example.

b) What is the power of y in 3y+2=14?
c) What is the value of x such that x+7=19 is a true statement?

3. Solve the following equations and check your answer:

a) x-5=6 b) y+9=21 c) 6x+2=9x+14
3
d) 5z-3=22 1 e)2a+8=3a+7 x f) 5 x = 6 + 7x
2 + 15.5
g) 4.5a - 3 = a + 1 h) 6.5x - 3 = 3

4. Solve the following equations. 3 5
a) 2(x-3) = -x + 12 b) 5(x-2) = 3(7-x) - 7 c) 17y - 5 = 3

d) x - 1 = x - 1 e) 3m-1 = 2m-1 + 3 f) 10x - 2 = x 1
2 3 4 3 +4
3 6 4 3 2

g) 4x-1 - 4x-2 = 3 12 h) 0.6(2n-9) = 3.4 - 2 (3n-1)

2 3

14.3.2 Verbal problems of Linear Equation in one variable:

We use an equation with one variable to solve the verbal (word) problems stating the
relationship between numbers which is known and variable which is unknown. We can
follow the following points (steps) in order to solve the verbal problems related to
equation in one variable.

(i) Read the given problem very well and identify the given item and unknown item
which we need to be found.

(ii) Supposing the unknown quantity as a variable like x, y, z etc. and translate the
statement of the problem into the mathematical statement.

(iii) The mathematical statement is an equation of one variable.
(iv) Solve the equation for unknown.
(v) At last check the value of the variable for the condition in the problems.
For the above steps, study and learn the following examples.

Prime Mathematics Book - 7 283

Example 1: There are 73 students in a class 7 of a school. If the number of girls is
15 less than the number of boys, write an equation to represent the
students and also find the number of boys and girls .

Solution.

Here, total students = 73.

Suppose, the number of boys be x.
Then, the number of girls is (x-15).

Now, Number of boys + number of girls = total students

or, x + x - 15 = 73
or, 2x - 15 = 73, which is the required equation.

To solve this equation, we write

2x - 15 + 15 = 73 + 15 [Add 15 in the both sides]
[Divide both sides by 2]
or, 2x = 88
8844
or, 2x = 2

2

\ x = 44

Hence, the number of boys = x = 44

and the number of girls = x - 15 = 44 - 15 = 29

Example 2: A number is thrice the other and their sum is 120.
(a) Write a linear equation to represent the given statement.
(b) Solve the equation to find the numbers.

Solution:
Here, Let a number be x. Then, thrice this number is 3x
(a) The sum of these two numbers is 120.

\ x + 3x = 120
or, 4x = 120, which is the required equation.

(b) To solve this equation, we write

4x = 12030 [ divide both sides by 4]
4
4

or, x = 30 = x = 30
= 3x = 3 × 30 = 90
Hence, one number
other number

284 Algebra

Example 3: The sum of two consecutive numbers is 43. Find the numbers.

Solution:

Here, Let a number be x. Then, next number is x +1.
The sum of these two numbers = x +(x+1).

According to question,

the sum of the numbers = 43.

\ x + (x+1) = 43.

or, 2x + 1 - 1 = 43 - 1

or, 2x = 42 21

or, 2x = 42
2
2

\ x = 21

Hence,

One number = x = 21

Other number = x + 1 = 21 + 1 = 22

Example 4: The length of a rectangle is 3 cm more than its breadth. If the
perimeter of the rectangle is 34cm,

(a) write down an equation to represent the length and breadth of the rectangle.
(b) find the length and breadth of the rectangle.
(c) also find the area of the rectangle.

Solution:
Here,

(a) Let the breadth of a rectangle be x cm.
Then, the length of the rectangle is (x+3)cm.

Now, perimeter of the rectangle = 34 cm.
or, 2(length + breadth) = 34

or, 2(x + 3 + x) = 34
or, 2(2x +3) = 34
or, 4x + 6 = 34, which is the required equation.

