Sequence and Series
b) Draw a square having each side of length 20 cm on a chart
paper. Mark the mid-points of each side and join them to form
a new square. Also, mark the mid-points of each side of this
new square and join them to form another square. Continue this
process till to get the fourth square. Find the area of each square
and explain the sequence formed by these areas.
c) Assume that you were symptomatic and positive patient of COVID-19. You forgot
to cover your mouth when two of your friends came to visit you at home. They left
and on the next day, they also became positive with virus. Each infected friend in
turn spread the virus to two other friends on the following day and continued this
pattern. In this scenario, how can we determine the number of persons infected?
What are the possible ways to detect the infected persons? Form a sequence and
justify your answer.
d) Visit the available website and search the population of Nepal in B.S. 2078. Also,
find the annual growth rate of population. Estimate the projected population of
Nepal in B.S. 2088.
OBJECTIVE QUESTIONS
1. The fifth term of the sequence 2, 5, 8, 11, … is
(A) 11 (B) 14 (C) 15 (D) 19
2. Which term comes next in the sequence 1, 2, 4, 8,..?
(A) 8 (B) 12 (C) 16 (D) 20
3. The series associated to the sequence 1, 3, 5, 7 is
(A) 1 + 3 + 5 + 7 (B) –1 – 3 –5 –7 (C) 1 – 3 + 5 – 7 (D) –1+3–5+7
4. What will the 6th term of a sequence whose nth term is given by tn = {1 + (-1)n}?
(A) -1 (B) 0 (C) 1 (D) 2
5. The 8th term of the sequence 1, 1, 2, 3, 5, 8, 13, … is
(A) 21 (B) 18 (C) 16 (D) 34
6. Given a1= 1 , a2 = 3 and an = 2an-1 + an-2 then a3 is
(A) 1 (B) 4 (C) 5 (D) 7
7. The general term of the sequence 1, 4, 9, 16, 25, … is
(A) n2 (B) (n + 1)2 (C) (n2 – 1) (D) 2n
8. The first term of an arithmetic progression is unity and the common difference is 3.
Which of the following will be a term of this A.P.?
(A) 8 (B) 14 (C) 19 (D) 24
9. If 5 times of 5th term of an A.P. is equal to 8 times the 8th term, then the 13th term of the
A.P. is
A) 0 (B) 5 (C) 8 (D) 13
10. How many terms are there in the series 1 + 3 + 9 + … + 243?
(A) 5 (B) 6 (C) 7 (D) 8
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 147 Vedanta Excel in Mathematics - Book 9
Unit 9 Factorisation of Algebraic Expressions
9.1 Factorisation –Looking back
Classwork-Exercise
Let’s say and write the factors of the expressions as quickly as possible.
1. a) ax + ay = ……………. b) 3x – 12y = …………….
c) y2 – y = ……………. d) x2y + xy2 = …………….
2. a) a (x + y) + b (x + y) = ……………. b) x (a – b) – y (a – b) = …………….
c) 2p (p – x) – 3 (p – x) = ……………. d) y (y – 3) – (y – 3) = …………….
3. a) x2 – a2 = ……………. b) p2 – 4 = …………….
c) 9a3 – 16a = ……………. d) 100m3 – m = …………….
9.2 Factors and Factorisation - Review x2 x
In the given figure, x unit is the length of a square.
x
So, the area of the square = x2 sq. unit. x+2
Let, the length of the square is increased by 2 units.
Then, the length of the rectangle = (x + 2) units.
Now, the area of the rectangle = x × (x + 2) sq. units x2 2x x
= (x2 + 2x) sq. units
Here, x2 + 2x is the product of x and (x + 2). x2
Therefore, x and (x + 2) are the factors of the expression x2 + 2x.
Thus, factorisation is the process of expressing a polynomial as the product of two or
more polynomials.
When we factorise an expression, we write it as the product of its factors. The process
of getting the factors becomes easier if we apply the selected method of factorisation
for the particular type of expression. So, it is important and very much useful to know
about the types of expressions which are to be factorised.
a) Expression having a common factor in each of its term
Let's take an expression ax + ay. Here, both terms contain a common term, a.
In such an expression, the factor which is present in all terms of the expression
is taken out as common and each term of the expression should be divided by the
common factor to get another factor.
Vedanta Excel in Mathematics - Book 9 148 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Factorisation of Algebraic Expressions
Worked-out Examples
Example 1: Factorise (i) 4ab + 6ac (ii) 2x2y – 4xy2 + 6xy
(iii) 2a (x – y) + 7b (y – x) (iv) 3p2q (a – b) – 6pq2 (b – a)
Solution:
(i) 4ab + 6ac = 2a (2b + 3c)
(ii) 2x2y – 4xy2 + 6xy = 2xy (x – 2y + 3)
(iii) 2a (x – y) + 7b (y – x) = 2a (x – y) – 7b (x – y)
= (x – y) (2a – 7b)
(iii) 3p2q (a – b) – 6pq2 (b – a) = 3p2q(a – b) + 6pq2(a – b)
= 3pq (a – b) (p + 2q)
b) Expression having common factors in the groups of terms
Let’s take an expression ax + by – ay – bx. It can be regrouped as ax – ay – bx + by
(or ax – bx – ay + by). In ax – ay – bx + by, the group ax – ay has the common factor
a and the group – bx + by has the common factor b.
Then, ax – ay – bx + by = a (x – y) –b (x – y) = (x – y) (a – b)
Similarly, ax – bx – ay + by = x (a – b) –y (a – b) = (a – b) (x – y)
In this way, in such expressions, the terms are arranged in suitable groups such that
each group has a common factor.
Example 2: Factorise a) x2 – ax + ab – bx b) ac (b2 + 1) + b (a2 + c2)
Solution:
a) x2 – ax + ab – bx = x2 – ax – bx + ab = x (x – a) – b (x – a) = (x – a) (x – b)
b) ac (b2 + 1) + b (a2 + c2) = ab2c + ac + a2b + bc2
= ab2c + a2b + bc2 + ac
= ab (bc + a) + c (bc + a) = (bc + a) (ab + c)
c) Expression of the form ax2 + bx + c, where a ≠ 0.
x2 + 5x + 6, 2x2 + x – 28, etc. are the trinomial expressions of the form
ax2 + bx + c. To factorise such expressions, we need to find the numbers p and q such
that p + q = b and pq = ac. Then, the trinomial expression is expanded to four terms
and factorisation is performed by grouping.
Example 3: Factorise: a) x2 + 7x + 12 b) 2x2 – 5x + 2 c) 2x2 – x – 6
x2
Solution: 12 × 1 = 12 x
4 × 3 = 12
x2 x x x x x + 3
a) x2 + 7x + 12 = x2 + (4 + 3) x + 12 x
= x2 + 4x + 3x + 12 xxx 12 34
56 78
= x (x + 4) + 3 (x + 4) 9 101112
= (x + 4) (x + 3) x+4
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 149 Vedanta Excel in Mathematics - Book 9
Factorisation of Algebraic Expressions
2×2=4 2x 2x – 1
4×1=4 x2 x2
xx
b) 2x2 – 5x + 2 = 2x2 – (4 + 1) x + 2 xx x2 x2 x
= 2x2 – 4x – x + 2 x–2 x–22 xx xx 12
= 2x (x – 2) – 1 (x – 2)
= (x – 2) (2x – 1) x–22x – 1
x–2
x–2x–2
2 × 6 = 12 2x + 3 2x + 3
4 × 3 = 12 x2 x2 x x x
c) xx3 x(x–2) x(x–2)
2x2 – x – 6 = 2x2 – (4 – 3) x – 6 2 xx xx
= 2x2 – 4x + 3x – 6 1 23
= 2x (x – 2) +3 (x – 2) 4 56
= (x – 2) (2x + 3)
2x + 3
x(x–2) x(x–2) 3x–6
Example 4: Resolve into factors:
a) a2 – 3 + 2b2 b) 9 (x + y)2 + x + y – 8
Solution: b2 a2
a) a2 – 3 + 2b2 = a2 – 3. a . b + 2b2
b2 a2 b2 b a a2
= a2 – a . b – 2 ab . ab + 2b2
b2 b a a2
= a (ab – ab ) – 2b (ab – b ) Vedanta ICT Corner
b a a Please! Scan this QR code or
browse the link given below:
=( – b ) ( a – 2b )
a b a
b) Let, x + y = a https://www.geogebra.org/m/kthvykmx
Then, 9 (x + y)2 + x + y – 8 = 9a2 + a – 8
= 9a2 + 9a – 8a – 8
= 9a (a + 1) – 8 (a + 1)
= (a + 1) (9a – 8)
= (x + y + 1) [9 (x + y) – 8]
= (x + y + 1) (9x + 9y – 8)
EXERCISE 9.1
General section
1. In each of the following figures, write the polynomial as the product of its factors.
a) b) 2 c) 1 d) 3
xx x a
x1 x x2 a3
Vedanta Excel in Mathematics - Book 9 150 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Factorisation of Algebraic Expressions
2. Factorise: b) 4p2 – 6p c) 6a2b + 9ab2
a) 2ax + 4bx e) 6x3y2 + 9x2y3 – 3x2y2 f) 2x (a + b) + 3y (a + b)
d) 2px2 – 4px + 6p2x
i) 2t (t – 1) – t + 1
g) 3a (x – y) – (x – y) h) x (x + 2) + 3x + 6
3. In each of the following figures write the polynomial as the product of its factors.
a) x3 b) x3 c) x
x 2 x xx 4 3
2 x x
e) x f)
d) x1
x x
x
x x2 x2 2 1
3 3 3
2
Creative section
4. Resolve into factors:
a) ax + by + ay + bx b) pm – qn + pn – qm c) a2 + ab + ca + bc
d) mx2 + my2 – nx2 – ny2 e) xy – 2y + 3x – 6 f) x2 + 4x + 3x + 12
g) p2 – 8p – p + 8 h) 16x2 – 4x – 4x + 1 i) a2 – a (b + c) + bc
j) x2 – (y – 3) x – 3y k) pq (r2 + 1) – r (p2 + q2) l) y (x + z) + z (x + y) + y2 + z2
5. Factorise: b) x2 – 7x – 8 c) a2 – 27a + 180
a) x2 + 4x + 3
d) 2x2 + 7x + 6 e) 3p2 – 7p – 6 f) 2x2 + 3xy – 5y2
h) 9a3bx + 12a2b2x – 5ab3x
g) 3a2 – 16ab + 13b2 k) 2 (x + y)2 + 9 (x + y) + 7 i) 12 a2 + a – 20
b2 b
x2 3y2
j) y2 – 2 – x2 l) 3 (x – y)2 – 10 (x – y) + 8
6. a) The area of a rectangular field is (x2 + 8x + 15) sq. m.
(i) Find the length and breadth of the field.
(ii) Find the perimeter of the field.
b) The area of a rectangular plot of land is (x2 + 13x + 40) sq. m.
(i) Find the length and breadth of the land.
(ii) If the length and breadth of the land are reduced by 2/2 m respectively, find the
new area of the land.
c) A rectangular ground has area (2x2 + 11x + 12) sq. m. If the length of the ground
is decreased by 2 m and the breadth is increased by 2 m, find the new area of the
ground. (Take a longer side of the rectangles as length)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 151 Vedanta Excel in Mathematics - Book 9
Factorisation of Algebraic Expressions
Project work and activity section
7. a) Write three different expressions of your own each in the form of a3 + b3 and a3 – b3.
Then, factorise your expressions.
b) Write an expression in each of the following forms, then, factorise your expressions
(i) x2 + ax + b (ii) x2 – ax – b (iii) x2 + ax – b (iv) x2 – ax + b
c) Write an expression in each of the following forms, factorise your expressions.
(i) ax2 + (ii) ax2 – bx – c (iii) ax2 + bx – c (iv) ax2 – bx + c
d) Expression of the form (a + b)2 and (a – b)2
(i) Expression of the form (a + b) 2
Geometrical interpretation: D ab C
When the length a of the smaller square is increased by a
b, each side of the bigger square ABCD is (a + b) a2 ab a+b
Now, area of ABCD = a2 + ab + ab + b2 = (a + b) 2 b b
= a2 + 2ab + b2 A ab b2 B
By usual multiplication: b
(a + b)2 = (a + b) (a + b) = a (a + b) + b (a + b) a+b
= a2 + ab + ab + b2 = a2 + 2ab + b2
Facts to remember
1. (a + b) 2 = (a + b) (a + b) = (a2 + 2ab + b2)
2. a2 + b2 = (a + b) 2 – 2ab
(ii) Expression of the form (a – b) 2
Geometrical interpretation: D a–b b C
Let each side of bigger square ABCD be a. R
When the length a of the smaller square is decreased (a – b)2 b(a – b) a–b
by b, each side of the bigger square PQRD is (a – b) a
Now, area of ABCD = (a – b) 2 + b (a – b) + b (a – b) + b2 P b(a – b) Q b2 b
or, a2 = (a – b)2 + ab – b2 + ab – b2 + b2
or, a2 = (a – b) 2 + 2ab – b2 A a–b bB
or, (a – b2) = a2 – 2ab + b2
By usual multiplication:
(a – b)2 = (a – b) (a – b) = a (a – b) – b (a – b) = a2 – ab – ab + b2 = a2 – 2ab + b2
Facts to remember
1. (a – b) 2 = (a – b) (a – b) = (a2 – 2ab + b2)
2. a2 – b2 = (a – b) 2 + 2ab
Vedanta Excel in Mathematics - Book 9 152 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Factorisation of Algebraic Expressions
e) Expression having the difference of two squared terms (Expression of the
form a2 – b2)
Let’s take a square sheet of paper of length a units.
