Green C Math Class 10

EXAMPLE 5 Simplify: 1 1 + 1 1
Solution : – xa – b – xb – a

11
= 1 – xa – b + 1 – xb – a

11

= xa + xb
1 – xb 1 – xa

xb xa
= xb – xa + xa – xb

xb xa
= xb – xa – xb – xa

xb – xa
= xb – xa

= 1

EXAMPLE 6 Simplify: oxxbapa2 + ab + b2 . oxxbcpb2 + bc + c2 . oxxacpc2 + ca + a2
Solution :
o xa pa2 + ab + b2 . o xb pb2 + bc + c2 . o xc pc2 + ca + a2
xb xc xa

= x(a – b) (a2 + ab + b2) . x(b – c) (b2 + bc + c2) . x(c – a) (c2 + ca + a2)

= xa3 – b3 . xb3 – c3 . xc3 – a3

= xa3 – b3 + b3 – c3 + c3 – a3

= x°
= 1

on + 1 m+n om – 1 m+n
mp np
EXAMPLE 7
Simplify: 1m 1n
om2 – n2p on2 – m2p

Solution : on + 1 m+n om – 1 m+n
mp np

= om2 – 1m on2 – 1 n
n2 p m2
p

GREEN Mathematics Book-10 95

on + 1 m + n om – 1 m+n
m np
p

= om + 1 m om – 1 m . on + 1 n on – 1n
n n m mp
p p p

on + 1 m+n–n om – 1 m+n–m
mp np

= om + 1 m . on – 1n
n mp
p

mn + 1 m . mn – 1 n
m n
o p o p

= mn + 1 m mn – 1 n
n m
o p . o p

mn + 1 m mn – 1 n

= o mnm+ 1 p .o n 1 p
mn –
nm

n m mn
omp o n p
.=

m –m m n
onp onp
.=

= m n–m
onp

qr p
xr xp xq
EXAMPLE 8 Simplify: qr r × pr p × pq q

xq xr xp

qr p
xr xp xq
Solution : qr r × pr p × pq q
xq xr xp

q1 r1 p1
x r × qr xp × pr
= × ×r 1 xq × pq
p1 q1

xq × qr x r × pr xp × pq

111

x r2 x p2 x q2
= 1 × 1 × 1

x q2 x r2 x p2

= x 1 + 1 + 1 – 1 – 1 –p12
r2 p2 q2 q2 r2


= x°

= 1

96 GREEN Mathematics Book-10

EXAMPLE 9 Simplify : 1 + 1 + 1
Solution : 1 + xl – m + xn – m 1 + xm – n + xl – n 1 + xn – l + xm – l

111
1 + xl – m + xn – m + 1 + xm – n + xl – n + 1 + xn – l + xm – l

11 1
= + +
1 + xl + xn 1 + xm + xl 1+ xn + xm
xm xm xn xn xl xl

111
= xm + xl + xn + xn + xm + xl + xl + xn + xm

xm xn xl

xm xn xl
= xl + xm + xn + xl + xm + xn + xl + xm + xn

xl + xm + xn
= xl + xm + xn

= 1

EXAMPLE 10 1 11
Solution : If a+b+c = 0, then prove that, 1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a =

11 1 1
1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a

x–a xc 1
= x–a (1 + xa + x–b) + xc (1 + xb + x–c) + 1 + xc + x–a

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

x–a xc 1
= x–a +1 + xc + xc + x–a + 1 + 1 + xc + x–a ( a + b + c = 0)

x–a + xc + 1
= x–a + xc + 1

= 1

GREEN Mathematics Book-10 97

EXAMPLE 11 If ax = by = cz and b2 = ac, then prove that 1 + 1 – 2 = 0
z x y

Solution : Here,

ax = by Raising power 1 on both sides.
x
11

or, (ax)x = (by)x

y

or, a = bx

Also,

by = cz Raising power 1 on both sides.
z
11

or, (by)z = (cz)z

y

or, c = bz

Now,

b2 = ac

yy

or, b2 = bx . bz

yy

or, b2 = bx + z

or, y + y
z
2= x

or, y2 = 1 + 1
z x

or, 1x + 1 = 2
z y

or, 1z + 1 – 2 = 0 Proved.
x y

EXERCISE 6.1

A. Very Short Questions

1. Evaluate: b. 38 ÷ 36 c. 100 d. 3 × 40 e. 10–3 f. 27–1
a. 56 ÷ 54
g. 2–2 h. 49 1 i. 8 2 j. 32 3 k. (54) 1 l. 25 3
2 3 5 4 2

2. Simplify: b. m × m2 c. x4 × x2 d. y2 × yb e. am × an f. p9 ÷ p2
a. x5 × x3 h. a7 ÷ a–2 i. (ab)0 j. a × c0
g. b4 ÷ b2 k. x + y0 l. (a3)4
n. (x–1y5)­–1 o. am × an p. x4 ÷ x–3 1
m. (x2y4)2 q. p–1 r. y–3
oz

98 GREEN Mathematics Book-10

B. Short Questions

3. Simplify:

a. 3n × 32n × 33n b. a3 × a5 × a–2 c. (x2y)4 × (y2x)5 d. a3b–2 × (a2b2)4

e. (a2)0 × (a 1 )4 f. 150mm––24mm g. x–1 + y–1 h. p+q
2 x+y p–1 + q–1

4. Simplify:

a. 3 m7 4 m–3 m–2 b. 3 (16)–1 c. 4 x3 3 x2 x2

5. Simplify:

a. 215199 ×× 355 b. 122166 ×× 248933 c. 134555××14033
255

6. Simplify:

a. o 12275 p 1 × qo 3125 p 1 ÷ o 3 p–2r b. o 81 p– 3 × qo 25 p– 3 o 5 p–3r
3 243 5 5 16 4
9 2÷ 2

c. o 2156 p– 1 × qo 125 p 1 o 8 p– 1
2 64
3÷ 27 3r

7. Simplify:

a. 3 3–1a14 × 3 81a–2 b. 4 8a5b6 ÷ 4 2–1ab–2
c. 4 8x5y6 × 4 2x3y–2 d. 3 9x6y4 × 3 3x6y8
e. 3 27x6y3 × 4 16x12y4 f. 9a4b6 × 3 8a9b6
g. 3 54x8y16 ÷ 3 2x2y7 h. 3 27x6y7 × 3 8x–3y8