Prime Mathematics Book - 7 285

(b) To solve this equation, we write

4x + 6 - 6 = 34 - 6

or, 4x = 28

or, 4x = 287
4
4
\ x =7

\ length (l) = x + 3 = 7 + 3 = 10 cm.

breadth (b) = x = 7cm.

(c) Area of the rectangle (A) = l × b = 10 cm × 7cm= 70 cm2

Exercise 14.3.2
1. There are 38 students in class six of a school. If the number of girls is 6 more than

the number of boys,
a) write down an equation representing the students.
b) find the number of boys and girls.
2. There are 356 students in a school. If the number of boys is 84 more than the
number of girls, find the number of boys and girls.

3. A number is twice the other and their sum is 87.
a) Form a linear equation in one variable.
b) Find the numbers.

4. A number is 3 more than twice the other and their sum is 39. Find the numbers.

5. The sum of two consecutive numbers is 57. Find the numbers.

6. The sum of three consecutive numbers is 87. Find the numbers.

7. The breadth of a rectangular garden is 3m less than its length. If the perimeter of
the garden is 54m,
a) write down an equation for its perimeter.
b) find the length of breadth of the garden.
c) find the area of the garden.

8. The perimeter of a rectangular book is 100cm. If the length is 8cm more than its
breadth, find the area of the book.

9. Mohan bought two packet of chocolates to distributes among his friends on the
occasion of his birthday. The number of chocolates in one packet is three times
the number of chocolates in others of packet. If the total number of chocolates of
both packets is 120,
a) write down an equation representing the total chocolates.
b) find the number of chocolates in each packet.

286 Algebra

14.3.3 Representing of Inequality in a number line.

Introduction of Inequality.

Let us consider two integers a and b.
What are the mathematical relation between a and b ? Either a and b are equal (a=b)

or a and b are not equal (a≠b).

If a and b are not equal ( a≠b), then either a is greater than b (a>b) or a is less than

b(a<b).

Therefore, a open mathematical sentence which shows the unequal relation
between two or more quantities is called an inequality. In the inequality we use the

symbols of trichotomy like grater than (>), less than (<), greater than or equal to (≥ )
and less than or equal to (≤).

For example, x<3 means x is less than 3. y > 2 means y is greater than 2. x m-2 means
the value of x is greater than or equal to -2.

Solutions of inequalities:
Consider an inequality as x > 3, where x is an integer. When we take the values of
x are 4, 5, 6 ,7 etc, then x > 3 is a true sentence. Therefore, the solution of this

inequality is 4,5,6,7 etc and the solution set of this inequality is {4, 5, 6,7,.....} .

The solution of the inequality x > 3, where x being an integer, can be shown on a

number line as follows.

Let us take an example of an inequality x + 2 > 7.
Then adding -2 to both sides of x + 2 > 7. We get

x+2-2>7-2
or x > 5
x > 5 means the values of x is greater than 5. Therefore, the value of x are 6, 7, 8 etc.

and the solution set is {6,7,8, .......}. The solution of this inequality in the number line
as follows,

Important statements:

If a and b are two integers where a>b and c is another integer, then
a+c>b+c (addition axim)
a-c>b-c (subtraction axim)

Prime Mathematics Book - 7 287

}ac>bc for c>0 (multiplicative axiom)

ac<bc for c<0

a > b for c>0 }(division axiom)
c c
a b
c < c for c<0

Note: If we multiply or divide both sides of the inequality containing trichotomy
symbols like <, >, ≤ or ≥ by a negative integer, then the symbol of trichotomy will

be reversed.

Example 1. Solve x - 2 ≥ 3 and show the solution in a number line.

Solution

Here, x - 2 ≥ 3
or, x - 2 + 2 ≥ 3 + 2 (Adding 2 on both sides)
\ x ≥ 5
x ≥ 5 means the values of x are 5, 6, 7, 8 etc and the solution set of x ≥ 5 is {5, 6, 7,

8, ...}. The solution set of this inequity in the number line as shown.