From a corner of the sheet, another square of length b units is cut out.
a aa ba ba
DC
a–b
ab a bb a–b
b b2
a a2 A a+b B
Area = a2 a–b Area = (a + b) (a – b)
Area of shaded region is = a2 – b2
Here, area of the rectangle ABCD is a2 – b2, which is the product of length (a + b) and
breadth (a – b)
∴ length × breadth = area of rectangle
i.e., (a + b) (a – b) = a2 – b2
Here, the expression a2 – b2 is the difference of two squared terms a2 and b2 and it is the
product of (a + b) and (a – b). So, (a + b) and (a – b) are the factors of a2 – b2. Thus, to
factorise an expression of the form a2 – b2, we should use the formula,
a2 – b2 = (a + b) (a – b)
Worked-out Examples
Example 1: Factorise a) 8x3y – 18xy3 b) 81ax5 – 16ax
Solution:
a) 8x3y – 18xy3 = 2xy (4x2 – 9y2) = 2xy[(2x)2 – (3y)2]
= 2xy (2x + 3y) (2x – 3y)
b) 81ax5 – 16ax = ax (81x4 – 16)
= ax [(9x2)2 – (4)2]
= ax (9x2 + 4) (9x2 – 4)
= ax (9x2 + 4) [(3x2) – 22]
= ax (9x2 + 4) (3x + 2) (3x – 2)
Example 2: Resolve into factors. a) 1 – 9 (a – b)2 b) a2 – b2 + 2b – 1
Solution:
a) 1 – 9 (a – b)2 = 12 – [3 (a – b)]2
= [1 + 3 (a – b)] [1 – 3 (a – b)]
= (1 + 3a – 3b) (1 – 3a + 3b)
b) a2 – b2 + 2b – 1 = a2 – (b2 – 2b + 1)
= a2 – (b – 1)2 Using (a – b)2 = a2 – 2ab + b2
= (a + b – 1) [a – (b – 1)]
= (a + b –1) (a – b + 1)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 153 Vedanta Excel in Mathematics - Book 9
Factorisation of Algebraic Expressions
Example 3: Factorise x2 – 6x – 40 + 14b – b2
Solution:
x2 – 6x – 40 + 14b – b2 = (x2 – 6x) – 40 + 14b – b2
= (x2 – 2.x.3 + 32 – 32) – 40 + 14b – b2
= (x – 3)2 – 9 – 40 + 14b – b2
= (x – 3)2 – (49 – 14b + b2)
= (x – 3)2 – (72 – 2.7.b + b2)
= (x – 3)2 – (7 – b)2 Using a2 – 2ab + b2 = (a – b)2
= (x – 3 + 7 – b) (x – 3 – 7 + b)
= (x – b + 4) (x + b – 10) Vedanta ICT Corner
Factorise (w2 – x2) (y2 – z2) – 4wxyz Please! Scan this QR code or
Example 4: browse the link given below:
Solution:
https://www.geogebra.org/m/cr273mr4
(w2 – x2) (y2 – z2) – 4wxyz = w2y2 – w2z2 – x2y2 + x2z2 – 4wxyz
= (wy)2 – 2.wxyz + (xz)2 – (wz)2 – 2wxyz – (xy)2
= (wy)2 – 2.wy.xz + (xz)2 – [(wz)2 + 2.wz.xy + (xy)2]
= (wy – xz)2 – (wz + xy)2
= (wy – xz + wz + xy) (wy – xz – wz – xy)
= (wy + wz + xy – xz) (wy – wz – xy – xz)
EXERCISE 9.2
General section
1. Factorise:
a) 9x2 – 4 b) 25a2b2 – 1 c) 48ax2 – 75ay2 d) x4 – y4
e) 16x5y – 81xy5 f) 625a4 – 256b4 g) 4 – (m – n)2 h) 1 – (a – b)2
i) 16 – 25(p – q)2 j) (2a – b)2 – (a – 2b)2 k) a2 + 2ab + b2 – c2 l) p2 – q2 – r2 – 2qr
m) a2 – b2 + 4b – 4 n) 16a4 – 4a2 – 4a – 1 o) ax2 – ay2 – x – y p) a2 – (a – b)x – b2
2. Let’s find the area of the shaded region using a2 – b2 = (a + b) (a – b).
10 cm
15 cm 18 m
a) b) c)
10 cm
15 cm 18 m 5m
2 cm 3 cm 5m
2 cm
21 m 3 cm f) 12 m
d) e) 25 ft
21 m 30 m
7m7m 9 ft 25 ft 12 m
9 ft 30 m
Vedanta Excel in Mathematics - Book 9 154 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Factorisation of Algebraic Expressions
Creative section
3. Factorise:
a) x2 + 6x + 5 – 4y – y2 b) a2 – 10a + 16 - 6b – b2
c) p2 – 12p – 28 + 16q – q2 d) x4 + 8x2 – 65 + 18y – y2
e) 9a2 – 30a + 24 – 8x – 16x2 f) 625y2 + 400y – 36 + 20z – z2
g) 16p2 – 72pq + 80q2 – 6qr – 9r2 h) 25x2 – 20xy – 21y2 + 10yz – z2
4. Resolve into factors.
a) (a2 – b2) (c2 – d2) + 4abcd b) (x2 – 1) (y2 – 1) – 4xy
c) (p2 – 4) (9 – q2) + 24pq d) (9 – x2) (100 – y2) – 120xy
5. a) A square sheet of paper is 25 cm long. A small square portion of length 9 cm is cut
out from it. Find the area of the remaining portion of the paper.
b) A farmer has a square field of length 150 m. He separates a small square portion of
length 60 m from it to cultivate vegetables and he cultivates crops in the remaining
portion. Find the area of the crops cultivated portion of the field.
Project work and activity section
6. a) Take a squared chart paper of length ‘a + b’. Then, show (a + b)2 in chart paper and
present in class room.
b) Show (a – b)2 geometrically in chart paper and present in class.
c) Write any three expressions of your own in the form of a2 – b2. Then, factorize your
expressions.
f) Expression of the form a4 + a2b2 + b4
The expressions of the form a4 + a2b2 + b4 are also factorised by using the similar
method of factorisation of the expression of a2 – b2 form. Following formulae are
useful while factorising these types of expressions.
a2 + 2ab + b2 = (a + b)2 a2 + b2 = (a + b)2 – 2ab
a2 – 2ab + b2 = (a – b)2 a2 + b2 = (a – b)2 + 2ab
a2 – b2 = (a + b) (a – b)
Example 1: Factorise a) a4 + a2b2 + b4 b) x4 – 3x2y2 + y4
Solution:
a) a4 + a2b2 + b4 = (a2)2 + (b2)2 + a2b2
= (a2 + b2)2 – 2a2b2 + a2b2 Using a2 + b2 = (a + b)2 – 2ab
= (a2 + b2)2 – (ab)2
= (a2 + b2 + ab) (a2 + b2 – ab) Using a2 – b2 = (a + b) (a – b)
= (a2 + ab +b2) (a2 – ab + b2)
b) x4 – 3x2y2 + y4 = (x2)2 + (y2)2 – 3x2y2
= (x2 – y2)2 + 2x2y2 – 3x2y2 Using a2 + b2 = (a – b)2 + 2ab
= (x2 – y2)2 – (xy)2
= (x2 – y2 + xy) (x2 – y2 – xy)
= (x2 + xy – y2) (x2 – xy – y2)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 155 Vedanta Excel in Mathematics - Book 9
Factorisation of Algebraic Expressions
Example 2: Resolve into factors a) 9x4 + 14x2 + 25 b) x4 + x2 +1 c) a4 + 4b4
Solution: y4 y2
a) 9x4 + 14x2 + 25 = (3x2)2 + (5)2 + 14x2
= (3x2 + 5)2 – 2.3x2.5 + 14x2
= (3x2 + 5)2 – 16x2
= (3x2 + 5)2 – (4x)2
= (3x2 + 4x + 5) (3x2 – 4x + 5)
b) x4 + x2 + 1 = x2 2 + (1)2 + x2 c) a4 + 4b4 = (a2)2 + (2b2)2
y4 y2 y2 y2
= x2 + 1 2 – 2. x2 .1 + x2 = (a2 + 2b2)2 – 2a2.2b2
= y2 y2 y2 = (a2 + 2b2)2 – (2ab)2
=(a2 + 2ab + 2b2) (a2 – 2ab + 2b2)
x2 + 1 2 x2
y2 y
–
= x2 + x +1 x2 – x +1
y2 y y2 y
EXERCISE 9.3
General section b) 4a4 + 1 c) p4 + 64q4 d) 81m4 + 4n4
1. Factorise:
a) x4 + 4
e) 64a4 + 81x4 f) 625b4 + 4d4 g) m4 + 1 h) 100c4 + t4
Creative section 4 25
2. Resolve into factors
a) x4 + x2 y2 + y4 b) a4 + 4a2 + 16 c) 36p4 + 3p2 + 1
f) 225x4 – x2y2 + 16y4
d) 9m4 + 14m2n2 + 25n4 e) 4x4 – 13x2y2 + 9y4 i) 48m5n –3m3n3 + 27mn5
g) 32x4 +14x2y2 + 2y4 h) 20b4 + 55b2y2 + 45y4 c) 25a4 – 19a2 + 1
f) 121p4 – 133p2q2 + 36q4
3. Find the factors of the following expressions. i) 60a5b – 240a3b3 + 135ab5
a) 4x4 – 16x2 + 9 b) 9p4 – 28p2 + 16 c) 25m4 – 29m2 + 4
f) 169p4 – 365p2q2 + 196q4
d) y4 – 13y2z2 + 4z4 e) 81a4 – 99a2x2 + 25x4
g) 4x5 – 56x3y2 + 25xy4 h) 27b4p – 21b2p3+ 3p5
4. Factorise the following expressions.
a) 9x4 – 25x2 + 16 b) 4y4 – 5y2 + 1
d) 9a4 – 34a2b2 + 25b4 e) 100c4 – 325c2d2 + 225d4
5. Factorise.
a) x4 + x2 + 1 b) a4 + b4 + 1 c) x2 + 1 + 1
y4 y2 b4 a4 x2
d) x4 – 7x2 + 1 e) m4 – 29m2 + 4 f) b4 – 21db22 + 100
y4 y2 n4 n2 d4
Vedanta Excel in Mathematics - Book 9 156 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Factorisation of Algebraic Expressions
Project work and activity section
6. a) Write any three expressions of your own in the form of a4 + b4. Then, factorize your
expressions.
b) Write any three expressions of your own in the form of a4 + a2b2 + b4. Then, factorize
your expressions.
c) Write any three expressions of your own in the form of a4 – a2b2 + b4. Then, factorize
your expressions.
g) Expression of the form (a + b)3 and (a – b)3 i.e., cubes of binomials
(i) Expression of the form (a + b) 3
Geometrical interpretation:
Let’s take a cube such that its each side is (a + b). Then, the volume of the cube is
(a + b)3 which is shown below.
a+b aa a b
a+b bb bb
a+b aa a
b b3
(a + b)3 = a3 + 3a2b + 3ab2 +
Thus, (a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + 3ab (a + b) + b3
By usual multiplication:
Let’s find the product of (a + b)3.
(a + b)3 = (a + b) (a + b) (a + b)
= (a + b) ( )2
= (a + b) (a2 + 2ab + b2)
= a (a2 + 2ab + b2) + b (a2 + 2ab + b2)
= a3 + 2a2 b + ab2 + a2 b + 2ab2 + b3
= a3 + 3a2 b + 3ab2 + b3
∴(a + b) 3 = a3 + 3a2 b + 3ab2 + b3 = a3 + 3ab (a + b) + b3
Facts to remember
1. The expanded form of (a + b)3 = a3 + 3a2 b + 3ab2 + b3
2. The factors of (a + b)3 are (a + b), (a + b) and (a + b).
(ii) Expression of the form (a – b) 3
Let’s take a cube such that its each side is (a – b). Then, the volume of the cube is
(a – b)3 which is shown below.
a
a–b b
a–b
a–b b b a–b a–bbb bb
a–b a–b a–b b
From the above model, the volume of cube = volume of parts of the cube
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 157 Vedanta Excel in Mathematics - Book 9
Factorisation of Algebraic Expressions
or, a3 = (a – b)3 + 3 (a – b)2b + 3 (a – b) b2 + b3
or, a3 – 3 (a – b)2b – 3 (a – b) b2 – b3 = (a – b)3
or, a3 – 3 (a2 –2ab +b2) b – 3 (a – b) b2 – b3 = (a – b)3
or, a3 – 3a2 b + 6ab2 – 3b3 – 3ab2 + 3b3 – b3 = (a – b)3
or, a3 – 3a2 b + 3ab2– b3 = (a – b)3
Thus, (a – b)3 = a3 – 3a2b + 3ab2 – b3 = a3 – 3ab (a – b) – b3
Facts to remember
1. The expanded form of (a – b)3 = a3 – 3a2 b + 3ab2 – b3
2. The factors of (a + b)3 are (a – b), (a – b) and (a – b).
Worked-out Examples
Example 1: Express 8x3 + 36x2 + 54x + 27 as a cube of a binomial.
Solution:
Here, 8x3 + 36x2 + 54x + 27
= (2x)3 + 3. (2x)2. 3 + 3. 2x. 32 + 33
= (2x + 3)3
Example 2: Factorise: 64a3 – 240a2b + 300ab2 – 125b3
Solution:
Here, 64a3 – 240a2b + 300ab2 – 125b3
= (4a)3 – 3. (4a)2. 5b + 3. 4a. (5b)2 – (5b)3
= (4a – 5b)3
= (4a – 5b) (4a – 5b) (4a – 5b)
h) Expression of the form (a3 + b3) and (a3 – b3)
Let’s find the product of the expressions (a + b) and (a2 – ab + b2).
(a + b) (a2 – ab + b2) = a (a2 – ab + b2) + b (a2 – ab + b2)
= a3 – a2 b + ab2 + a2 b – 2 + b3
= a3 + b3
Also, we have
(a + b)3 = a3 + 3ab (a + b) + b3
or, (a + b)3 – 3ab (a + b) = a3 + b3
or, (a + b) {(a + b)2 – 3ab} = a3 + b3
or, (a + b) (a2 + 2ab + b2 – 3ab) = a3 + b3
or, (a + b) (a2 - ab + b2) = a3 + b3
So, (a + b) and (a2 – ab + b2) are the factors of a3 + b3
Similarly, (a – b) (a2 + ab + b2) = a3 – b3 and (a – b) and (a2 + ab + b2) are the factors of a3 – b3.
Facts to remember 2. a3 – b3 = (a – b) (a2 + ab + b2)
1. a3 + b3 = (a + b) (a2 – ab + b2)
Example 3: Factorise: a) 8a4 + 125a b) x6 – y6 c) a7 + 1
a5
Vedanta Excel in Mathematics - Book 9 158 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Factorisation of Algebraic Expressions
Solution:
a) 8a4 + 125a = a (8a3 + 125)
= a [(2a)3 + 53]
= a (2a + 5) [(2a)2 – 2a.5 + 52] Using a3 + b3 = (a + b) (a2 – ab + b2)
= a (2a + 5) (4a2 – 10a + 25)
b) x6 – y6 = (x3)2 – (y3)2
= (x3 + y3) (x3 – y3)
= (x + y) (x2 – xy + y2) (x – y) (x2 + xy + y2)
= (x + y) (x – y) (x2 – xy + y2) (x2 + xy + y2)
c) a7 – 1 = a7 – a × 1 = a(a6 – a16) = a[(a3)2 – 1 2 a3 + 1 a3 – 1
a5 a6 a3 a3 a3
]=a
1 11 1 11
= a a + a (a2 – a × a + a2) a – a (a2 + a × a + a2)
11 1 1
= a a + a a – a a2 – 1+ a2 a2 + 1+ a2
11 1 12 1
=a a + a a – a a2 – 1+ a2 a+ a –2×a×a+1
=a a + 1 a – 1 a2 – 1+ 1 a+ 1 2
a a a a
– (1)2
=a a + 1 a – 1 a2 – 1+ 1 a + 1+ 1 a – 1+ 1
a a a a a
Example 4: Resolve into factors: 8x3 – 20x2y + 30xy2 – 27y3.