2

i. 3 27a6b9 ÷ 4 81a8b–12 j. 3 (a+b)–5 . (a + b) 3

8. Simplify:

a. 3n × 5 – 3n b. 2n × (2–1)n × 1
2n + 1 × 2n – 1 4–n
3n + 2 – 3n + 1

c. 273n + 1 . 81–n d. 3n + 4 – 6 × 3n + 1

32n + 5 × 33n – 7 3n + 2

e. 3 × 3n + 1 f. 3.2n – 4.2n – 1
(n – 1) (n + 1)
2n –2n – 1
3n(n – 1) × 9n + 1

g. 210 + n × 43n – 5 h. (27)3n + 1 . – 4n
5
24n + 1 × 23n – 1 (243)

i. 11n + 2 – 55.11n – 1 9n + 5 . 33n – 7

11n × 116 j. 7n + 1 – 28 × 7n – 1

7n × 45

GREEN Mathematics Book-10 99

C. Long Questions

9. Simplify:

a. 1 + 1 b. 1 +1
1 – ax – y 1 – ay – x 1 + pq –r
1 + pr – q

c. oa x–1y p 1 oa 1 p 1 oa 1 p 1
y–z y–z z–x z–x x–y

10. Simplify:

a. (ay – z)x × (az – x)y × (ax – y)z

b. o xp pp +q × o xq pq + r × o xr pr + p
xq xr xp

c. oxx–pq pp – q × o xq pq – r × o xr pr – p

x–r x–p

d. oxl 2x+–lmm2 l – m × oxmx2–m+nn2pm – n × o xn2 + l 2n – l
x–nl
p p

e. (xa ÷ xb)a2 + ab + b2 (xb ÷ xc)b2 + bc + c2 (xc ÷ xa)c2 + ca + a2

f. o xxmn pm – p + n × o xn pn – m + p × o xp pp – n + m
xp xm

g. o ay py + z – x × o az pz + x – y × o ax px + y – z
ay
az ax

h. s(xa.xb)a – b . o xb pb + ct × o xc pc + a
xc xa

i. o xxaa + b pc – a o xb + c pa – b o xc + a pb – c
– b xb – c xc – a

11. Simplify:

1m 1n ab
bp bp a a–b b a–b
oa + × oa – b. ooba2+–b1a1p2ap+bb××ooab2 – 1 a+ b o1 + bp × o1 – ap
– ap
a. 1m 1n 1a c. a a b
ap ap b2 p b ob
ob + × ob – oa + 1p a – b × – 1p a – b

12. Simplify:

a. ab xxba xb xc abc
xc xa xb xc xa
× bc × ca b. ab b × bc c × ca a
xa xb xc

c. x1y aayx11 1 1 1 1
yz zx az
× ay × 1 d. x + y zx2 – y2 ×y+z zy2 – z2 × z + x zz2 – x2
1 ax

az

100 GREEN Mathematics Book-10

13. Show that:

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

b. 1 + ax 1 + ax – z + 1 + 1 =1
1 + ay – x + ay – z 1 + az – x + az – y
–y

14. If a = bx, b = cy and c = az, then prove that xyz = 1.

15. If a + b + c = 0, then prove that: 1 + 1 + 1 =1
1 + xa + x–b 1 + xb + x–c 1 + xc + x–a

16. If x = aq + r × bp, y = ar + p ×bq, z = ap + q × br, then prove that:

xq – r × yr – p × zp – q = 1

17. If a = 5x, b = 5y and ay.bx = 25, then prove that: xy = 1.

6.3 Exponential equations

Let's take an equation 3x = 27. In this case base is 3 and its exponent is unknown variable.
Such type of equation is exponential equation.
We can define this as : an exponential equation is one in which a variable occurs in the
exponent.
We should follow the following rules for solving the exponential equations.

Steps of solving the exponential equations
i. Reduce the given exponential equation in the simplest form i.e. a single term to

both sides of the equation.
ii. Use the law of the indices.
iii. Compare the base of both sides of the equation and equate them.
iv. Simplify the remaining equation and calculate the unknown value.

GREEN Mathematics Book-10 101

Worked Out EXAMPLES

EXAMPLE 1 Solve : 3x = 1
27

Solution : Here,

3x = 1
27

or, 3x = 1
33

or, 3x = 3–3

∴ x = –3

EXAMPLE 2 Solve : 2x+2 + 2x = 80
Here,
Solution :

2x+2 + 2x = 80
or, 2x × 22 + 2x = 80

or, 2x (22 + 1) = 80

or, 2x × 5 = 80

or, 2x = 80
5

or, 2x = 16

or, 2x = 24

∴ x = 4

EXAMPLE 3 Solve : 3x+2 × 5x–1 = 405
Here,
Solution :

3x+2 × 5x–1 = 405

or, 3x × 32 × 5x = 405
5

or, 3x × 9 × 5x = 405 × 5

or, (3 × 5)x = 405 × 5
9

or, (15)x = 225

or, 15x = 152

∴ x =2
102 GREEN Mathematics Book-10

EXAMPLE 4 Solve : 3x + 2 – 3x + 1 = 6
Here,
Solution :