Example 2. Solve 3x + 2 ≤ 8 and show the solution in a number line.
Solution:
Here, 3x + 2 ≤ 8
or, 3x + 2 - 2 ≤ 8 - 2 (Subtracting 2 on both side)
or, 3x ≤ 6
3x 6
or, ≤ 3 (Divide both side by 3)
3

\x≤2

The solution set of x ≤ 2 is {2, 1, 0, -1, -2, ......}. This solution shows in the number

line in the adjoining figure.

Example 3: If 2 is subtracted from four times a number, the result will be less than
or equal to 10. Write down an inequality to express the statement and
solve it. Also show the solution in a number line.
Solution:
Here, Let the number be x. Then, four times the number means 4x. 2 is subtracted
from 4x, we get 4x-2.
Now, 4x-2 ≤ 10, which is the required inequality.
or, 4x - 2 + 2 ≤ 10 + 2
or, 4x ≤ 12
4x 12
or ≤ 4 \x≤3
4
The solution set of x ≤ 3 is {3, 2, 1, 0, -1,......}. This solution shows in the

number line as follows.

288 Algebra

Exercise 14.3.3
1. Fill in the blanks by using the suitable symbols =, > or < .

a) -2 -1 b) 0 -3 c) 2 × (-4) 2×(-3)

d) -4 ×(-2) 4×2 e) 5×(-3) 4 ×(-3)

2. Identify whether each of the following statements are true or false by using the
rules of trichotomy. If 2, -3 and 5 are integers, then

a) 2 + (-3) = 5 + (-3) b) 2×(-3) = 5 × (-3) c) 5 + (-3) > 2 + (-3)

d) 2 + (-3) < 5 + (-3) e) 2 -(-3) < 5 -(-3) f) 2 × (-3) > 5 × (-3)

3. Write the inequalities represented by the following number lines.
a) b)

c) d)

4. Draw separate number lines to show each of the following inequalities.

a) x > 2 b) x ≤ -1 c) x ≥ - 2

d) -3 ≤ x < 5 e) -2 ≤ x ≤ 4 f) 3 ≥ x > -2

5. Solve the following inequalities and show the solution in the number line.

a) x + 3 > 5 b) 2x - 4 < 6 c) 2x + 6 ≤ 14

d) 2x + 5 > -1 e) 3x + 2 ≥ x + 6 f) 15 - 8y ≤ 2y - 15

6. Write down an inequality to express the given statement and solve them. Also
show the solution in the number lines.

a) If 3 is subtracted from twice of a number, the remainder will be less than 5.
b) If 5 is added to the twice of a number, then the result will be greater than or

equal to 9.
c) If a number is multiplied by 2 and added to 5, the result will be less than or

equal to 13.

Prime Mathematics Book - 7 289

14.3.4 Relation between variables in linear equations of Two
variables by a Function Machine.

We know that a flour grinding machine produce the flour
after grinding the grains which is shown in a diagram
alongside. The amount of grain put into the machine is the
input and the amount of the flour which we received from
the machine is the output .

Similarly, we can also make a machine in Mathematics.
That machine is called function machine. In the figure
alongside, we put the cards numbering 1,2,3 in the function
machine, then we receive the same cards numbering 3,4,5
by the processing of the machine from the other side of the
machine.

The numbers which we put inside the machine are called input.
So, in the above figure, the numbers 1,2,3 are called input.
The numbers which we receive after processing by the machine are called output. So,
in the above figure, the numbers 3,4,5 are called output.

Let us we discuss the process of the machine so that the input 1,2,3 are converted to
the output 3,4,5.

Input Process of the machine Output
1 +2 = 3
2 +2 = 4
3 +2 = 5

In the above discussion, 2 is added in all input numbers by the machine, then we
receive the output. So, this addition process is called the function of the machine.

We can use the symbol (variable) x and y for the input and output. Then, the above

function of the machine can be written as

y = x + 2, where x and y are variables.

The function of the machine can be shown by making table also.

The above process of the machine is shown in the following table.