Solution:
8x3 – 20x2y + 30xy2 – 27y3 = (2x)3 – (3y)3 – 20x2y + 30xy2
= (2x – 3y) [(2x)2 + 2x.3y + (3y)2] – 10xy (2x – 3y)
= (2x – 3y) (4x2 + 6xy + 9y2 – 10xy)
= (2x – 3y) (4x2 – 4xy + 9y2)
Example 5: Factorise: 2a6 – 19a3 + 24
Solution:
2a6 – 19a3 + 24 = 2a6 – 16a3 – 3a3 + 24
= 2a3 (a3 – 8) – 3 (a3 – 8)
= (a3 – 8) (2a3 – 3)
= (a3 – 23) (2a3 – 3)
= (a – 2) (a2 + 2a + 4) (2a3 – 3) = (a – 2)(2a3 – 3) (a2 + 2a + 4)
EXERCISE 9.4
General Section
1. Let’s say and write the expanded forms of the following cubes.
a) (a + x)3 = ............................................ b) (a – x)3 = ............................................
c) (p + q)3 = ............................................ d) (p – q)3 = ............................................
e) (x+1)3 = .............................................. f) (x – 1)3 = .............................................
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 159 Vedanta Excel in Mathematics - Book 9
Factorisation of Algebraic Expressions
2. Let’s say and write the factors of the expressions.
a) m3 + n3 = ............................................ b) m3– n3 = ..............................................
c) x3 + y3 = ............................................. d) x3 – y3 = ...............................................
e) a3 + 1 = .............................................. f) a3 – 1 = .............................................
Creative section-A
3. Express the following expressions as cubes of binomials
a) 8x3 + 60x2 + 150x + 125 b) 27a3 + 108a2 + 144a + 64
c) 64y3 – 240y2 + 300y – 125 d) 343m3 – 882m2 + 756m – 216
e) 1000a3 – 900a2 b + 270ab2 – 27b3 f) 8x3 +132x2 y + 726xy2 + 1331y3
4. Find the factors of the following expressions.
a) 8a3 + 36a2 + 54a + 27 b) 125x3 + 75x2 + 15x + 1
c) 27p3 – 108p2 + 144p – 64 d) 729w3 – 1944w2 + 1728w – 512
e) 216m3 – 540m2n+ 450mn2 – 125n3 f) 2197e3 +3549e2g+1911eg2+ 343g3
5. Resolve into factors.
a) 8x3 + y3 b) 1 + 27a3 c) 128t 4 – 2t d) x3y – 64y4
e) a6 + b3 f) 64x6 y3 – 125 g) a6 – 64 h) 729x6 – y6
Creative section-B b) (x + 2)3 – 27 c) (x – y) 3 – 8(x + y) 3 d) p3 + 1
6. Factorise. p3
a) (a + b) 3 + 1
e) a3 – ab33 f) x4 + 1 g) y7 – 1 h) pq4 – pq22
b3 x2 y5
7. Resolve into factors. b) 8 – 6a – 9a2 + 27a3
a) 27x3 + 30x2y + 40xy2 + 64y3 d) 2p6 – 19p3 + 24
c) 64m3 – 28m2n + 35mn2 – 125n3 f) 8x6 – 9x3 + 1
e) 40a6 + 11a3 – 2
8. a) The volume of a rectangular tank is (2a3 + a2 – 2a – 1) cu. ft. where a>4 ft., the
length is longer than its breadth and the breadth is longer than its height.
(i) Find its length, breadth and height.
(ii) Find the area of the its floor.
b) From a cubical log of each side x cm, a small cube of each side 3 cm is removed.
(i) What is the volume of removed part of the log?
(iii) What is the volume of remaining part of the log?
Project work and activity section
9. a) Write any three expressions of your own in the form of a3 + b3. Then, factorize your
expressions.
b) Write any three expressions of your own in the form of a3 – b3. Then, factorize your
expressions.
Vedanta Excel in Mathematics - Book 9 160 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Factorisation of Algebraic Expressions
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. The factorisation of 16 – p4 is
(A) (4 + p2) (2 + p) (2 – p) (B) (4 + p2) (2 – p) (2 – z)
(C) (4 – p2) (2 – p) (2 – p) (D) (4 + p2) (2 + p) (2 + p)
2. The factors of x2 – x – 12 are
(A) (x+4), (x – 3) (B) (x – 4), (x – 3) (C) (x – 4), (x + 3) (D) (x+4), (x + 3)
3. a2 + b2 = (a + b)2 + x, what is the value of x?
(A) 2ab (B) –2ab (C) 4ab (D) –4ab
4. a + b is a factor of ….
(A) a2 – b2 (B) a2 + b2 (C) (a – b)2 (D) a3 – b3
5. (x + y) (x2 – xy + y2) is equal to
(A) (x + y)3 (B) (x – y)3 (C) x3 + y3 (D) x3 – y3
6. p3 – q3 is equal to
(A) (p + q) (p2 – pq + q2) (B) (p + q) (p2 – pq + q2)
(C) (p – q) (p2 – pq + q2) (D) (p + q) (p2 + pq + q2)
7. The factors of x3 – 64 are
(A) (x – 4), (x2 + 4x + 16) (B) (x + 4), (x2 + 4x + 16)
(C) (x + 4), (x2 – 4x + 16) (D) (x – 4), (x2 – 4x + 16)
8. The factors of 125x3 + 1 are
(A) (5x – 1), (25x2 + 5x + 1) (B) (5x + 1), (25x2 – 5x + 1)
(C) (5x – 1), (25x2 – 5x + 1) (D) (25x + 1), (25x2 – 5x + 1)
9. The factorisation of x4 + x2y2 + y4 is
(A) (x2 + xy + y2) (x2 – xy + y2) (B) (x2 – xy + y2) (x2 – xy + y2)
(C) (x2 + xy + y2) (x2 + xy + y2) (D) (x2 – xy + y2) (x2 + xy – y2)
10. The factors of a4 + 4 are
(A) (a2 + 2a + 2), (a2 + 2a – 2) (B) (a2 + 2a + 2), (a2 + 2a + 2)
(C) (a2 – 2a + 2), (a2 + 2a + 2) (D) (a2 – 2a + 2), (a2 + 2a – 2)
Vedanta ICT Corner
Please! Scan this QR code or
browse the link given below:
https://www.geogebra.org/m/hgfn6vtt
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 161 Vedanta Excel in Mathematics - Book 9
Unit 10 Highest Common Factor (H.C.F.) and
Lowest Common Multiples (L.C.M.)
10.1 H.C.F. and L.C.M. –Looking back
Classwork-Exercise
Let’s answer the following questions as quickly as possible.
1. a) For any two expressions 2a – 4 and a2 – 4,
(i) What are the factors of 2a– 4?
(ii) What are the factors of a2 – 4?
(iii) what is the common factor among the factors of these expressions?
(iv) what is the common factor of these expressions called?
b) ax + ay and bx + by are two given expressions.
(i) What are the factors of ax + ay?
(ii) What are the factors of bx + by.
(iii) What is the common factor among the factors of these expressions?
(iv) What are the remaining factors of these expressions?
(v) What is the product of common factor and remaining factors of these
expressions called?
10.2 H.C.F. of algebraic expressions
Let's take any two monomial expressions 4a2 and 6a3.
Here, all the possible factors of 4a2 are a, a2, 2a, 2a2, 4a, 4a2.
Also, all the possible factors of 6a3 are a, a2, a3, 2a, 2a2, 2a3, 3a, 3a2, 3a3.
Now, the common factors of 4a2 and 6a3 are a, a2 2a and 2a2. Among these common
factors, 2a2 is the highest one.
So, the Highest Common Factor (H.C.F.) of 4a2 and 6a3 is 2a2.
Thus, to find the H.C.F. of the monomial expressions, at first we should find the
H.C.F. of the numerical coefficients. Then, the common variable with the least power
is taken as the H.C.F. of the expressions. For example,
4 is the H.C.F. of 12 and 8.
In 12x4y3 and 8x3y2 x3 is the H.C.F. of x4 and x3.
y2 is the H.C.F. of y3 and y2.
So, the H.C.F. of 12x4y3 and 8x3y2 is 4x3y2.
Vedanta Excel in Mathematics - Book 9 162 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
H.C.F. and L.C.M.
We can find the H.C.F. of polynomial expressions either by the process of factorisation
or by division process. In case of factorisation process, the given polynomials are to
be factorised, then a common factor or the product of common factors is obtained as
their H.C.F.
Facts to remember
1. The H.C.F. of two or more algebraic expressions is an expression of the highest
degree which exactly divides the given expressions.
2. If there is no common factor other than one, then H.C.F. is 1.
Worked-out Examples
Example 1: Find the highest common factors (H.C.F.) of each pair of expressions. Also,
represent the H.C.F. in Venn-diagram.
a) x3 – 4x, x2 + 7x + 10
b) a2 – 6a + 6b – b2, b2 + ab – 6b
c) m3 + 8, m3 – 8
d) (p – q)2 + 4pq, p2 – pq – 2q2
Solution:
Here,
a) The 1st expression = x3 – 4x
= x (x2 – 4)
= x (x + 2) (x – 2)
The 2nd expression = x2 + 7x + 10 1st expression x(x – 2) 2nd expression
(x+2)
= x2 + (5 + 2) x + 10 (x + 5)
= x2 + 5x + 2x + 10
= x (x + 5) + 2 (x + 5)
= (x + 5) (x + 2)
Hence, the H.C.F. = common factor = (x + 2)
b) The 1st expression = a2 – 6a + 6b – b2 1st expression (a – b) 2nd expression
= a2 – b2 – 6a + 6b (a+b–6)
= (a + b) (a – b) – 6 (a – b)
= (a – b) (a + b – 6) b
The 2nd expression = b2 + ab – 6b
= b (b + a – 6)
Hence, the H.C.F. = common factor = (a + b – 6)
c) The 1st expression = m3 + 8
= m3 + 23
= (m + 2) (m2 – m.2 + 22)
= (m + 2) (m2 – 2m + 4)
The 2nd expression = m3 – 8 1st expression m3 + 8 2nd expression
= m3 – 23 m3 – 8
1
= (m – 2) (m2 + m.2 + 22)
= (m – 2) (m2 + 2m + 4)
Hence, the H.C.F. = common factor = 1
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 163 Vedanta Excel in Mathematics - Book 9
H.C.F. and L.C.M.
d) The 1st expression = (p – q)2 + 4pq
= p2 – 2pq + q2 + 4pq
= p2 + 2pq + q2
= (p + q)2
= (p + q) (p + q) 1st expression (p + q) 2nd expression
(p + q)
The 2nd expression = p2 – pq – 2q2 (p – 2q)
= p2 – (2 – 1) pq – 2q2
= p2 – 2pq + pq – 2q2
= p(p – 2q) + q (p – 2q)
= (p – 2q) (p + q)
Hence, the H.C.F. = common factor = (p + q)
Example 2: Find the highest common factors (H.C.F.) of each pair of expressions.
a) 18(2x3 – x2 – x), 20 (24x4 + 3x)
b) 16a4 + 4a2 b2 + b4, 8a3 + b3
c) x4 + 3x2 + 196, x3 + x (x + 14) + 4x2
d) (1 – m2) (1 – n2) + 4mn, 1 – 2m + n – m2 n + m2
Solution:
Here,
a) The 1st expression = 18(2x3 – x2 – x)
= 18x (2x2 – x – 1)
= 2 × 3 × 3x (2x2 – 2x + x – 1)
= 2 × 3 × 3x {2x (x – 1) + 1 (x – 1)}
= 2 × 3 × 3x (x – 1) (2x + 1)
The 2nd expression = 20 (24x4 + 3x)
= 20 × 3x (8x3 + 1)
= 2 × 2 × 5 × 3x {(2x)3 + (1)3}
= 2 × 2 × 3 × 5x (2x + 1) (4x2 – 2x + 1)
Hence, the H.C.F. = common factor = 2 × 3x (2x +1) = 6x (2x + 1)
b) The 1st expression = 16a4 + 4a2 b2 + b4
= (4a2)2 + (b2)2 + 4a2b2
= (4a2 + b2)2 – 2. 4a2.b2 + 4a2b2
= (4a2 + b2)2 – 4a2b2
= (4a2 + b2)2 – (2ab)2
= (4a2 + b2+2ab) (4a2 + b2 – 2ab)
= (4a2 +2ab + b2) (4a2 – 2ab + b2)
The 2nd expression = 8a3 + b3
= (2a)3 + b3
= (2a + b) [(2a)2 – 2a. b + b2]
= (2a + b) (4a2 – 2ab + b2)
Hence, the H.C.F. = common factor = (4a2 – 2ab + b2)
Vedanta Excel in Mathematics - Book 9 164 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
H.C.F. and L.C.M.
c) The 1st expression = x4 + 3x2 + 196
= (x2)2 + (14)2 + 3x2
= (x2 + 14)2 – 2. x2.14+ 3x2
= (x2 + 14)2 – 25x2
= (x2 + 14)2 – (5x)2
= (x2 +5x +14) (x2 – 5x +14)
The 2nd expression = x3 + x (x + 14) + 4x2
= x3 + x2 + 14x + 4x2
= x3 + 5x2 + 14x
= x (x2 + 5x + 14)
Hence, the H.C.F. = common factor = (x2 + 5x + 14)
d) The 1st expression = (1 – m2) (1 – n2) + 4mn
= 1 – n2 – m2 + m2n2 + 4mn
= m2n2+ 2mn + 1 – m2 + 2mn – n2
= (mn)2 + 2.mn. 1 + (1)2 – (m2 – 2mn + n2)
= (mn + 1)2 – (m – n)2
= {(mn + 1) + (m – n)} {(mn + 1) – (m – n)}
= (mn + 1 + m – n) (mn + 1 – m + n)
= (mn + m – n + 1) (mn – m + n + 1)
The 2nd expression = 1 – 2m + n – m2 n + m2
= m2 – 2m + 1 – m2n + n
= (m – 1)2 – n (m2 – 1)
= (m –1)2 – n (m + 1) (m – 1)
= (m – 1) {m – 1 – n (m + 1)}
= (m – 1) (m – 1 – mn – n)
= (1 – m) (mn – m + n + 1)
Hence, the H.C.F. = common factor = (mn – m + n + 1)
Example 3: Find the highest common factors (H.C.F.) of the following expressions.