3x+2 – 3x+1 =6
or, 3x. 32 – 3x . 31 =6

or, 3x(9 – 3) =6

or, 3x (6) =6

or, 3x = 1

or, 3x = 30

∴ x =0

EXAMPLE 5 Solve : 16x – 5 × 4x + 1 + 64 = 0
or, 42x – 5 × 4x × 4 + 64 = 0
Solution :

or, (4­x­)² – 20 × 4x + 64 = 0

or, (4x)² – 16x4x – 4 × 4­x + 64 = 0

or, 4x (4x – 16) – 4 (4x – 16) = 0

or, (4x – 16) (4x – 4) = 0

Either, 4x – 16 = 0

or, 4x – 4 = 0

When 4x – 16 = 0

or, 4x = 16

or, 4x = 4²

or, x = 2
When, 4x – 4 = 0
or, 4x = 4
or, 4x = 41

or, x = 1

∴ x = 1, 2

GREEN Mathematics Book-10 103

EXAMPLE 6 Solve : 5a + 1 = 25 1
5a 25

Solution : Let 5a = x
Then
x + 1 = 626
x 25

or, x² + 1 = 626
x 25

or, 25x² + 25 = 626x

or, 25x² – 626x + 25 = 0

or, 25x² – 625x – x + 25 = 0

or, 25x (x – 25) – 1 (x – 25) = 0

or, (x – 25) (25x – 1) = 0

Either,

x – 25 = 0

or, x = 25

Substituting x = 5a

then 5a = 25

or, 5a = 5²

or, a = 2

Again,

25x –1 = 0

or, 25x = 1

or, x = 1
25

Substituting value of x, we get,
5a = 5–2

or, a = – 2

∴ a = ± 2

104 GREEN Mathematics Book-10

EXAMPLE 7 Solve : 4 × 3x + 1 = 27 + 9x

Solution : 4 × 3x + 1 = 27 + 9x

or, 9x – 4 × 3x + 1 + 27 = 0

or, (3x)² – 4 × 3x × 3 + 27 = 0

or, (3x)2 – 12 × 3x + 27 = 0

or, (3x)² – 9 × 3x – 3 × 3x + 27 = 0

or, 3x (3x – 9) – 3(3x – 9) = 0

or, (3x – 9) (3x – 3) = 0

Either

3x – 9 = 0

or, 3x – 3 = 0

When, 3x – 9 = 0 or 3x – 3 = 0

or, 3x = 9 3x = 3

or, 3x = 3² 3x = 31

or, x = 2 or, x = 1

∴ x = 1 and 2

EXAMPLE 8 Solve : 73a – 4 × 52a – 3 = 72a + 2 × 5a + 3
73a – 4 × 52a – 3 = 72a + 2 × 5a + 3
Solution :

or, 73a - 4 × 52a–3 = 1
72a + 2 × 5a + 3

or, 73a - 4 - 2a - 2 × 5 2a-3-a-3 = 1
or, 7a-6 × 5a-6 = 1
or, (7 × 5)a-6 = 1
or, (35)a-6 = (35)0
or, a – 6 = 0
∴ a = 6

GREEN Mathematics Book-10 105

EXERCISE 6.2

A. Very Short Questions

1. Solve: b. 101 – x = 104 c. 3x = 9
a. 5x = 53

d. 3x–4 = 9x–6 e. 32x–1 = 27 f. 2x = 2

g. ( 2 )3x–1 = ( 2 )x–2 h. (0.5)x = (0.25)2 i. 3x = 1
27
1
j. yx ÷ y2 = 1 k. 3m+2 = 27m

B. Short Questions

2. Solve:

a. 4x–1 = 1 b. 8x–2 = 8 c. 3x+1 – 3x = 18
64

d. 2x + 2x–2 = 5 e. 3x+1 × n1–x = 32x–1­ × n3–2x f. 54a+4.m1–a = 25a+3.ma–1

g. 2x+3 . 3x+4 = 18 h. 2x–5 . 5x–4 = 5 i. 3x–2 + 3x = 10
9

j. 72x+1 ÷ 52x+1 = 7 k. 73 . 49x+1 = 494x
5 7

C. Long Questions

3. Solve:

a. 4x + 1 = 16 1 b. 4x – 6 × 2x + 1 + 32 = 0
4x 16
1
c. 5a + 1 + 52 – a = 126 d. 3x + 3 + 3x = 28

e. 5 × 4x + 1 – 64 = 16x f. 2x + 1 = 17
2x 4

g. 2x + 3 + 1 = 9 h. 4x + 4–x = 257
2x 16

i. 9a – 10 × 3a + 9 = 0 j. 5t2 . 5–3t = 1
25

k. 7x + 343 = 56 l. 3 × 2x+1 – 4x = 8
7x

4. a. If xa = y, yb = z and zc = x, prove that abc = 1.

b. If a = bc, b = ca and c = ab, prove that abc = 1.

c. If 2x = 3y = 12z, prove that 2 + 1 – 1 = 0.
x y z

d. If a x = b y = c z and xyz = 1, prove that a + b + c = 0.

e. If 2a = 5b = (100)c, prove that 2(a + b) = acb.

106 GREEN Mathematics Book-10

f. If a3 + b3 + c3 = 1, prove that xa a2 –ab+b2 xb .b2 –bc+c2 xc = xc2 –ca+a2 2
x-b x-c x-a
.

g. If y = 1 – -1 show that y3 + 3y = ab – a1b .

(ab)3 (ab) 3 ,

21

h. If c – 2 = 23 + 23, find the value of c3 – 6c2 + 6c – 2.

6.4 Surd

Rational Number For example :
5
The set of numbers which can be expressed in the
p i. 5 is a rational number.
20
form of q where p and q are integers and q ≠ 0 is
called rational number which is denoted by 'Q'. ii. 9 is also rational number.

Irrational Number For example :

The numbers except rational numbers are called 2 = 1.4114213562 .........
irrational numbers. It is neither terminating nor
repeating decimal. A number which cannot be 3 = 1.7327 ...............
Thus, 4 3 , 5 2 and 5 3 4 , etc.
p are irrational numbers.
ceaxlpleredsasendirirnatthioenfaolrnmumq bwehr.ere p, q ∈ z and q ≠ 0 is

The irrational number is non-terminating and
non-recurring decimal. Thus, a non-terminating
and non-recurring decimal is called an irrational
number. It is denoted by Q.