Input (x) 1 2 3 y=x+2
Output (y) 3 4 5 13
2 54
The function of the machine can also show clearly the relation between 3

two set where input is denoted by the set x and output is denoted by set
y with the help of an arrow diagram as shown in the figure alongside.

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Example 1. What are the output in the adjoining function machine ?
Solution :
In the adjoining machine, input are 3, 4, 5. The processing
of the machine is multiple by 2. So, 3, 4, 5 are multiplied by
2, then the product are 6, 8, 10 respectively.

Hence, the output by the machine is 6, 8, 10.

Example 2. A function machine is given in the adjoining figure. The function of the machine
is to send the number outside after processing (+3) to each number put into it.

(a) Write down the set of numbers in output in the form of
a table.

(b) Write down an equation for the relation of output (y)
and input (x).

(c) Draw an arrow diagram to show the relation between x and y.

Solution:
The function machine gives a set of output after adding 3 to each input numbers is
{4,5,6,7,8}.

These are shown in the following table.

(a) Input (x) 1 234 5
8
Output (y) 4 5 6 7

(b) The relation between input (x) and output (y) is output = input + 3
\y =x+3

(c) We may draw an arrow diagram to show the relation between

input (x) and output (y).

Example 3 : Write down the relation between the input (x) and output (y) of the

process of a function machine for the adjoining arrow diagram.

Solution:
In the given arrow diagram, the output of input 2 is 8, the output of input 4 is 14 and
the output of input 5 is 17.
Let us see the relation between them.
8=2×3+2
14 = 4 × 3 + 2

17 = 5 × 3 + 2

All the input numbers are multiplied by 3 and added 2 in the product.

Therefore the relation between input (x) and output (y) is y = 3x + 2.

Prime Mathematics Book - 7 291

Exercise 14.3.4
1. The input numbers from 0 to 5 are used in each of the following function machines.

Express the values of input and output in a table for each machine.
a) b) c)

3

d) e) f)

2. For each of the function machines in Q. No.1, the numbers in input and output are
denoted by x and y. Write down the relation between x an y in the mathematical
equation.

3. Draw an arrow diagram to show the relation of input and output of Q. No.1, for
each function machine.

4. Write down the relation between the input (x) and output (y) of the process of
each function machine from the following arrow diagram.

a) b) c)

5. In the table below, x show the number of cycles and y shows the number of
wheels.

Input (x) 1 2 34 5 6 7 ?
Output (y) 2 4 ? 8 10 ? ? 16

a) Complete the above table by putting the correct values of x and y.

b) Write down the relation between input (x) and output (y) in the
mathematical language.

c) Draw an arrow diagram to show the relation of x and y.

292 Algebra

6. Complete the following tables on the basis of the input and output.
(a)

Input 2 3 4 5 ? ?

Output 1 2 ? ? 8 11
(b)

Input 2 3 4 ? 8 10 ?

Output 7 10 13 19 ? ? 25
(c)

Input 4 5 ? ? 12 15 18 ?

Output 10 12 16 20 ? ? 38 44

7. a) Write down the relation between input and output in the mathematical
equation for each table of Q. No. 6

b) Draw an arrow diagram to show the relation of input and output for
each table of Q. No. 6.

14.3.5 Graph of Linear Equation of Two variables

We already discuss about a linear equation of two variables by a function
machine.
Let the processing of a function machine be addition with number 3. If the input
numbers in the machine are 1, 3 and 5, then the output after processing by the machine
are 4, 6 and 8.
We already discuss that the function of the machine can be
shown by making table also. So, the above process of the
function machine is shown on the following table.

Input (x) 135
Output(y) 468

The above relation between input (x) and output (y) is

output = input + 3

i.e. y = x + 3

The relation y = x + 3 is an equation of two variables x and y. In this equation, the
value of y depends on the value of x. So, we say that y is a dependent variable and x

is an independent variable.

In the relation y = x + 3, when the value of x is 1, then the value of y is 4. It means for
the value of y is 4, we first put the value of x is 1. Otherwise we don’t get the value
of y at first.

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