a) 9p2 – 4q2 – 4qr – r2, r2 – 4q2 – 9p2 – 12pq and 9p2 + 6pr + r2 – 4q2
b) 27x3 + 18x2 + 6x +1, 81x4 – 9x2 – 6x – 1 and 81x4 + 9x2 + 1
c) x2 – 12x – 28 + 16y – y2, x2 + 2x – y2 + 2y and x2 – y2 + 4y – 4
Solution:
Here,
a) The 1st expression = 9p2 – 4q2 – 4qr – r2
= (3p)2 – (4q2 + 4qr + r2)
= (3p)2 – {(2q)2 + 2. 2q. r + (r)2}
= (3p)2 – (2q + r)2
= (3p + 2q + r) (3p – 2q – r)
The 2nd expression = r2 – 4q2 – 9p2 – 12pq
= (r)2 – (4q2 + 12pq + 9p2)
= (r)2 – {(2q)2 + 2. 2q. 3p + (3p)2}
= (r)2 – (2q + 3p)2
= (r + 2q + 3p) (r – 2q – 3p)
The 3rd expression = 9p2 + 6pr + r2 – 4q2
= (3p)2 + 2. 3p.r + (r)2 – (2q)2
= (3p + r)2 – (2q)2
= (3p + r + 2q) (3p + r – 2q)
Hence, the H.C.F. = common factor = (3p + 2q + r)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 165 Vedanta Excel in Mathematics - Book 9
H.C.F. and L.C.M.
b) The 1st expression = 27x3 + 18x2 + 6x +1
= (3x)3 +(1)3 + 6x (3x +1)
= (3x +1) (9x2 – 3x + 1) + 6x (3x +1)
= (3x + 1) (9x2 – 3x + 1 + 6x)
= (3x + 1) (9x2 + 3x + 1)
The 2nd expression = 81x4 – 9x2 – 6x – 1
= (9x2)2– (9x2 + 6x + 1)
= (9x2)2 – {(3x)2 + 2. 3x .1 + (1)2}
= (9x2)2– (3x +1)2
= (9x2 + 3x + 1) (9x2 – 3x – 1)
The 3rd expression = 81x4 + 9x2 + 1
= (9x2)2 + (1)2 + 9x2
= (9x2 + 1)2 – 2. 9x2.1+ 9x2
= (9x2 + 1)2 – 9x2
= (9x2 + 1)2 – (3x)2
= (9x2 +3x +1) (9x2 – 3x +1)
Hence, the H.C.F. = common factor = (9x2 +3x +1)
c) The 1st expression = x2 – 12x – 28 + 16y – y2
= x2 – 2.x.6 + 62 – 62 – 28 + 16y – y2
= (x – 6)2 – (64 – 16y + y2)
= (x – 6)2 – (82 – 2.8.y + y2)
= (x – 6)2 – (8 – y) 2
= (x – 6 + 8 – y) (x – 6 – 8 + y)
= (x – y + 2) (x + y – 14)
The 2nd expression = x2 + 2x – y2 + 2y
= x2 – y2 + 2(x + y)
= (x + y) (x – y) + 2(x + y)
= (x + y) (x – y + 2)
The 3rd expression = x2 – y2 + 4y – 4
= x2 – (y2 – 2. y. 2 + 22)
= x2 – (y – 2)2
= (x + y – 2) (x – y + 2)
Hence, the H.C.F. = common factor = (x – y + 2)
EXERCISE 10.1
General section
1. a) Define the H.C.F. of algebraic expressions.
b) What is the H.C.F. of 4ax and 6xy?
c) What will be the H.C.F. of x and y if x is a factor of y?
d) Write the condition under which the H.C.F. of the given algebraic expressions is 1.
2. Find the H.C.F. of the following expressions:
a) 2x2 (x + 2) (x – 2) and 4x (x + 2) (x + 3)
b) 4xy2 (x – 1) (x + 2) and 6x2 y (x – 1) (x – 4)
c) 6a2 b2 (a – b) (2a + 3b) and 9a3 b3 (2a + 3b) (a + b)
d) 12a2 b3 (a – 3b) (a + b – 2) and 16a3 b2 (a + 3b) (a + b – 2)
e) (p + q) (p – q), (p – q) (p2 + pq + q2) and (p – q) (p + q) (p2 + q2)
Vedanta Excel in Mathematics - Book 9 166 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
H.C.F. and L.C.M.
3. Find the H.C.F. of the following expressions:
a) a2 – b2 and a3 + b3 b) a2 – 2a and a4 – 8a
c) x2 – 9 and 3x + 9 d) x2 – y2 and x2 + 2xy + y2
e) 4x2 – 100 and 4x + 20 f) 4x3 + 8x2 and 5x3 – 20x
g) a2 – b2 – 2a + 1 and a2 – ab – a h) x3 – 8y3 and x2 + 2xy + 4y2
i) p4 – 16 and p2 – p – 6 j) x4 + 4y4 and 2x3y + 4xy3 + 4x2 y2
Creative section
4. Find the H.C.F. of the following expressions: b) p3 + 8 and p4 + 4p2 + 16
a) a3 – 1 and a4 + a2 + 1
c) 16x4 – 4x2 – 4x – 1, 8x3 – 1 d) 81a4 – 9a2 + 6a – 1, 27a3 + 1
e) 36(x3+x2y –12xy2), 54(x3y+4x2y2 –21xy3) f) 6(5x3y–500xy3), 20(x3 – 1000y3)
g) x4 – x2 + 144, x3 + x (x + 12) + 4x2 h) a4 + 4a2+100, a3 + a (a+10) + 3a2
i) 2a3 – a2 – a + 2, a3 + a2 – a – 1 j) 3x3 – x2 + x – 3, x3 – x2 + x –1
5. Find the highest common factors (H.C.F.) of the following expressions:
a) a2 – 4, a3 + 8, a2 + 5a + 6 b) x2 – 9, x3 – 27, x2 + x – 12
c) 4x2 – 9, 2x2 + x – 3, 8x3 + 27 d) 3x2 – 8x + 4, 2x2 – 5x + 2, x4 – 8x
e) 5a3 – 20a, a3 – 3a2 – 10a, a3 – a2 – 2a + 8
f) m3 – m2 – m + 1, 2m4 – 2m, 3m2 – 3m – 6
g) x3 – 64y3, x2 – 6xy + 8y2, x2 – 16y2
h) 4x4 + 16x3 – 20x2, 3x3 + 14x2 – 5x, x4 + 125x
i) 8x3 + 27y3, 16x4 + 36x2 y2 + 81y4, 4x3 – 6x2 y + 9xy2
j) x3 y + y4, x4 + x2 y2 + y4, 2ax3 – 2ax2 y + 2axy2
6. Find the highest common factors (H.C.F.) of the following expressions:
a) x3 – 1, x4 + x2 + 1, x3 + 2x2 + 2x + 1
b) 5p3 – 20p, p3 – p2 – 6p, p3 – p2 – 2p + 8
c) (a + b) 2 – 4ab, a3 – b3, a2 + ab – 2b2
d) (a – b)3 + 3ab (a – b), a4 + a2b2 + b4, a3 + a2b + ab2
e) a2 + 2ab + b2 – c2, b2 + 2bc + c2 – a2, c2 + 2ca + a2 – b2
f) 9x2 – 4y2 – 8yz – 4z2, 4z2 – 4y2 – 9x2 – 12xy, 9x2 + 12xz + 4z2 – 4y2
g) x2 – 18x – 19 + 20y – y2, x2 + x – y2 – y, x2 – y2 + 2y – 1
h) a2 – 6a – 7 + 8b – b2, a2 + a – b2 + b, a2 - b2 + 2b – 1
i) 1+4x+4x2 – 16x4, 1+2x – 8x3 – 16x4, 1 + 4x2 + 16x4
j) 81x4 – 9x2 – 6x – 1, 81x4 + 27x3 – 3x – 1, 81x4 + 9x2 + 1
7. a) Sunayana and Bishwant have rectangular plots of land of same width. The area of
Sunayana’s and Bishwant’s plots are (x2 + 35x + 300) sq. m. and (x2 + 27x + 180)
sq. m. respectively. Find the width of their plots.
b) The area of four walls of a rectangular room is (4x2 + 20x – 56) sq. m and the volume
of the room is (x3 – 8) cu. m, find the height of the room.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 167 Vedanta Excel in Mathematics - Book 9
H.C.F. and L.C.M.
Project work and activity section
8. a) Write a pair of expressions of your own in the forms of a2 – b2 and a2 + bx + c = 0,
a ≠ 0. Then, factorize your expressions and find their highest common factor.
b) Write a pair of expressions of your own in the forms of a3 + b3 and a4 + a2b2+b4.
Then, factorize your expressions and find their H.C.F.
3.3 L.C.M. of algebraic expressions
Let's take any two monomial expressions x2 and x3.
A few multiplies of x2 are x2, x3, x4, x5, x6,.....
A few multiplies of x3 are x3, x4, x5, x6, x7, ....
Here, the common multiplies are x3, x4, x5, x6, ...
Among these common multiplies, x3 is the lowest one.
So, the Lowest Common Multiple (L.C.M.) of x2 and x3 is x3.
Thus, to find the L.C.M. of the monomial expressions, at first we should find the
L.C.M. of the numerical coefficients. Then, the common variable with the highest
power is taken as the L.C.M. of the expressions. For Example,
24 is the L.C.M. of 6 and 8.
In 6x4y2 and 8x3y3 x4 is the L.C.M. of x4 and x3.
y3 is the L.C.M. of y2 and y3.
So, the L.C.M. of 6x4y2 and 8x3y3 is 24x4y3.
In case of polynomial expressions, their L.C.M. is obtained by the process of
factorisation. By this process the product of common factors and the factors which or
not common is taken as the L.C.M. of the polynomials
Facts to remember
1. The L.C.M. of two or more algebraic expressions is an expression of the lowest
degree which is exactly divisible by the given expressions.
2. L.C.M. of three expressions is the product of factors common to at least two and the
remaining factors of the given expressions.
3. For any two algebraic expressions, the product of the expressions is equal to the
product of the H.C.F. and L.C.M. of the expressions i.e., if H and L be the H.C.F. and
L.C.M. of two algebraic expressions A and B, then A×B = H×L
Example 1: Find the lowest common multiple (L.C.M.) of each pair of expressions. Also,
represent the L.C.M. in Venn-diagram.
a) a3 – a and a4 – a
b) 6x6 + 6x4 + 6x2 and 4x6 – 4x3
c) (m + 5)3 – 15m (m + 5) and (m – 5)2 + 20m
1st expression 2nd expression
Solution:
(a + 1)
Here, a (a – 1)
a) The 1st expression = a3 – a a2 + a + 1
= a (a2 – 1)
= a (a + 1) (a – 1)
Vedanta Excel in Mathematics - Book 9 168 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
H.C.F. and L.C.M.
The 2nd expression = a4 – a
= a (a3 – 1)
= a (a – 1) (a2 + a + 1)
Hence, the L.C.M. = common factors × remaining factors
= a (a – 1) (a + 1) (a2 + a + 1) = a (a +1) (a3 – 1)
b) The 1st expression = 6x6 + 6x4 + 6x2
= 6x2 (x4 + x2 + 1)
= 2×3x2 [(x2) 2 + 12 + x2]
= 2×3x2 [(x2 + 1)2 – 2x2 + x2] 1st expression 2nd expression
= 2×3x2 [(x2 + 1)2 – x2] 3(x2 – x+1)
= 2×3x2 (x2 + x + 1) (x2 – x + 1)
2x2(x2+x+1)
The 2nd expression = 4x6 – 4x3
2x(x – 1)
= 4x3 (x3 – 1)
= 2×2x3 (x – 1) (x2 + x + 1)
∴ The L.C.M. = common factors × remaining factors
= 2×3×2x3 (x2 + x + 1) (x – 1) (x2 – x + 1)
= 12x3 (x3 – 1) (x2 – x + 1)
c) The 1st expression = (m + 5)3 – 15m (m + 5)
= (m + 5) {(m + 5)2 – 15m}
= (m + 5) (m2 + 10m+ 25 – 15m)
= (m + 5) (m2 – 5m + 25) 1st expression 2nd expression
The 2nd expression = (m – 5)2 + 20m (m2–5m+25)
= m2 – 10m + 25 + 20m (m + 5)
= m2 + 10m + 25
= m2 + 2×m× 5 + 52 (m + 5)
= (m + 5)2 = (m + 5) (m + 5)
∴ The L.C.M. = common factors × remaining factors
= (m + 5) (m2 – 5m + 25) (m + 5) = (m +5) (m3 + 125)
Example 2: Find the L.C.M. of following expressions.
a) x2 + 2xy + y2 – z2, y2 + 2yz + z2 – x2 and z2 + 2xz + x2 – y2
b) 2a4 – 54a, 2a5 + 18a3+ 162a and 4a3 + 12a2 + 36a
c) 1+4p+4p2 – 16p4, 1+2p – 8p3 – 16p4 and 1 + 4p2 + 16p4
Solution:
Here,
a) The 1st expression = x2 + 2xy + y2 – z2
= (x + y)2 – z2
= (x + y + z) (x + y – z)
The 2nd expression = y2 + 2yz + z2 – x2
= (y + z)2 – x2
= (y + z + x) (y + z – x)
The 3rd expression = z2 + 2zx + x2 – y2
= (z + x)2 – y2
= (z + x + y) (z + x – y)
∴ The L.C.M. = common factors × remaining factors
= (x + y + z) (x + y – z) (y + z – x) (x – y + z)
b) The 1st expression = 2a4 – 54a
= 2a (a3 – 27)
= 2a (a3 – 33)
= 2a (a – 3) (a2 + 3a + 9)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 169 Vedanta Excel in Mathematics - Book 9
H.C.F. and L.C.M.
The 2nd expression = 2a5 + 18a3+ 162a
= 2a (a4 + 9a2 + 81)
= 2a {(a2)2 + 92 + 9a2}
= 2a {(a2 + 9)2 – 2.a2.9 + 9a2}
= 2a {(a2 + 9)2 – 9a2}
= 2a {(a2 + 9)2 – (3a)2}
= 2a (a2 + 3a + 9) (a2 – 3a + 9)
The 3rd expression = 4a3 + 12a2 + 36a
= 4a (a2 + 3a + 9)
∴ The L.C.M. = common factors × remaining factors
= 4a (a2 + 3a +9) (a – 3) (a2 – 3a + 9)
= 4a (a – 3) (a4 + 9a2 + 81)
c) The 1st expression = 1 + 4p + 4p2 – 16p4
= (1)2 + 2×1×2p + (2p)2 – 16p4
= (1 + 2p)2– (4p2)2
= (1 + 2p + 4p2) (1 + 2p – 4p2)
The 2nd expression = 1+2p – 8p3 – 16p4
= 1 (1 + 2p) – 8p3 (1 + 2p)
= (1 + 2p) (1 – 8p3)
= (1 + 2p) {1 – (2p)3}
= (1 + 2p) (1 – 2p) (1 + 2p + 4p2)
The 3rd expression = 1 + 4p2 + 16p4
= 1 + (4p2)2 + 4p2
= (1 + 4p2)2 – 2 × 1 × 4p2 + 4p2
= (1 + 4p2)2 – 4p2
= (1 + 4p2)2 – (2p)2
= (1 + 2p + 4p2) (1 – 2p + 4p2)
∴ The L.C.M. = common factors × remaining factors
= (1 + 2p + 4p2) (1 + 2p – 4p2) (1 + 2p) (1 – 2p) (1 – 2p + 4p2)
= (1 + 8p3) (1 – 8p3) (1 + 2p – 4p2) = (1 – 64p6) (1 + 2p – 4 2)
Example 3: The H.C.F. and L.C.M. of any two expressions are (a – 9) and (a + 9) (a – 9)2
respectively. If the first expression is (a2 – 81), find the second expression.