Real Numbers

The numbers formed with rational R Set of irrational numbers (Q)
and irrational numbers together are
called set of real numbers. Therefore, Q
the union of rational numbers and
irrational numbers is called set of real Q Set of rational numbers (Q)
numbers. The set of real numbers is
denoted by 'R'. N WZ Set of integers numbers (Z)
Set of whole numbers (W)

Set of natural numbers (N)

In the given figure, every rational number is a real number and every irrational
number is a real number.

So we can write N⊂W⊂Z⊂Q⊂R. The relation between set of real numbers is shown
in the following charts.

GREEN Mathematics Book-10 107

Real Number (R)

Rational Number (Q) Irrational Number (Q)

Integers (Z) Fraction (F)

Negative Integers (Z) Whole Number (W)

Zero Natural Numbers (N)

Odd Numbers Even Numbers

Surd

If n b (read as nth root of b) is an irrational For example:
number, then n is called radical (or order),
n is called radical sign and ‘b’ is called 4 2 is an irrational number in which 4
radicand of the irrational number. is radical sign, 4 is called radical and 2 is
called radicand of 4 2.
Now, nth root of a rational number may be
rational or irrational number. 4 16 = 2 which is a rational number.

Here, 3 6 is a surd but 4 16 is not a surd. 3 6 which is an irrational number.

Let n be a natural number. If nth root of a 3, 3 2, 4 , 3 –2 etc. are surds because
rational number is an irrational number, 5
then it is called a surd. So, if the value of 4
nth root of a rational number cannot be 3, 2, 5 , – 2 are rational numbers and 3,
determined exactly, then it is said to be a
surd. 3 2, 45, 3 –2 are irrational numbers.
1
If the surd is expressed in the form n b
where n ∈ N and b ∈ Q, then ‘n’ is called A surd n b is also written as (b) n .
order, is called root sign and ‘b’ is
called radicand of the surd.

108 GREEN Mathematics Book-10

Laws of surd

Since n b can be written as 1 so laws of indices can be applied to the surds.

bn,

Let m, n, p ∈ N and a, b ∈ Q.

Laws Examples =4
1. n (a) n = a
3

3 (4)3 = 43

2. n a . n b = n ab 3 2 . 3 5 = 3 10

3. na =n a 4 2 4 2
nb b 4 3
3=

4. m n a = mn a = n m a 4 3 2 = 12 2

5. n am = np amp 3 24 = 3×2 24×2 = 6 28

Like Surds For example :
2 , 7 2 , 32 , etc. are similar surds.
The surds are said to be like (or
similar) if they have same irrational
factors. Thus, two surds are similar
if they can be reduced so as to have
same order and same radicand.

Unlike Surds For example :
3 and 3 7 are unlike surds.
The surds are said to be unlike (or
dissimilar) if they cannot be reduced
so as to have the same irrational
factors. Thus, two surds are unlike if
they cannot be reduced so as to have
same order and same radicand.

GREEN Mathematics Book-10 109

Addition or subtraction rule

Like surds can be added or subtracted. For example :

i. 3 3 + 2 3 ii. 8 2 – 3 2

= 5 3 = 5 2

Thus, like surds are to be added or subtracted by adding or subtracting the rational
coefficients and the common irrational factor is kept the same.

∴ an x ± b n x = (a ± b) n x

For example : 2 2 + 7 2 = (2 + 7) 2 = 9 2

4 3 4 – 2 3 4 + 3 3 4 = (4 – 2 + 3) 3 4 = 5 3 4

Multiplication and Division rule

Rule (1)

If the surds are of same order, then multiply or divide rational and irrational factors

separately. Then, reduce it to the simplest form. We have.

m n a × p n b = (mp) n ab and mn a ÷ p n b = m n a
p b

Multiply: 6 4 2 by 2 4 8 Divide: 5 3 18 by 10 3 9
3
2 5 3 18
64 2 × 3 4 8 10 3 9

= 6 × 2 4 2.8 5 18
3 10 9
= 3
= 4 4 16

= 4 4 24 = 1 3 2
= 4×2 2
= 8

Rule (2)

If the surds are of different orders, then reduce them into equivalent surds of the same
lowest order, then proceed as rule (1).

We have,

Multiply: p. n a × q m b = p mn am × q mn bn = pq mn am bn

Divide: p na = p mn am = p mn am
q mb q mn bn q bn

110 GREEN Mathematics Book-10

Multiply: 4 2 × 5 3 2 Divide: 6 2 by 3 3 2

L.C.M. of radical 2 and 3 = 6 62
4 2 × 5 3 2 = 20 6 23 × 6 22 33 2

= 20 6 23×22 = 2. 6 23
6 22

= 20 6 32 = 2. 6 23 = 2.6 2
22

Methods for solving surds

1. Reduce the surds by using four fundamental operations in the simplest form.

2. Add or subtract the rational coefficients of the like surds keeping the irrational
factor same.

For example: 7x – 2x = 5x

7 3 –2 3 =5 3

3. Connect the like and unlike surds with appropriate + or – sign.

Properties of surds

i. The sum or difference of a rational number For example: ii. 4 – 2
and an irrational number is an irrational i. 3 + 5
number.
For example:
ii. Product of a rational number and an
irrational number is always an irrational i. 3 × 2 = 3 2
number. It is also called mixed surd.
ii. 5 ÷ 32 5
= 32
iii. If a and b are two different rational numbers

a rational number between them = a+ b . We find the other rational number
If a, b ∈ Q and a < b, then a < a+b < 2 between two rational numbers
adopting this process.
2 b.