Solution:
Here, the 1st expression = a2 – 81 = (a + 9) (a – 9)
H.C.F. = (a – 9) and L.C.M. = (a + 9) (a – 9)2
We have, the first expression × the second expression = H.C.F. × L.C.M.
or, The second expression = H.C.F. × L.C.M.
First expression
or, The second expression = (a – 9) (a + 9) (a – 9)2 =(a – 9)2
(a + 9) (a – 9)
Hence, the second expression is (a – 9)2.
Vedanta Excel in Mathematics - Book 9 170 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
H.C.F. and L.C.M.
EXERCISE 10.2
General section
1. a) Define the L.C.M. of algebraic expressions.
b) What is the L.C.M. of a2b3 and a3b2?
c) If p is the multiple of q, what is the L.C.M. of p and q?
d) Write down the relation of any two algebraic expression and their H.C.F. and L.C.M.
2. Find the L.C.M. of the following expressions:
a) 3x (x +1) (x – 1) and 2x2 (x – 1) (x + 3)
b) 4x3 (x – 3) (x + 2) and 6x2 (x + 2) (x + 3)
c) 8a2 b (a – b) (a2 + ab + b2) and 12ab2 (a – b) (a + b)
d) (x + 2) (x + 3), (x + 3) (x – 2), (x – 2) (x – 3)
e) (a – 3) (a – 4), (a – 4) (a – 5), (a – 5) (a – 3)
3. Find the L.C.M. of the following expressions:
a) 3x2 + 6x, 2x3 + 4x2 b) ax2 + ax, ax2 – a c) 4x5 y4 + 2x4 y5, 10x4 y3 + 5x3 y4
d) x2 – xy, x3 y – xy3 e) 4x2 – 2x, 8x3 – 2x f) a3 – b3 , a2 + ab + b2
g) x2 + 5x + 6, x2 – 4 h) x2 – 9, 3x3 + 81 i) a4 b – ab4 , a4 b2 – a2 b4
j) x4 + x2 y2 + y4, x3 – y3 k) a4 + a2 b2 + b4, a3 + b3 l) 6x2 – x – 1, 54x4 + 2x
Creative section
4. Find the L.C.M. of the following expressions:
a) 4x3 + 500y3, 2x4 + 50x2y2 + 1250y4 b) 24a3 – 81b3 , 32a5 + 72a3b2 + 162ab4
c) 1 + 4x + 4x2 – 16x4 , 1 + 8x3 d) 16a4 – 4a2 – 4a – 1 , 16x4 + 8x3 – 2x – 1
e) x4 – 7x2 + 81, x3 + 5x (x + 2) – x f) 100y4 – 36y2 +1, 10y3 – y (3y + 1) – y2
5. Find the lowest common multiple (L.C.M.) of the following expressions:
a) a2 – 4, a3 – 8, (a + 2)2 b) (a – 3)2, a2 – 9, a3 + 27
c) a3 – 2a2 + a, a3 + a2 – 2a, a3 – 4a d) x4 – y4, x2 – y2, x3 – y3
e) 4x3 – 10x2 + 4x, 3x4 – 8x3 + 4x2, x4 – 8x f) x3 – 9x, x4 – 2x3 – 3x2, x3 – 27
g) a3 – 4a, a4 – a3 – 2a2, a3 – 8 h) a4 + a2 + 1, a3 – 1, a3 – a2 + a
i) x3 – 2x2y + 2xy2 – y3, x4 – y4, x3 + y3 j) x2 + 3x + 2, x2 + 5x + 6, x2 + 4x + 3
6. Find the L.C.M. of the following expressions:
a) a2 + 2ab + b2 – c2, b2 + 2bc + c2 – a2 and c2 + 2ca + a2 – b2
b) x2 – y2 – 2yz – z2, y2 – z2 – 2zx – x2 and z2 – x2 – 2xy – y2
c) 9x2 - 4y2 - 8yz – 4z2, 4z2 - 4y2 - 9x2 – 12xy, 9x2 + 12xz + 4z2 - 4y2
d) x2+(a + b) x + ab, x2 + (a – b) x – ab and x2 – (a + b) x + ab
e) x3 + 2x2 – x – 2, x3 + x2 – 4x – 4 and x4 – 5x2 + 4
f) a2 – 4a – 5 + 6b – b2, a2 + 2ab + b2 – 25 and a2 + ab – 5a
g) x2 + 8x – 33 – 42y – 9y2, x2 – 6xy + 9y2 – 121 and x2y + 11xy – 3xy2
7. a) The H.C.F. and L.C.M. of any two expressions are 2ab (a – b) and
12a2b2 (a + b) (a2 – b2) respectively. If the first expression is 6a4b + 12a3b2 + 6a2b3,
find the second expression.
b) The H.C.F. and L.C.M. of any two expressions are x2 – x + 1 and (x3 +1) (x2 + x + 1)
respectively. If the second expression is x4 + x2 +1, find the first expression.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 171 Vedanta Excel in Mathematics - Book 9
H.C.F. and L.C.M.
Project work and activity section
8. a) Write a pair of expressions of your own in the forms of a3 + b3 and a4 + a2b2+b4.
Then, factorize your expressions and find their lowest common multiple.
b) Write a pair of expressions of your own, then show that the product of the expressions
is equal to the product of their H.C.F. and L.C.M.
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. The H.C.F. of a3bx and abx3 is
(A) x (B) ab (C) abx (D) a3bx3
2. The expression of highest degree which exactly divides the given expression is called
(A) H.C.F. (B) L.C.M. (C) Factor (D) Multiple
3. What is the H.C.F. of (x – y)3 and x3 – y3?
(A) x – y (B) x3 – y3 (C) (x – y)3 (D) x2 – y2
4. The L.C.M. of a2 + ay and a3y – ay3 is
(A) a (a + y) (B) ay (a + y) (C) a (a2 – y2) (D) ay (a2 – y2)
5. If p is a factor of q, what is the H.C.F. of p and q?
(A) p (B) q (C) 1 (D) pq
6. If x and y are H.C.F. and L.C.M. of two expressions respectively, what is the relation
between x and y?
(A) x is always a multiple of y (B) x is always a factor of y
(C) y is always a factor of x (D) H.C.F. of x and y is always 1.
7. The highest common factor of a3 + 1 and a4 + a2 + 1 is
(A) a+1 (B) a2 + a + 1 (C) a2 – a + 1 (D) (a + 1)3
8. If H and K be the H.C.F. and L.C.M. of the expressions P and Q respectively, what is the
relation among H, K, P and Q?
(A) PK = HQ (B) HP = KQ (C) HK = PQ (D) H + K = P + Q
9. If the product of two expressions is m (m2 – 9), what is the product of their H.C.F. and
L.C.M.?
(A) m (m – 3) (B) m (m + 3) (C) m (m2 – 9) (D) m (m – 9)
10. The greatest common divisor of x4 + x2y2 + y4 and x3 + y3 is
(A) (x2 + xy + y2) (B) (x2 – xy + y2) (C) x3 + y3 (D) 1
Vedanta ICT Corner
Please! Scan this QR code or
browse the link given below:
https://www.geogebra.org/m/jbytnwfv
Vedanta Excel in Mathematics - Book 9 172 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit 11 Indices
11.1 Indices – review
An index is a number that shows how many times a base is multiplied by itself. An
index is also called an exponent.
Let's take a few examples of repetitive multiplication of the same base.
a = a1, a × a = a2, a × a × a = a3, a × a × a × a = a4, etc.
Here, 1 is the index of a1, 2 is the index of a2, 3 is the index of a3, 4 is the index of a4,
and so on.
Thus, the index refers to the power to which a number is raised. For example, in
23 the base 2 is raised to the power 3. Indices is the plural of index.
11.2 Laws of Indices
While performing the various operations of indices, we apply different proven rules
like product rule, quotient rule, power rule, etc. These rules are well known as laws
of indices.
The table given below shows the laws of indices at a glance.
Name of laws Rules Examples
Product law am × an = am + n 23 × 25 = 23+5 = 28
Quotient law am ÷ an = am – n when m > n 37 ÷ 33 = 37 – 3 = 34 and
am ÷ an = 1 when m < n 11
an – m 33 ÷ 37 = 37 – 3 = 34
Power law (am)n = am × n, (ab)m = ambm, (24)2 = 24 × 2 = 28,
am (2x)5 = 25 × x5, …
Negative index law a m = bm
Zero index law b
Root laws of indices 1 1
a– m = am or am = a–m 5–3 = 1 or 53 = 1
53 5–3
a0 = 1, (ab)0 = 1, (a + b)0 = 1 20 = 1, 30 = 1, 990 = 1
1 = 2 31 or 3,
m n am or n am m 32
an = = an 2 = 3 32
33
(i) Product law of indices
If am and an are the two algebraic terms, where m and n are the positive integers,
then am × an = am + n
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 173 Vedanta Excel in Mathematics - Book 9
Indices
Proof
We know that,
a2 = a × a (two factors)
a3 = a × a × a (three factors)
am = a × a × a × ... ('m' factors)
an = a × a × a × ... ('n' factors)
\ am × an = (a × a × a × ... 'm' factors) × (a ×a × a × ... 'n' factors)
= a × a × a × ... ('m + n' factors)
= am + n
Thus, am × an = am + n
Similarly, if m, n, p, q, r, ... are the positive integers, then
am × an × ap × aq × ar × ... = am + n + p + q + r + ...
(ii) Quotient law of indices
If am and an are the two algebraic terms, where m and n are the positive integers,
then am ÷ an = am – n when m > n
am ÷ an = 1 m when n > m
an –
Proof
am a × a × a × ... 'm' factors
am ÷ an = an = a × a × a × ... 'n' factors
= a × a × a × ... ('m – n' factors), where m > n
= am – n
Thus, am ÷ an = am – n, when m > n
But, in the case of n > m,
am a × a × a × ... 'm' factors
am ÷ an = an = a × a × a × ... 'n' factors
= a ×a × a × 1 – m' factors
... 'n
= 1 m
an –
Thus, am ÷ an = 1 m, when n > m
an –
(iii) Power law of indices
If am be an algebraic term, then (am)n = am × n = amn, where m and n are the
positive integers.
Proof
(am)n = am × am × am × ... (n) factors
= am + m + m + ... 'n' times 'm'
= amn
Thus, (am)n = amn
cor 1. amn = (am)n cor 2. amn = (an)m
cor 3. am = a m cor 4. a m = am
bm b b bm
cor 5. (ab)m = ambm cor 6. ambm = (ab)m
Vedanta Excel in Mathematics - Book 9 174 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Indices
(iv) Law of negative index
If a–m is an algebraic term, where m is a negative integer,
a–m = 1 or, 1 = a–m or, am = 1
am am a–m
Proof
Here, a–m = am – 2m = am ÷ a2m = am = am = 1
a2m am × am am
Thus, a–m = 1
am
Similarly, 1 = a–m and am = 1
am a–m
(v) Law of zero index
If a0 is an algebraic term, where a ≠ 0, then a0 = 1.
Proof am
am
Here, a0 = am – m = am × a–m = =1
Thus, a0 = 1, where a ≠ 0
(vi) Root law of indices
If amn is an algebraic term, where m and n are the positive integers, then
amn = n am
Proof n
nth order of root in represented as 2
In this way the 2nd order of root is represented as or only
3
The 3rd order of root is represented as and so on.
The square root of 32 = 3 = 322 = 2 32 or 32
The cube root of 53 = 5 = 533 = 3 53
Thus, 322 = 2 32
3 = 3 53
53
In general, amn = n am
Worked-out Examples
Example 1: Find the value of: 243 – 52
32
a) (16)43 b) (9–3)16 c) d) (160.5)23 e) 3 729 f) 3 64–1
Solution:
a) (16)34 = (24) 34 = (2)4 × 34 = 23 = 8
b) (9–3)61 = (32)–3 × 16 = 3–6 × 16 = 3–1= 31
c) 243 – 52 = 35 – 25 = 25 52 = 2 =5 × 52 2 2= 4
32 25 35 3 3 9
d) (160.5)23 = (24)0.5 × 32 = 24 × 0.5 × 32 = 23 = 8
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 175 Vedanta Excel in Mathematics - Book 9
Indices
e) 3 729 = 3 93 = 933 = 9 = 32 = 3
1 1 2= 3 1 3 1 3 1
64 8 8 2 2
f) 3 64–1 = 3 = 3 = =
Example 2: Evaluate: a) 25 – 21 125 13 ÷ 16 – 14 b) 127 × 286
16 64 81 217 × 166
1 1 1
c) (1 – 6–9)–1 + (1 – 69)–1 d) (a – b)–1 + (b – c)–1 + (c – a)–1
Solution:
a) 25 – 12 125 13 ÷ 16 – 14
16 64 81
= 16 21 125 31 ÷ 81 14
25 64 16
= 4 2 × 21 5 3× 31 ÷ 3 4× 14
5 4 2
= 4 5 ÷ 3 = 4 5 × 2 = 4 × 5 = 2
5 4 2 5 4 3 5 6 3
b) 127 × 286 = (22 × 3)7 × (22 × 7)6 = 214 × 37 × 212 × 76 = 214 + 12 × 76 = 226 – 24 = 4
217 × 166 (3 × 7)7 × (24)6 37 × 77 × 224 224 × 77 27 – 6 7
c) (1 – 6–9)–1 + (1 – 69)–1
= 1 – 1 –1 + (1 – 69)–1
69
= 69 – 1 –1 + (1 – 69)–1 = 69 + 1 = 69 – 1 = 69 – 1 = 1
69 69 – – 69 69 – 69 – 69 – 1
1 1 1 1
d) (a 1 + (b 1 + (c 1 = (a – b) + (b – c) + (c – a) = 0
– b)–1 – c)–1 – a)–1
Example 3: Prove that a) 7m + 2 + 4 × 7m =1 b) 5x – 5x – 1 =1
7m + 1 × 8 – 3 × 7m 4 × 5x – 1
c) 273n + 1 × (243)–45n = 1 d) 3–p × 92p – 2 = 2
9n + 5 × 33n – 7 33p – 2 × (3 × 2)–1 3
Solution: 7m + 2 + 4 × 7m 7m × 72 + 4 × 7m 7m (49 + 4) 53
7m + 1 × 8 – 3 × 7m 7m × 71 × 8 – 3 × 7m 7m (7 × 8 – 3) 53
a) LHS = = = = = 1 = RHS
5x – 5x – 1 5x – 5x 5x 1 – 1 4
4 × 5x – 1 4× 5 5 5
b) LHS = = 5x = 5x = 4 = 1= RHS
5 4 × 5
5
c) LHS = 273n + 1 × (243)–45n = (33)3n + 1 × (35)–45n = 39n + 3 × 3–4n
9n + 5 × 33n – 7 (32)n + 5 × 33n – 7 32n + 10 × 33n – 7
= 39n + 3 – 4n = 35n + 3 = 35n + 3 – (5n + 3) = 30 = 1 = RHS
32n + 10 + 3n – 7 35n + 3
3–p × 92p – 2
d) LHS = 33p – 2 × (3 × 2)–1
3–p × (32)2p – 2 3–p + 4p – 4 33p – 4 33p – 4 – 3p + 3 3–1 2
= 33p – 2 × 3–1 × 2–1 = 33p – 2 –1 × 2–1 = 33p – 3 × 2–1 = 2–1 = 2–1 = 3 = RHS
Vedanta Excel in Mathematics - Book 9 176 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Indices
Example 4: Simplify:
a) 3 50ba5c–2 × 3 20ab5c8 b) 3 27x3y6 ÷ 4 81x4y8 c) 3 (a + b)–8 × (a + b)23
Solution:
a) 3 50ba5c–2 × 3 20ab5c8 = 3 1000a5+1 b1+ 5 c–2 + 8 = 3 1000a6 b6 c6 = (103a6b6c6)31 = 10a2b2c2
b) 3 27x3y6 ÷ 4 81x4y8 = (33x3y6)31 ÷ (34x4y8)14 = 3xy2 ÷ 3xy2 = 1.
c) 3 (a + b)–8 × (a + b)23 = (a + b)– 83 × (a + b)32 = (a + b)– 38 + 23 = (a + b) – 83+ 2
= (a + b)– 63 = (a + 1
b)–2 = (a + b)2
Example 5: Simplify: xa a2 + ab + b2 × b b2 + bc + c2 c c2 + ca + a2
Solution: xb a
xx × xxc
xa a2 + ab + b2 × xb b2 + bc + c2 × xc c2 + ca + a2 = (xa – b)a2 + ab + b2 × (xb – c)b2 + bc + c2 × (xc – a)c2 + ca + a2
xb xc xa
= xa3 – b3 × xb3 – c3 × xc3 – a3
= xa3 – b3 + b3 – c3 + c3 – a3 = x0 = 1
Example 6: Simplify: a+b xa2 × b+c xb2 × c+a xc2
xb2 xc2 xa2
Solution:
a2 – b2 b2 – c2 c2 – a2
a+b xa2 × b+c xb2 × c+a xc2 = x a+b
xb2 xc2 xa2 × x b+c × x c+a
= xa – b × xb – c × xc – a = xa – b + b – c + c – a = x0 = 1
111
Example 7 : Simplify: 1 + xa – b + xc – b + 1 + xb – c + xa – c + 1 + xc – a + xb – a
Solution:
111
1 + xa – b + xc – b + 1 + xb – c + xa – c + 1 + xc – a + xb – a
= 1 + 1 + 1
1+ xa + xc 1+ xb + xa 1+ xc + xb
xb xb xc xc xa xa
1 1 1
= xb + xa + xc + xc + xb + xa + xa + xc + xb
xb xc xa
= xb + xc + xa = xb + xc + xa =1
xa + xb + xc xa + xb + xc xa + xb + xc xa + xb + xc
Example 8 : Simplify m + (mn2)31 + (m2n)13 × 1 – n31
m–n m13
Solution:
m + (mn2)31 + (m2n)13 × 1 – n13 = m +m13 n32 + m23 n31 × m13 – n13
m– n m13 m–n m31
= m13 (m32 + n23 + m31 n31 ) × m13 – n31
m– n m13
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 177 Vedanta Excel in Mathematics - Book 9
Indices
= (m31 – n13) [(m13)2 + m31.n13+ (n31)2] = (m31)3 – (n13)3 = m–n =1
m–n m – n m–n
x2 – 1 x× x – 1 y–x
y2 y
Example 9 : Show that = x x+y
Solution: 1 y× 1 x–y y
y2 – x2 y + x
x2 – 1 x× x – 1 y–x x + 1 x x – 1 x x – 1 y–x
y2 y y y y
Here, LHS = =
1 y× 1 x–y 1 y 1 y 1 x–y
y2 – x2 y + x y + x y – x y + x
x + 1 x x – 1 x+y–x x + 1 x x – 1 y
y y y y
= =
1 y+x–y 1 y 1 x 1 y
y + x y – x y + x y – x
x + 1 x x – 1 y xy + 1 x xy – 1 y
y y yy
= ×
y + 1 y – 1 = xy + 1 × xy – 1
x x xx
= xy + 1 × xy x 1 x× xy – 1 × x y= x x× x y= x x + y = RHS
y + y xy – 1 y y y
111
Example 10 : If pqr = 1, prove that 1 + p + q–1 + 1 + q + r–1 + 1 + r + p–1 = 1.
Solution:
1
Here, pqr = 1, then qr = p = p–1
111
Now, LHS = 1 + p + q–1 + 1 + q + r–1 + 1 + r + p–1
qr r 1
= qr(1 + p + q–1) + r(1 + q + r–1) + 1 + r + p–1
qr r 1
= qr + pqr + r + r + qr + 1 + 1 + r + qr
qr r 1 qr + 1 + r
= qr + 1 + r + qr + r + 1 + qr + r + 1 = qr + 1 + r = 1 = RHS
Example 11: If x – 1 = 223 + 231 , show that x3 – 3x2 – 3x = 1.
Solution:
Here, x – 1 = 223 + 213
21
or, (x – 1)3 = (23 + 23 )3 [Cubing on both sides]
3 231 3 + 3. 232 . 213
or, x3 – 3x2 + 3x – 1 = 223 232 + 231
+
or, x3 – 3x2 + 3x – 1 = 4 + 2 + 6 (x – 1)
or, x3 – 3x2 + 3x – 1 = 6 + 6x – 6
or, x3 – 3x2 + 3x – 6x = 1
or, x3 – 3x2 – 3x = 1
Hence, x3 – 3x2 – 3x = 1 proved.
Vedanta Excel in Mathematics - Book 9 178 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Indices
EXERCISE 11.1
General section
1. a) Express am × an as a single base. b) What is the value of (5a)0, a ≠ 0?
c) Find the value of am – n × an – m. d) What is the value of (p + q)0 + 1p + q?
2. Evaluate: b) 5–7 × 58 c) 11–5 ÷ 11–3 d) (25) 3
a) 29 × 2-6 2
e) (64)– 2 f) 1 1 g) 8 – 4 h) 169 – 1
3 64 3 2
6 27 196
i) (32–1)5–1 j) a0 – 2 k) (70.5)2 l) 9 0.5 × 32 0.2
125 3 25 243
m) 3 64–1 n) 729 –1 3 1
64 4 100
3 o) 100 × 4 p) 3 9 3 9 9
Creative section - A
3. Find the value of:
a) 8 – 1 ÷ 4 – 1 b) 125 – 2 ÷ 625 – 1
27 3 9 2 64 3 256 2
c) 27 – 13 81 41 ÷ 4 – 12 d) 25 – 21 125 31 ÷ 8 – 13
8 16 25 16 64 27
4. Simplify: 1 2
3 3
a) (8a3 ÷ 27x–3)– b) (125p3 ÷ 64q–3)– c) 146 × 155 d) 409 × 498
356 × 65 569 × 358
5. Simplify:
a) (xa)b – c × (xb)c – a × (xc)a – b b) (ax + y)x – y × (ay+z)y – z × (az – x)z + x
c) x2a + 3b × x3a – 4b d) 1 1 + 1 1 e) 1 1 + 1 1
xa + 2b × x4a – 3b + ax – y + ay – x – xm – n – xn – m
f) (1 – 3–5)–1 + (1 – 35)–1 g) (a + b)–1.(a–1 + b–1) h) y–1 + x–1 –1
x–1 y–1
6. Prove that:
a) 3x + 1 + 3x = 1 b) 5n + 2 – 5n = 1 c) 72p +1 –3× 49p = 1
4 × 3x 24 × 5n × 49p
4
d) 6m + 2 – 6m = 7 e) 7n + 2 + 4 × 7n = 1 f) 5a + 3 – 55 × 5a – 1 = 1
6m+1 – 6m 7n + 1 × 8 – 3 × 7n 5a + 2 + 89 × 5a
7. Simplify:
a) 3p – 3p –31p b) 5x – 5x – 1 c) 5 × 2m – 4 × 2m – 2
3p + 1 + 4× 5x – 1 3 × 2m + 2 – 5 × 2m + 1
d) 2n +2 × (2n – 1)n + 1 ÷ 4n e) 5–n × 625n –21)–1 f) 9x × 3x – 1 – 3x
2n(n – 1) 53n – 2 × (5 × 32x + 1 × 3x – 2 – 3x
8. Simplify:
a) 25a2b2 × 3 27a3 b) a6b–2c4 ÷ 4 a4b–4c8
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 179 Vedanta Excel in Mathematics - Book 9
Indices
c) 4 16x8y4 ÷ 3 8x6y3 d) 3 56p7q4
3 7p4q7
e) 4 216m7n5 ÷ 4 6–1m–1n f) 3 (a + b)–7 × (a + b) 1
3
g) 3 (2x – y)–8 ÷ (2x – y)– 2 h) (a + b)–1 × (a – b) (a2 – b2)
3
9. Simplify:
a) xa a+b × xb b+c × xc c+a b) ax x–y × ay y–z × az z – x
xb xc xa a–y a–z a–x
c) xa ×a2 + ab + b2 xb ×b2 + bc + c2 xc d)c2 + ca + a2 xl2+m2 l – m× xm2+n2 m–n× xn2+l2 n–l
xb xc xa x–lm x–mn x–nl
e) xa + b c–a × xb + c a–b × xc + a b–c f) xm + n m–n× xn + p n–p × xp + m p – m
xa – b xb – c xc – a xp xm xn
11 1 1 1 1
p + (pq2) 3 + (p2q) 3 q 3 a–c
g) p–q × 1 – h) b b–a × c c–b × a
p 1
3 xb–c xc–a xa–b
x+ 1 a× 1 – x a a2 – 1 a× a – 1 b–a
y y b2 b
i) j)
1 a× 1 a 1 b× 1 a–b
y+ x x – y b2 – a2 b + a
10. Simplify:
a) yz ay × zx az × xy ax b) pq xr – p × pr xq – r × qr xp – q
az ax ay xr – q xq – p xp – r
1 1 1
c) x+y ax2 × y+z ay2 × z+x az2 d) xb xc xa ab
ay2 az2 ax2 bc × ca ×
xc xa xb
111
e) 1 + xa – b + xc – b + 1 + xb – c + xa – c + 1 + xc – a + xb – a
111
f) 1 + ax – y + ax – z + 1 + ay – z + ay – x + 1 + az – x + az – y
Creative section - B
11. a) If a3 + b3 + c3 = 0, prove that (xa + b)a2 – ab + b2 × (xb + c)b2 – bc + c2 × (xc + a)c2 – ca + a2 = 1
b) If a = xq + r.yp, b = xr + p.yq and c = xp + q.yr, prove that aq – r × br – p × cp – q = 1
c) If xyz = 1, prove that 1 + 1 + 1 = 1.
1 + x + y–1 1 + y + z–1 1 + z + x–1
1 1 1
d) If a + b + c = 0, prove that 1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a = 1.
12. a) If x = 1 + 2– 1 , prove that 2x3 – 6x = 5.
3
23
b) If a = 1 – p– 1 , prove that a3 + 3a = p – 1 .
3 p
p3
12
c) If x – 2 = 3 3 + 3 3 , show that x(x2 – 6x + 3) = 2.
Vedanta Excel in Mathematics - Book 9 180 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Indices
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. How many times has x been multiplied in 4x3?
(A) 4 (B) 3 (C) 7 (D) 12
(D) amn
2. For any two positive integers m and n, am × an is equal to (D) 0
(A) a2mn (B) am + n (C) am – n
3. What is the value of (9x) o, x≠0?
(A) 1 (B) 9 (C) 9x
4. The simplified value of 24 × 42÷83 is
(A) 2 (B) 4 (C) 1 (D) 41
2
5. If m + n = m, what is the value of mn?
(A) 0 (B) 1 (C) m (D) n
6. What is the value of ab when a = –3 and b = –2?
(A) 9 (B) –9 (C) 1 (D) – 1
9 9
7. What is the value of a° – 2 ?
64 3
(A) 4 (B) 16 (C) 64 (D) 512
8. What is the simplified form of 1 + 1 1 ?
1 + xa – b + xb – a
(A) 1 (B) x (C) a (D) b
9. The value of 3 99 is
(A) 3 (B) 9 (C) 27 (D) 81
10. The simplified form of 5m+2 – 5m is
5m+1 – 5m
(A) 2 (B) 4 (C) 6 (D) 8
Vedanta ICT Corner
Please! Scan this QR code or
browse the link given below:
https://www.geogebra.org/m/xffnt9b3
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 181 Vedanta Excel in Mathematics - Book 9
Unit 12 Linear Equations
12.1 Simultaneous equations - review
Let’s take a linear equation, y = x + 4.
It is a first degree equation with two variables x and y because each variable is raised
to the first power, i.e. 1.
The standard form of linear equations in two variables is ax + by = c, where a, b, and
c are constants, and a or b both are not zero.
The equation, y = x + 4 has as many pairs of solutions as we wish to find. The table
given below shows a few pairs of solutions.
x 0 1 2 3 –1 –2 –4
y4567320
Thus, (0, 4), (1, 5) (2, 6), (3, 7), (–1, 3), (–2, 2), (–4, 0), … are a few pairs of solutions
that satisfy the equation y = x + 4.
Again, let’s consider another linear equation y = 2x + 1
A few pairs of solutions of this equations are shown in the table below:
x 0 1 3 4 –1 –2
y 1 3 7 9 –1 –3
Thus, (0, 1), (1, 3), (3, 7), (4, 9), (–1, –1), (–2, –3) are a few pairs of solutions that satisfy
y = 2x + 1.