GREEN Mathematics Book-10 111

Worked Out EXAMPLES

EXAMPLE 1 Simplify: 4 12 – 50 + 7 48
4 12 – 50 + 7 48
Solution : = 4 22×3 – 52×2 + 7 24×3
= 8 3 – 5 2 + 28 3
EXAMPLE 2 = (8 + 28) 3 – 5 2
= (– 20 3 – 5 2 )
Solution : = – (36 3 – 5 2 )
Simplify: 3 2 + 4 2500 + 4 64 + 6 8

3 2 + 4 2500 + 4 64 + 6 8
= 3 2 + 4 54 × 22 + 4 26 + 6 23
= 3 2 + 5 4 22 + 2 4 22 + 12 2
= 3 2 + 5 4 4 + 2 4 4 + 12 2
= (3 + 12) 2 + (5 + 2)4 4
= 15 2 + 7 4 4

EXAMPLE 3 Find the square of : 2 3 – 3 2

Solution : The square of (2 3 – 3 2) = (2 3 – 3 2)2
= (2 3)2 – 2 × 2 3 × 3 2 + (3 2)2
= 4 × 3 – 12 3 × 2 + 9 × 2
= 12 – 12 6 + 18
= 30 – 12 6
= 6(5 – 2 6)

EXAMPLE 4 Simply: a. 4 7 × 6 10 b. 4 3 192 – 4 3 375 + 2 2 24

Solution : a. 4 7 × 6 10
Taking L.C.M. of 4 and 6 = 12

4 7 = 12 7³ = 12 343
6 10 = 12 10² = 12 100
∴ 4 7 × 6 10 = 12 343×100 = 12 34300

112 GREEN Mathematics Book-10

b. = 5 3 192 – 4 3 375 + 2 3 24
= 5 3 64 × 3 – 4 3 125 × 3 + 2 3 8 × 3
=5×4 3 3 –4×5 3 3 +2×2 3 3
= 20 3 3 – 20 3 3 + 4 3 3
= (20 – 20 + 4) 3 3
=43 3

EXAMPLE 5 Simplify: (1 + 3 ) (1 + 3 )

Solution : (1 + 3 ) (1 + 3 )

= 1(1 + 3 ) + 3 (1 + 3 )

= 1 + 3 + 3 + 3 × 3

= 1 + 2 3 + 3

= 4 + 2 3

EXERCISE 6.3

A. Very Short Questions

1. Simplify :

a. 32 b. 48 c. 125 d. 325 e. 20 × 70

f. 32 + 72 g. 48 – 75 + 27 h. 6x 5 – 80x² + 45x²
5 5x²

2. Simplify the following:

a. 2 ( 6 + 54) b. 6 ( 2 + 8 ) c. ( a – x ) ( a + x )
d. (2 + 3 ) (1 + 3 ) e. (3 – 5 ) (3 – 2 5 ) f. (5 – 2 ) (5 + 2 )
g. ( 3 + 5 ) ( 3 + 5 )

3. Add the following:

a. 12 + 27 b. 80 + 45 c. 5 3 7 + 3 3 7
d. 6 2 + 7 2 + 3 2 e. 4 2 + 2 32 f. (2 a + b ) + ( a – b )

GREEN Mathematics Book-10 113

4. Subtract the following:

a. 54 – 24 b. 2 3 4 – 6 3 4 c. 10 3 – 7 3

d. 7 12 – 27 e. 5 27 – 6 3

B. Short Questions
5. Simplify:

a. 8 + 3 245 – 125 b. 125 – 45 + 5

c. 4 45 – 3 80 + 4 125 d. 4 3 – 3 12 + 2 75

e. 3 128 + 2 3 54 – 2 3 250 f. 2 3 16 – 2 3 250 + 4 3 54

g. 2 3 4 + 7 3 4 + 7 3 32 – 3 500 h. 4 81 – 8 3 216 + 15 5 32 + 225

i. 3 4 162 – 7 4 32 + 4 1250 j. 5 3 54 – 2 3 16 + 4 3 686

k. 3 625 –243–435 + 6 4 243 l. 5 3 1 – 2 1 + 2 18
27 4 3

6. Multiply:

a. 6 × 150 b. 8 × 50 c. 3 32 × 3 250

d. 2 3 2 × 33 4 e. 3 7 × 2 f. 3 2 × 4 3

g. 2 3 × 5 3 × 48 h. 3 2x5y4 × 3 4x7y5 i. 3 4a5b–2 × 3 54a–2b11

j. 4a2b6 × 3 8a9b12

7. Divide:

a. 242 ÷ 72 b. 6 12 × 3 c. 25 3 25 ÷ 3 3 15

d. 12 96 by 3 24 e. 21 384 ÷ 8 98 f. 3 24a7b11 ÷ 3 3a4b9

g. 4 32x7y11 ÷ 4 2x–1y–1 h. 3 27a6b6 ÷ 4 81a8b8

8. Simplify:

a. (3 2 – 5) × 2 2 b. (2 3 – 5 2 ) (2 3 + 5 2 ) c. (3 5 – 2 3 ) (2 3 + 3 5 )

d. (3 2 + 3 )2 e. (4 2 – 3 5 )2 f. ( a + a – 1 ) × a – 1

g. a + b ( a + b + a – b ) h. ( 18 × 6 ) ÷ 75

9. Simplify:

a. 3 128 – 3 16 b. 18 – 3 32 c. 6 4 32 +2441262 – 5 4 2
23 2
2

114 GREEN Mathematics Book-10

6.5 Rationalization of Surds

If the product of two surds is a rational number, then each surd is called the rationalizing
factor (abbreviated R.F.) of the other surd.

The process of reducing a surd to a rational number by multiplying it with a suitable
rationalizing factor is called the rationalization of the surd.

Example

1. 2 3 × 3 = 2 × 3 = 6, which is a rational number. So, 3 is R.F. of 2 3 .

2. ( 3 – 5 ) ( 3 + 5 ) = 3 – 5 = = – 2 which is a rational number. So, 3 + 5 is R.F. of 3 – 5

Rationalization of a Monomial Surd

The simplest rationalizing factor of a monomial surd n a (i.e. 1 where n is a natural

an)

number is a– 1 .
n

Example:

The R.F. of 2 is 2 .

The R.F. of 3 2 is 2 .

The R.F. of 34 1 z = 4

is 4 3 = 43

Rationalization of Binomial Surds

Two binomial surds are said to be conjugate of each other if they differ only in sign +
or – between the two terms.

Thus, a + b and a – b are conjugate of each other. Also, a c + b d and a c – b d are
conjugate to each other. Here, the simplest rationalizing factor of a binomial surd of
order 2 is its conjugate.

Example:

The rationalizing factor of 2 – 3 is 2 + 3 .