Here, (3, 7) is the common pair of solution that satisfies both the equations
simultaneously. Such pair of equations that have only one pair of solution which
satisfies both the equations simultaneously at a time are called simultaneous
equations.
12.2 Method of solving simultaneous equations
There are various methods of solving simultaneous equations. Here, we discuss only
two methods:
(i) Elimination method (ii) Substitution method
(i) Elimination method
In this method, we add or subtract the given equations to eliminate one of the
two variables by making their coefficients equal. Then, a single equation with
one variable so obtained is solved to find the value of the variable. The value of
the variable is substituted to any one equation to find the value of the eliminated
variable.
Vedanta Excel in Mathematics - Book 9 182 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Linear Equations
Worked-out examples
Example 1: Solve: x + 2y = 11 and x + y = 7.
Solution:
x + 2y = 11 ....................... (i)
x + y = 7 .......................... (ii) The coefficients of x are the
Subtracting equation (ii) from (i) same in both equations. So,
x + 2y = 11 we subtract one equation
+– x +– y =+– 7 from the other to eliminate x.
y =4
Now, substituting the value of y in equation (ii), we get,
x + 4 = 7 Checking:
or, x = 3 When x = 3 and y = 4,
\ x = 3 and y = 4 From equation (i): LHS = x + 2y = 3 + 2 × 4 = 11 = RHS
From equation (ii): LHS = 3 + 4 = 7 = RHS
Example 2: Solve: 2x + 5y = 9 and x – y = 1
Solution:
2x + 5y = 9 ....................... (i)
x – y = 1 .......................... (ii)
Multiplying equation (ii) by 5 and adding to (i), Here we are eliminating y.
2x + 5y = 9 So, to make the coefficient
5x – 5y = 5 of y same, equation (ii) is
multiplied by 5.
7 = 14
or, x = 2
Now, substituting the value of x in equation (ii), we get,
2 – y = 1 Checking:
or, y = 1 When x = 2 and y = 1,
\ x = 2 and y = 1 From equation (i): LHS = 2x+5y = 2×2 + 5×1 = 9 = RHS
From equation (ii): LHS =x – y = 2 – 1 = 1 = RHS
Example 3: Solve: 3 + 2 = 1 and 4 + 3 = 17
Solution: x y x y 12
3 + 2 = 1 ........................ (i)
x y
4 + 3 = 17 ....................... (ii)
x y 12
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 183 Vedanta Excel in Mathematics - Book 9
Linear Equations
Multiplying equation (i) by 3 and (ii) by 2 and subtracting (ii) from (i),
9 + 6 = 3 ........................ (i) Checking:
x y When x = 6 and y = 4,
From equation (i);
± 8 ± 6 = – 34 ....................... (ii)
x y 12
3x+2y 3 2
1 = 1 LHS = = 6 + 4
x 6
= 21+12 = 1 = RHS
or, x = 6
Now, substituting the value of x in equation (i), we get, From equation (ii);
3 + 2 =1 LHS =4x+3y = 4 + 3
6 y 6 4
or, y = 4 = 32+43 = 17 = RHS
\ x = 6 and y = 4 12
(ii) Substitution method
In this method, a variable is expressed in terms of another variable from one
equation and it is substituted in the remaining equations.
Example 5: Solve x + 3y = 1700 and 7x – y = 900
Solution:
x + 3y = 1700 ... (i) and 7x – y = 900 ... (ii)
From equation (i),
x = 1700 – 3y ... (iii)
Substituting for x from equation (iii) in equation (ii), we get,
7(1700 – 3y) – y = 900
or, 11900 – 21y – y = 900
or, 11000 = 22y
∴ y = 500
Now, substituting the value of y in equation (iii), we get,
x = 1700 – 3 × 500 = 1700 – 1500 = 200
\ x = 200 and y = 500.
Example 6: Solve the given system of equation by substitution method.
x–1 = 3 and x+ 2 = 4
Solution: y+1 4 y– 2 3
x–1 = 3
y+1 4
or, 4x – 4 = 3y + 3
or, 4x – 3y = 7
or, 4x = 7 + 3y
7 + 3y
∴ x = 4 ... (i)
Vedanta Excel in Mathematics - Book 9 184 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Linear Equations
Also, x + 22 = 4 Checking:
y – 3 When x = 10 and y = 11,
or, From equation (i);
or, 3x + 6 = 4y – 8
3x – 4y = –14 ... (ii)
Now, substituting the value of x in equation (ii), we get, LHS = yx+–11= 10–1 =192= 3 RHS
11+1 4
3 7 + 3y – 4y = –14
4 From equation (ii);
or, 21 + 9y – 16y = –14 LHS = xy+–22= 10+2 =192= 4 RHS
4 11–2 3
or, 21 – 7y = –56
or, 77 = 7y ∴ y = 11
Again, substituting the value of y in equation (i), we get,
x= 7 + 3× 11 = 10
4
Hence, x = 10 and y = 11.
EXERCISE 12.1
General section
1. Find the value of the variables as indicated.
a) If x = 2 in 2x + y = 5, find the value of y.
b) If x = –1 in x – y = 3, find the value of y.
c) If x = –y in x + 2y = –2, find the values of y and x.
d) If y = 3 in 2x – y = 5, find the value of x.
x
e) If y = 2 in x + y = 9, find the value of x.
2. Select the correct pair of solutions of the given pair of equations.
a) x + y = 7 and x – y = 3 (i) (4, 3) (ii) (6, 3) (iii) (5, 2)
b) 2x – y = –2 and x + 2y = 9 (i) (1, 4) (ii) (4, 6) (iii) (3, 3)
c) y – 3x = 10 and 3y – x = 6 (i) (7, –1) (ii) (–3, 1) (iii) (–1, 1)
d) y = x + 2 and y = 3y – x = 6 (i) (5, 3) (ii) (4, 6) (iii) (3, 5)
e) y = x – 1 and y = x + 1 (i) (5, 2) (ii) (2, 5) (iii) (2, 1)
2 3
Creative section - A
3. Solve each pair of simultaneous equations by elimination method.
a) x + y = 5 b) 3x + y = 13 c) x + 2y = 7
x – y = 1 x–y=3 x+y=4
d) 3x + 5y = 11 e) 2x + 3y = 3 f) 6x + 7y = 5
4x – y = 7 x + 2y = 1 7x + 8y = 6
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 185 Vedanta Excel in Mathematics - Book 9
Linear Equations
g) 2x + 3y – 1 = 0 h) 2x – 3y – 12 = 0 i) 3x + 2y – 15 = 0
5x + 2y + 3 = 0 3x – 2y – 13 = 0 5x – 3y – 25 = 0
j) 4(x – 1) + 5(y + 2) = 10 k) 2(x + 2) – 3(y – 1) = 6 l) 2x + 5 = 4(y + 1) – 1
5(x + 1) – 2(y – 2) = 14 7(x – 1) + 4(y + 2) = 12 3x + 4 = 5(y + 1) – 3
m) 2x + 6 = 3 n) 9 – 4 = 5 o) x + 6 = 4
y x y 5 y
1x0 9 3 8 x 9
– y = 2 x + y =– 3 2 – y = 2
4. Solve each pair of simultaneous equations by substitution method.
a) y = 2x b) y = 5x c) y = –4x d) 3x + 2y = 26
x + y = 9 x + y = 12 x + 3y = – 11 x = y + 2
e) y = x – 3 f) y = 3x + 1 g) x = 3 – y h) x + 2y = 9
3x + 4y = 2 2x + y = 6 2x – y = 3 x = 2y + 1
i) 2x + y = 8 k) x + 3y = 7 j) 2x + y = 4 l) 2x – 3y = 6
x – y = 1 3x – y = 11 2x – y = 8 x + 3y = 3
m) yx +– 1 = 1 n) x–2 = 1 o) x+ y–2 =6
1 2 y–2 2 3
yx – 2 = 1 x + 2 = 3 y + x + 1 = 8
+ 2 3 y + 2 4 2
12.3 Application of simultaneous equations
Simultaneous equations are broadly used to find two unknown quantities. Considering
two unknown quantities which are asked to find in a word problem by any two
variables such as x and y, we can make a pair of simultaneous equations under the
two given conditions. By solving the equations, we can find the unknown quantities.
Worked-out examples
Example 1: The sum of two numbers is 45. If the greater number is 9 more than the
smaller one, find the numbers.
Solution:
Let the greater number be x and the smaller one be y.
From the first given condition, The sum of the number is 45.
x + y = 45 ∴ x + y = 45
or, x = 45 – y ………….. (i)
From the second given condition, The greater number is 9 more than smaller one.
x = y + 9 …………… (ii) ∴x= y+9
Vedanta Excel in Mathematics - Book 9 186 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Linear Equations
Substituting the value of x from equation (i) in (ii), we get,
45 – y = y + 9
or, – y – y = 9 – 45
or, – 2y = – 36
or, y = 18
Now, substituting the value of y in equation (i), we get,
x = 45 – 18 = 27
So, the required greater number is 27 and the smaller one is 18.
Example 2: The total cost of 3 kg of apples and 5 kg of oranges is Rs 1,155. If the cost
of 1 kg of apples is the same as the cost of 2 kg of oranges, find the rate of
cost of these fruits.
Solution:
Let the rate of cost of apples be Rs x per kg.
And, the rate of cost of oranges be Rs y per kg.
From the first given condition,
3x + 5y = 1155 ………… (i)
From the second given condition,
x = 2y
Substituting the value of x from equation (ii) in equation (i), we get,
3 × 2y+ 5y = 1155
or, 6y + 5y = 1155
or, 11y = 1155
or, y = 1155 = 105
11
Now, substituting the value of y in equation (ii), we get,
x = 2y = 2 × 105 = 210
So, the required rate of cost of apples is Rs 210 per kg and that of oranges is Rs 105 per kg.
Example 3: In a fraction, the numerator is 1 less than the denominator. If 1 is added to
21.
the numerator and 5 is added to the denominator, the fraction becomes
Find the original fraction.
Solution:
Let the numerator and the denominator of the original fraction be x and y respectively. Then,
the required fraction is x .
y
From the first given condition,
x = y – 1 .................. (i)
From the second given condition,
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 187 Vedanta Excel in Mathematics - Book 9
Linear Equations
x + 15 = 1
y + 2
or, 2x + 2 = y + 5
or, 2x – y = 3 ............... (ii)
Now, substituting the value of x from equation (i) in equation (ii), we get,
2(y – 1) – y = 3
or, 2y – 2 – y = 3
or, y = 5
Again, putting the value of y in equation (i) we get,
x=y–1=5–1=4
So, the required fraction is 4 .
5
Example 4: The monthly income of Ram and Hari are in the ratio of 3 : 2 and their
expenses are in the ratio of 5 : 3. If each of them saves Rs 4000 in a month,
find their monthly incomes.
Solutions:
Let the monthly income of Ram be Rs x.
And the monthly income of Hari be Rs y.
∴ The monthly expense of Ram = Rs (x – 1,000)
The monthly expense of Hari = Rs (y – 1,000)
From the first given condition,
x 3
y = 2
Forr,o m thexs=eco32nyd ……………………..….. (i)
given condition,
x – 4,000 = 5
y – 4,000 3
or, 3x – 12,000 = 5y – 20,000
or, 3x = 5y – 8,000 ………. (ii)
Substituting the value of x from equation (i) in equation (ii), we get,
3 × 3y = 5y – 8,000
2
or, 9y = 10y – 16,000
or, y = 16,000
Now, substituting the value of y in equation (i), we get.
x = 3 × 16000 = 24,000
2
Hence, the monthly income of Ram is Rs 24,000 and that of Hari is Rs 16,000.
Example 5: Two years ago, father was 6 times as old as his son was. Three years hence
he will be 5 years older than 3 times the age of his son. Find their present
ages.
Vedanta Excel in Mathematics - Book 9 188 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Linear Equations
Solution:
Let the present age of the father be x years.
And the present age of his son be y years.
From the first given condition, Two years ago, the age of the father was (x – 2)
x – 2 = 6 (y – 2) years and that of son was (y – 2) years.
or, x = 6y – 12 + 2 Father was 6 times as old as his son means
x – 2 = 6 (y – 2)
or, x = 6y – 10 ……… (i) 3 years hence, father will be (x + 3) years
From the second given condition, and the son will be (y + 3) years.
5 years older than 3 times the age of the son
x + 3 = 5 + 3 (y + 3) means x + 3 = 5 + 3 (y + 3)
or, x = 5 + 3y + 9 – 3
or, x = 3y + 11 ……………. (ii)
Substituting the value of x from equation (i) in equation (ii), we get,
6y – 10 = 3y + 11
or, 6y – 3y = 11 + 10
or, 3y = 21
or, y = 7
Now, substituting the value of y in equation (i), we get,
x = 6 × 7 – 10
or, x = 32
Hence, the present age of the father is 32 years and that of the son is 7 years.
Example 6: A year hence a father will be 5 times as old as his son. Two years ago,
he was three times as old as his son will be four years hence. Find their
present ages.
Solution:
Let the present age of the father be x years and the present age of the son be y years.
From the first condition,
x + 1 = 5 (y + 1)
or, x = 5y + 5 – 1
or, x = 5y + 4 …………. (i)
Substituting the value of x from equation (i) in equation (ii), we get,
5y + 4 = 3y + 14
or, 5y – 3y = 14 – 4
or, 2y = 10
or, y = 5
Now, substituting the value of y in equation (i), we get, x = 5 × 5 + 4 = 29
Hence, the present age of the father is 29 years and that of the son is 5 years.
Example 7: The sum of the digits of a two–digit number is 8. If 18 is subtracted from the
number, the places of the digits are reversed. Find the number.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 189 Vedanta Excel in Mathematics - Book 9
Linear Equations
Solution:
Let the digit at tens place of the number be x
and the digit at ones place of the number be y.
∴ The number so formed is 10x + y. Let's consider a number 26.
In 26, 2 is in tens place and 6 is in ones place.
From the first given condition, ∴ 26 = 10 × 2 + 6
x+y=8 Similarly, if x is in tens place and y in ones place, the
number is 10x + y.
x = 8 – y …………… (i)
From the second given condition, The original number is 10x + y. When the
10x + y – 18 = 10y + x places of the digits are reversed, the new
or, 10x + y – 10y – x = 18 number so formed is 10y + x.
or, 9x – 9y = 18
or, 9 (x – y) = 18
or, x – y = 2 …………. (ii)
Substituting the value of x from equation (i) in equation (ii), we get,
8–y –y =2
or, – 2y = 2 – 8
or, y = 3
Now, substituting the value of y in equation (i), we get, x = 8 – 3 = 5
Hence, the required number is 53.
Example 8 : A two digit number is 3 times the sum of its digits. The sum of the number
formed by reversing its digits and 9 is equal to 3 times the original number.
Find the number.
Solution:
Let the digit at tens place of the number be x and the digit at ones place of the number be y.