The rationalizing factor of 2 + 3 is 2 – 3 .

Note:

1. If a + b = c + d , where a and c are rational numbers and b and d are
irrational numbers then, a = c and b = d.

2. If a + m b = c + n b , where a, c, m, n are rational numbers and b is an irrational
number, then a = c and m = n.

GREEN Mathematics Book-10 115

Worked Out EXAMPLES

EXAMPLE 1 Rationalize the denominator of (i) 3 (ii) 5 3
5
72

Solution : i. 3 = 3 × 5 = 35
5 5 5 5

ii. 5 3

72

= 5 3 × 2

72 2

= 5 3×2

7×2

= 5 6

14

EXAMPLE 2 Rationalize the denominator of 3 – 2 2 .

3 +2 2

Solution : 3 –2 2

3 +2 2

= 3 – 2 2 × 3 – 2 2

3 +2 2 3 –2 2

= ( 3 – 2 2 )2

( 3 )2 – (2 2 )2

= 3 – 2× 3 ×2 2 + 8

9–8

= 11 – 4 6

1

= (11 – 4 6 )

EXAMPLE 3 Rationalize the denominator of 2+ 1 5 .
3+

Solution : 1
2+ 3+ 5

= ( 2+ 1 5 × ( 2+ 3 )– 5
3) + ( 2+ 3 )– 5

= 2 + 3 – 5

( 2 + 3 )2 – ( 5 )2

= 2 + 3 – 5

2 + 2× 2 × 3 + 3 – 5

= 2 + 3 – 5 × 6

26 6

= 12 + 18 – 30

2×6

= 2 3 + 3 2 – 30

12

116 GREEN Mathematics Book-10

EXAMPLE 4 Simplify: 5+ 3 + 5– 3
5– 3 5+ 3

Solution : 5+ 3 + 5– 3

5– 3 5+ 3

= ( 5 + 3 )2 + ( 5 – 3 )2

( 5 )2 – ( 3 )2

= 5 + 2 5× 3 + 3 + 5 – 2 5× 3 + 3
5–3
16
= 2

=8

EXAMPLE 5 Simplify: 32 – 43 + 6
6– 3 6– 2 2+ 3

Solution : 32 – 43 + 6
6– 3 6– 2 2+ 3

= 3 2 × 6+ 3 – 4 3 × 6+ 2 + 6 × 2– 3

6– 3 6+ 3 6– 2 6+ 2 2+ 3 2– 3

= 3 12 + 3 6 – 4( 18 + 6 ) + 12 – 18
6–3 6–2 2–3

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

3 4 –1

= 2 3 + 6 – (3 2 + 6 ) – (2 3 – 3 2 )

= 2 3+ 6 –3 2– 6 –2 3+3 2

=0

EXAMPLE 6 Simplify: 3+ 2 = a+ b 6 , find the value of a and b.
2 3– 2

Solution : 3+ 2 =a+b 6
o
2 3– 2

or, 3+ 2 × 2 3 + 2 = a + b 6
2 3– 2 2 3 + 2

or, 3 + 6 + 2 6 + 2 = a + b 6

(2 3 )2 – ( 2 )2

or, 3 + 3 6 = a + b 6

12 – 2

or, 3 + 3 6 = a + b 6

10

or, 15012+ 3 6 =a+b 6
10

Comparing both sides

a= 1 ,b= 3
2 10
1 3
∴ The value a and b are 2 and 10 respectively.

GREEN Mathematics Book-10 117

EXERCISE 6.4

A. Very Short Questions

1. Rationalize the denominator of:

a. 2 b. 1 c. 2 d. 23 e. 150
6 3 5 6

2. Rationalize the denominator of:

a. 1 b. 3 + 1 c. 2 + 3 d. 4 + 2 e. 2 3 1 5 f. 7 +3 5
3 –1 +3
3 –1 2– 3 2 +1 7–3 5

B. Short Questions

3. Rationalize the denominator of:

a. 3 – 2 2 b. 3– 1 5 c. a² + 1 + a² – 1 d. 5 + 2
5– 2+ a² + 1 – a² – 1 5– 2

4. Simplify:

a. 4– 15 + 4 1 b. 3+ 8+ 1 c. 5+ 24 + 5 1
– 15 3+ 8 + 24

5. Simplify:

a. 1 3 – 1 b. x+ a – x– a
6+ 6– 3
x– a x+ a

c. 5 + 3 + 5 – 3 d. 1 1 + 1 2 + 1 2
f. 2+ 3+ 3+
5– 3 5+ 3

e. 2 1– 3 – 2 3 + 3 2 32 – 43 + 23
5+ 5–
6– 3 6– 2 6+ 2

g. 2 +6 3 + 32 – 43 2 h. 1–x + 1–x
6+ 2 6+ 1– 1–x 1+ 1–x

i. 5 2 – 8 5 + 3 10

5 ( 2 + 1) 2 + 10 5+ 2

6. If a and b are rational numbers, find the values of a and b:

a. 3 – 1 = a + b 3 b. 2 3 + 5 = a + b 15 c. 6 – 1 = a + b 6 d. 3 + 2 = a + b 6
3 +1 2 3– 5 6 +1 3 3 –2 2

7. If x = 65 + 8, find the value of : 1 1
i. x – x ii. x + x

1 i. a– 1 ii. a2 – 1
8. If a = 3 – 1 , find the value of : 2a a2

9. If x = 3 – 2 and y = 3 + 2 , find the value of x2 + xy + y2.
3+ 2 3– 2

6.6 Equation Involving Surds

Equation containing the variable under a radical sign in the form of a surd is called an
equation involving surds. e.g. x + 3 = 4, x – 1 = x + 2 .

118 GREEN Mathematics Book-10

For solving such types of equations we have to raise even power like two or four or six or
eight, etc. on both sides then we may get some additional values including the solutions
of the given equation because square of positive or negative is positive. Thus, in such
cases we have to check the values, so obtained by substituting the values in the original
equation.

But in solving an equation, if we have to raise the odd power like three, five or seven, etc.
then it is not necessary to check the solution because in raising power by odd number of a
positive number is positive and negative number is negative, so we do not get additional
values.

Method to solve an equation involving surds

1. If there is only one term containing the surd in an equation, then keep the term
containing the surd on one side and remaining terms on the other side of the
equation.

Then, raise power on both sides of the equation by the order of the surd.

Again, solve the simple linear equation or quadratic equation. If we have raised
power by 2 or 4 or 6 or ......., then check the solutions whether they satisfy or not the
given equation. If the given equation is not satisfied by any values, then there is no
value for solution of the given equation. The values of the variables which satisfy
the given equation only are solution to the given equation.

2. If there are two terms containing the surds in an equation, then keep the one term
on one side and rest on the other side.

Do squaring on both sides of the equation and use the formula (a + b)2 = a2 + 2ab +
b2 and simplify.

Then, repeat the same process as in (1).

Worked Out EXAMPLES

EXAMPLE 1 Solve: 2x + 1 – 3 = 0

Solution : The equation is, Checking

2x + 1 – 3 = 0 putting x = 4 in the given equation,
2.4 + 1 – 3 = 0
or, 2x + 1 = 3 or, 9 – 3 = 0
or, 3 – 3 = 0
Squaring both sides, or, 0 = 0 which is true.
\ x = 4
( 2x + 1 )2 = (3)2
GREEN Mathematics Book-10 119
or, 2x + 1 = 9

or, 2x = 9 – 1

or, 2x = 8

or, x = 8
2

\ x = 4

EXAMPLE 2 Solve: 2 x + 1 = 4x + 1

Solution : 2 x + 1 = 4x + 1
Squaring both sides,
(2 x + 1 )2 = ( 4x + 1 )2
or, 4(x + 1) = 4x + 1
or, 4x + 4 = 4x + 1
or, 4 = 1 which is not true.
\ There is no solution to the given equation.

EXAMPLE 3 Solve: 4 – 3 3x + 4 = 0

Solution : 4 – 3 3x + 4 = 0 Checking

or, 4 = 3 3x + 4 Putting x = 20 in the given eqn.
4 – 3 3×20 + 4 = 0
Cubing both sides, 4 – 3 60 + 4 = 0
4 – 3 64 = 0
(4)3 = (3 3x + 4 )3
4–4=0
or, 64 = 3x + 4
0 which is true.
or, 3x = 64 – 4
\ x = 20
60
or, x = 3

\ x = 20

EXAMPLE 4 Solve: 4 2y + 1 = 1

Solution : 4 2y + 1 = 1

Raising power 4 on both sides,

(4 2y + 1 )4 = (1)4 Checking

or, 2y + 1 = 1 putting y = 0 in the given equa-
tion,
or, 2y = 1 – 1
4 2.0 + 1 = 1
or, y= 0 or, 1 = 1 which is true
2 \ y = 0.

\ y = 0

120 GREEN Mathematics Book-10

EXAMPLE 5 Solve: 2x – 1 – x + 2 = 0

Solution : 2x – 1 – x + 2 = 0 Checking

or, 2x – 1 = x – 2 putting x = 5 in the given equation,
Squaring both sides, 2.5 – 1 – 5 + 2 = 0
( 2x – 1 )2 = (x – 2)2 or, 3 – 3 = 0
or, 2x – 1 = x2 – 4x + 4 or, 0 = 0 which is true.
Again, putting x = 1 in the given

or, x2 – 6x + 5 = 0 equation,

or, x2 – 5x – x + 5 = 0 2.1 – 1 – 1 + 2 = 0

or, 1 – 1 + 2 = 0

or, x(x – 5) – 1(x – 5) = 0

or, 2 = 0 which is not true.

or, (x – 5) (x – 1) = 0 \ x = 5 or 1

either x – 5 = 0 or x – 1 = 0

\ x = 5 or x = 1

Rejecting x = 1, we have x = 5

EXAMPLE 6 Solve: x + 7 = 1 + x

Solution : x+7 =1+ x
Squaring both sides,
( x + 7 )2 = (1 + x )2 Checking
or, x + 7 = 1 + 2 x + x
or, 7 = 1+ 2 x putting x = 9 in the give equation,
9+7 =1+ 9
or, 4 = 4 which is true.
\ x=9

or, 2 x = 7 – 1

or, 2 x = 6

or, x = 3

Again, squaring both sides, we get,

x=9

GREEN Mathematics Book-10 121

EXAMPLE 7 Solve: 2 + x + 2 – x = 2

Solution : 2+x + 2–x =2 Checking
or, 2 + x = 2 – 2 – x
Squaring both sides, putting x = 2 in the given equation,
( 2 + x )2 = (2 – 2 – x )2 2+2 + 2–2 =2
2 + x = 4 – 4. 2 – x + 2 – x or, 2 + 0 = 2
or, 2x – 4 = – 4 2 – x or, 2 = 2 which is true.
or, 2(x – 2) = – 4 2 – x Again, putting x = – 2 in the given equn
or, x – 2 = – 2 2 – x 2–2 + 2+2 =2
Squaring both sides, or, 0 + 4 = 2
x2 – 4x + 4 = 4(2 – x) or, 2 = 2 which is true
or, x2 – 4x + 4 = 8 – 4x \ x = ±2
or, x2 = 4
\ x = ±2

EXAMPLE 8 Solve: x–1 = 4 + x –1
x +1 2

Solution : x–1 = 4 + x –1 Checking,
x +1 2

( x )2 – (1)2 x –1 Putting x = 81
x +1 2
or, =4+ 81 – 1 =4+ 81 – 1
81 + 1 2

or, ( x + 1) ( x – 1) =4+ x –1 80 9 – 1
x +1 2 9 + 1 = 4 + 2

or, x – 1 = 4 + x – 1 80 =4+ 8
10 2
2

or, x – 5 = x – 1 8 = 8 which is true

2

or, 2 x – 10 = x – 1 \ x = 81.