Then, the number so formed is 10x + y.
Again, the number formed by reversing the digits is 10y + x.
From the first given condition, From the second given condition,
10x + y = 3(x + y) (10y + x) + 9 = 3(10x + y)
or, 10x + y = 3x + 3y or, 10y + x + 9 = 30x + 3y
or, y = 7x .......................... (i) or, 7y – 29x = – 9 ...................... (ii)
2
Substituting the value of y from equation (i) in equation (ii), we get,
7 × 7x – 29x =–9
2
or, 49x – 58x = –18
Vedanta Excel in Mathematics - Book 9 190 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Linear Equations
or, –9x = – 18
or, x = 2 7x 7 × 2
2 2
Now, putting the value of x in equation (i), we get, y = = =7
Hence, the required number is 27.
Example 9: Upendra started his motorbike journey from Damak to Lahan at 7 a.m. with
an average speed of 40 km/hr. An hour later Tashi also started his journey
from Damak to Lahan with a uniform speed of 60 km/hr. At what time
would they meet each other?
Solutions:
Suppose Upendra meets Tashi after x hours.
But, Tashi meets Upendra after (x – 1) hour.
Now, the distance travelled by Upendra in x hours = 40x km.
Also, the distance travelled by Tashi in (x – 1) hours = 60(x – 1) km.
When they travelled the equal distance, they meet each other.
∴ 40x = 60(x – 1)
or, 2x = 3x – 3
or, x = 3
∴ They meet after 3 hours.
Since Upendra started his journey at 7 a.m., they meet at 10 a.m.
Example 10: Two buses were coming from two villages situated just in the opposite
direction. The uniform speed of one bus is 10 km/hr more than that of
another one and they had started their travelling in the same time. If the
distance between the villages is 450 km and they meet after 5 hours, find
their speeds.
Solutions:
Let the speed of the faster bus be x km/hr
and the speed of slower bus be y km/hr.
From the first given condition,
x – y = 10 ………..….. (i)
From the second given condition, 450 km
5x + 5y = 450 Distance travelled by faster bus = 5x
Distance travelled by slower bus = 5y
or, 5(x + y) = 450 When they meet, they travelled 450 km.
travelling from opposite direction.
or, x + y = 90 ……………... (ii) So, 5x + 5y = 450
Adding equations (i) and (ii),
x – y + x + y = 10 + 90
or, 2x = 100
or, x = 50
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 191 Vedanta Excel in Mathematics - Book 9
Linear Equations
Now, substituting the value of x in equation (ii), we get,
50 + y = 90
or, y = 40
So, the speed of the faster bus is 50 km/hr and the slower one is 40 km/hr.
Example 11: A few number of students are made to stand in rows. If 5 more students
were kept in each row, there would be 3 rows less. If 5 less students were
kept in each row, there would be 5 rows more. Find the number of students.
Solutions:
Let, the number of rows be x and the number of students in each row be y.
Then, the total number of students = xy
From the first condition, From the second condition,
(x – 3) (y + 5) = xy (x + 5) (y – 5) = xy
or, xy + 5x – 3y – 15 = xy or, xy – 5x + 5y – 25 = xy
or, 5x – 3y = 15 .......... (i) or, –5x + 5y = 25 .......... (ii)
Adding equation (i) and (ii), we get,
5x – 3y – 5x + 5y = 15 + 25
or, 2y = 40
or, y = 20
Now, putting the value of y in equation (i), we get,
5x – 3 × 20 = 15
or, x = 15
Then, the number of students = 15 × 20 = 300
Hence, there are 300 students.
EXERCISE 12.2
General section
1. a) The sum of two numbers is 60 and their difference is 10. Find the number.
b) A number is twice the other number. If their sum is 30, find the numbers.
c) The sum of two numbers is 59. If the smaller number is less than the bigger one by
7, find the number.
d) The difference of two numbers is 12. If the greater number is three times the smaller
one, find the numbers.
e) If three times the sum of two, numbers is 42 and five times their difference is 20,
find the numbers.
f) The total cost of a watch and a radio is Rs 5,000. If the watch is cheaper than the
radio by Rs 1,500, find their cost.
g) The sum of the age of a father and his son is 38 years. If the father is 22 years older
than his son, find their present ages.
Vedanta Excel in Mathematics - Book 9 192 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Linear Equations
h) 1 of the age of a mother is the age of her daughter. If the difference of their ages is
6
35 years, find the age of the mother.
i) Father is three times as old as his son. If the difference of their ages is 24 years, find
their present ages.
j) The perimeter of a rectangular field is 120 m. If the field is 10 m longer than its
breadth, find the length and the breadth of the field.
Creative section - A
2. a) The total cost of 4 kg of apples and 6 of oranges is Rs 1,540. If the cost of 1 kg of
apples is the same as the cost of 2 kg of oranges, find the rate of cost of apples and
oranges.
b) The cost of tickets of a comedian show of 'Gaijatra' is Rs 150 for an adult and Rs 50
for a child. If a family paid Rs 550 for 5 tickets, how many tickets were purchased
in each category?
c) A man buys a few number of books at Rs 72 each and a few number of pens at
Rs 25 each. If he buys 11 articles and pays Rs 510 altogether, find the number of
each article bought by him.
3. a) aIofdri4dgeiindsatlsoufrtbhatcertadicoetnen?do mfrionmatotrheofnthuemoerriagtionraloffraactfiroanct,iiotns ,vaitlsuevableuceombeecso21me. sW31ha. tIfis5thies
b) The numerator of a fraction is 5 less than its denominator. If 1 is added to each, its
value becomes 1 . Find the original fraction.
2
c) If the numerator of a fraction is multiplied by 4 and the denominator is reduced
by 2, the result is 2. If the numerator of the fraction is increased by 15 and 2 is
subtracted from the double of the denominator, the result is 9 . Find the fraction.
7
d) The ratio of the monthly income and expenditure of Sunayana is 5 : 3. If she
increases her income by Rs 5,000 and decreases the expenditure by Rs 3,000, the
new ratio becomes 5 : 2. Find her income and expenditure.
e) The monthly income of Rita and Bishwant are in the ratio of 4 : 3 and their expenses
are in the ratio 3:2. If each of them saves Rs 2000 in a month, find their monthly
income.
4. a) If twice the son's age in years is added to the father's age, the sum is 70. But twice
the father's age is added to the son's age, the sum is 95, find the present ages of the
father and his son.
b) The sum of the ages of the father and son is 44 years. After 8 years, the age of the
father will be twice the age of the son. Find their present ages.
c) Three years ago the sum of the ages of a father and his son was 48 years and three
years hence father's age will be three times that of his son's age. Find the present
ages of the father and his son.
d) Eight years ago, the daughter's age was thrice the son's age. Now the daughter's age
is 4 years more than the son's age. Find their present ages.
e) 2 years ago, father's age was nine times the son's age but 3 years later it will be 5
times only. Find the present ages of the father and the son.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 193 Vedanta Excel in Mathematics - Book 9
Linear Equations
f) 5 years ago a man's age was 5 times the age of his daughter's. 3 years hence, twice
his age will be equal to 6 times his daughter's age. Find their present ages.
g) A year hence a father will be 5 times as old as his son. Two years ago the father was
3 times as old his son will be 4 years hence. Find their present ages.
h) 14 years ago the age of the mother was 4 times the age of her daughter. The present
age of the mother is 2 times the age of her daughter will be 4 years hence. What are
their present ages?
i) 3 years later a mother will be 4 times as old as her son. 3 years ago, the mother's age
was two times her son's age will be 8 years hence. What are their present ages?
j) The present age of the father is three times the age of his son. If the age of the son
after 10 years is equal to the age of the father before 20 years, find the present ages
of the father and the son.
k) The sum of the ages of a father and his son is 40 years. If they both live on till the
son becomes as old as the father is now, the sum of their ages will be 96 years. Find
their present ages.
l) The sum of present ages of a father and his son is 80 years. When the father's age
was equal to the present age of the son, the sum of their ages was 40 years. Find
their present ages.
m) The ages of two girls are in the ratio of 5 : 7. Eight years ago their ages were in the
ratio 7 : 13. Find their present ages.
n) Three years ago, the ratio of the ages of A and B was 4 : 3. Three years hence, the
ratio of their ages will be 11 : 9. Find the present ages of A and B.
5. a) A number consists of two digits. The sum of its digits is 16. if 18 is subtracted from
the number, the digits interchange their place. Find the number.
b) The sum of the digits of a two-digit number is 10. When 18 is added to the number,
its digits are reversed. Find the number.
c) The digit at tens place of a two digit number is two times the digit at ones place.
When 27 is subtracted from the number, its digit are reversed. Find the number.
d) The digit at ones place of a two digit number is three times the digit at tens place.
When 54 is added to the number, its digits are reversed. Find the number.
e) A number consisting of two digits is three times the sum of its digits. If 45 is added
to the number, the digits will be interchanged. Find the number.
f) A certain number of two digits is seven times the sum of the digits. If 36 is subtracted
from the number, the digits will be reversed. Find the number.
g) The sum of the digits in a two-digit number is 11. The number formed by
interchanging the digits of the number will be 45 more than the original number.
Find the original number.
h) A number of two-digits exceeds four times the sum of its digits by 3. If 36 is added
to the number, the digits are reversed. Find the number.
i) A number consists of two digits. If the number formed by reversing its digits
is added to it, the sum is 143 and if the same number is subtracted from it the
remainder is 9. Find the number.
Vedanta Excel in Mathematics - Book 9 194 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Linear Equations
j) A number consists of two digits whose sum is 9. If three times the number is equal
to eight times the number formed by interchanging the digits, find the number.
k) A two digit number is 4 times the sum of its digits. The sum of the number formed
by reversing its digits and 9 is equal to 2 times the original number. Find the
number.
l) Five times the sum of the digits of a two digit number is 9 less than the number
formed by reversing its digits. If four times the value of the digit at ones place is
equal to half of the place value of the digit at tens place, find the number.
Creative section - B
6. a) In a city, the taxi charges consists of two types of a charges: a fixed charge together
with the charge for the distance covered. If a person travels 10 km, he pays Rs 180
and for travelling 12 km, he pays Rs 210. Find the fixed charges and the rate of
charge per km.
b) A lending library has a fixed charge for the first four days and an additional charge
for each day thereafter. Dorje paid Rs 50 for a book kept for 9 days, while Kajal paid
Rs 35 for the book she kept for 6 days. Find the fixed charge and the charge for each
extra day.
7. a) If Prakash gives one of the marbles from what he possesses to Kamala, then they
will have equal number of marbles. If Kamala gives one of the marbles from what
she possesses to Prakash, Prakash will have double number of marbles. Find the
number of marbles possessed by them initially.
b) When Gopal gives Rs 10 from his money to Laxmi her money becomes double than
that of the remaining money of Gopal. When Laxmi gives Rs 10 to Gopal his money
is still Rs 10 less than the remaining money of Laxmi. Find their original sum of
money.
8. a) Janak started his bicycle journey from Kohalpur to Dhangadhi at 6:00 a.m. with an
average speed of 20 km/hr. Two hours later Ganesh also started his journey from
Kohalpur to Dhangadhi with an average speed of 30 km/hr. At what time would
they meet each other if both of them maintain non-stop journey.
b) Two buses were coming from two different places situated just in the opposite
direction. The average speed of one bus is 5 km/hr more than that of another one
and they had started their journey in the same time. If the distance between the
places is 500 km and they meet after 4 hours, find their speed.
9. a) When the length of a rectangular field is reduced by 5 m and breadth is increased
by 3 m, its area gets reduced by 9 sq. m. If the length is increased by 3 m and
breadth by 2 m the area increases by 67 sq. m. Find the length and breadth of the
room.
b) A few number of students are made to stand in a certain number of rows equally.
If 3 more students were kept in each row, there would be 2 rows less. If 3 less
students were kept in each row, there would be 3 rows more. Find the number of
students.
Project work and activity section
10. a) Make a pair of simultaneous equations and solve them to get the following values
of the variables.
(i) x = 3 (ii) x = 5 (iii) x = 4 (iv) x = –2 (v) x = –3
y=2 y=7 y = –1 y=5 y = –2
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 195 Vedanta Excel in Mathematics - Book 9
Linear Equations
b) Think the values of x and y of your own choice. Then, complete the each pair of
equations and solve them. x y
2 3
(i) x + y = ....... (ii) 2x+y = ....... (iii) 3x–2y = ....... (iv) + = .......
x – y = ....... x +2y = ....... 2x – y = ....... x + y = .......
3 2
c) Make word problems reflecting to the real life situations as far as possible from the
given pairs of simultaneous equations, then solve them.
(i) x + y = 30 (ii) x+y = 36 (iii) 3x = y (iv) x + 2y = 70
x – y = 10 x = 2y y – x = 20 x:y=3:2
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. Which of the following is a solution of the equation 3x – y = 7?
(A) (2, 1) (B) (5, 8) (C) (–1, 4) (D) (3, –2)
2. 5 years ago, Mr. Rai was x years, what will be his age after 5 years?
(A) x + 5 (B) x + 10 (C) x – 5 (D) x – 10
3. The present ages of a father and his daughter are 40 years and 10 years respectively,
what will be their ages x years hence?
(A) 40 + x, 10 + x (B) 40 – x, 10 + x (C) 40+x, 10–x (D) 40 – x, 10 – x
4. If the cost of a dozen of bananas is Rs 45 more than that of 7 bananas, what is the cost
of a banana?
(A) Rs 5 (B) Rs 7 (C) Rs 9 (D) Rs 10
5. Mrs. Sharma has only one son. Last year, her son was 11 years old. Now, she is thrice
as old as her son. What was the age of Mrs. Sharma when her son was born?
(A) 23 years (B) 24 years (C) 25 years (D) 36 years
6. The sum of two numbers is 30. If twice the bigger number is thrice the smaller, the
smaller is
(A) 6 (B) 8 (C) 12 (D) 18
7. The sum of three consecutive numbers is 15, what is the greatest consecutive number?
(A) 4 (B) 5 (C) 6 (D) 7
8. Which of the two–digit number has sum of digits 9 and the digit at ten’s place is twice
the digit at one’s place?
(A) 24 (B) 36 (C) 63 (D) 48
9. If x and y are the digits at ten’s and one’s places of a number, what is the number
formed by reversing its digits?
(A) 10x + y (B) 10 (x + y) (C) 10y + x (D) 10xy
10. The corresponding linear equation for “the sum of digits of a two–digit number 10x +
y is 11” is
(A) x + y = 11 (B) 10x + y = 11 (C) 10y + x = 11 (D) xy = 11
Vedanta Excel in Mathematics - Book 9 196 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Vedanta Excel in Mathematics Book 9 Final Re_CTP
Discover the best professional documents and content resources in AnyFlip Document Base.
Search