or, 2 x – x = – 1 + 10

or, x = 9
Squaring both sides, we get,
x = 81

122 GREEN Mathematics Book-10

EXAMPLE 9 Solve: x+4 + x–4 =3
x+4 – x–4

Solution : x+4 + x–4 =3
x+4 – x–4

or, x + 4 + x – 4 = 3 x + 4 – 3 x – 4

or, x – 4 + 3 x – 4 = 3 x + 4 – x + 4

or, 4 x – 4 = 2 x + 4 Checking

or, 2 x – 4 = x + 4 putting x = 20 in the given equation,
3
Squaring both sides, 20 20
3 + 4 + 3 – 4

4(x – 4) = x + 4 =3

or, 4x – 16 = x + 4 20 – 20
3 +4 3 –4

or, 4x – x = 4 + 16 32 + 8 32 + 8
or, 32 – 8 × 32 + 8 = 3

or, 3x = 20 or, 32 + 2. 32 . 8 + 8 = 3
32 – 8
or, x = 20
3 40 + 32
20 24
x = 3 or, =3

or, 72 =3
24

or, 3 = 3 which is true.

EXAMPLE 10 Solve : 2+ 4m – 1 = 5 \ x= 20
2– 4m – 1 3 3

Solution : Let 4m – 1 = x Checking

Now, 2+ x = 5 4.5
2– x 3 2+ 16 + 1
4.5 = 5
3
or, 6 + 3 x = 10 – 5 x
2 – 16 + 1
or, 3 x + 5 x = 10 – 6
5-4
2+
8 x = 4 or, 4 = 5
5-4 3
4
x = 8 2– 4

x = 1 or, 2+ 1 = 5
2 2– 2 3
1
Substituting value of 'x' we get 2

4m – 1 = 1 ∴ 5 = 5 which is true
2 3 3

or, 2 4m – 1 = 1 ∴ x= 5 =
16

GREEN Mathematics Book-10 123

Squaring on both sides

4(4m – 1) = 1
or, 16m – 4 = 1

or, 16m = 5

or, m = 5
16
5
∴ m = 16

EXAMPLE 11 Solve: 2x² + 3x + 2 = 2x² – 5x + 3 + 3

Solution : 2x² + 3x + 2 = 2x² – 5x + 3 + 3

Squaring both sides
or, ( 2x² + 3x + 2 )² = ( 2x² – 5x + 3 + 3)²

or, 2x² + 3x + 2 = 2x² – 5x + 3 + 6 2x² – 5x + 3 + 9

or, 8x – 10 = 6 2x² – 5x + 3

or, 4x – 5 = 3 2x² – 5x + 3

Squaring both sides,

or, (4x – 5)² = (3 2x² – 5x + 3 )²

or, 16x² – 40x + 25 = 18 x² – 45x + 27

or, – 2x² + 5x – 2 = 0 Checking

or, – (2x² – 5x + 2) = 0 putting x = 2 in the given equation,

or, 2x² – 4x – x + 2 = 0 2.2² + 3.2 + 2 = 2.2² – 5.2 + 3 + 3

or, 2x(x – 2 ) – 1 (x – 2) = 0 or, 4 = 1 + 3

or, (x – 2) (2x – 1) = 0 or, 4 = 4 which is true

either, x – 2 = 0 or, 2x – 1 = 0 \ x = 2

1 1
Again, putting x = 2 in the given equation,
\ x = 2 , x = 2
Thus, x = 2 rejecting

1 2(1 )² + 3(1 )+ 2 = 2(1 )² – 5(1 )+ 3 + 3

∴ x = 2 22 22

or, 4 = 1 + 3

or, 2 = 1 + 3

2 = 4 which is not true

1

\ x = 2

124 GREEN Mathematics Book-10

EXERCISE 6.5

A. Very Short Questions b. x – 1 = 3 c. 3 . x + 1 = 5
1. Solve:
a. a + 2 = 2

d. 3x + 1 = x + 3 e. x – 3 = 5x + 7 f. 4x2 – 1 = 2x + 1

g. x + 1 = 3 h. x–1 = 1 i. x + 1 = 3 8
x 2 x +1

j. 3 x – 1 = 4

2. Solve: b. 3 x + 3 = 2 c. 3 x + 2 – 2 = 1
a. 3 2x + 7 = 3 e. 3 x – 1 – 5 = 0 f. 3 y – 2 = 0
d. 3 4x + 5 = 2
g. 4 – 3y – 2 = 0

3. Solve: b. 4 x + 1 = 2 c. 4 3 + x = 4
a. 4 2x – 3 = 1 e. 3 x + 1 – 4 16 = 0 f. x = 4 81
d. 4 3x + 1 – 2 = 0
3
B. Short Questions
4. Solve: b. x2 + 15 + x – 5 = 0 c. x2 + 5 – x = 1
a. x2 + 3 = 3 – x e. x – x = 2 f. x2 + x – 3 = x –1
d. x² + 3 = x + 2 h. x2 + 3 = x + 1
g. m2 – 3m + 2 = m

C. Long Questions b. x + 5 + x = 5 c. x + 13 + x – 13 = 0
e. x = 2 x – 1 f. y + 9 – 1 = y
5. Solve: h. x + 12 = 6 – x i. 2x + 1 – 5 + x = 0
a. x + 5 – x = 1
d. x = x – 7 + 1
g. x = x + 6

6. Solve:

a. x + 4 + 3 = x + 31 b. x – 8 + 2 = 2x – 9 c. x + 4 + x – 5 = 3

d. xx –9 = 1 e. 5x + 10 = 5x + 2 f. b + 8 – 2 = b
+3

g. 4a + 5 – a – 1 = 3 h. 2y + 4 – 1 = y + 3 i. a² – 3a + 5 = 1 + a² – a + 1

GREEN Mathematics Book-10 125

7. Solve:

a. xx –1 = 2 + x–5 b. 3x – 4 = 2 + 3x – 2
+1 3 3x + 2 2

c. x – 25 =2– x +5 d. mx – 1 = 4 – mx +1
x +5 mx + 1 2
5

e. 3m – 4 = 3 – 3m – 2 f. x + p + x² + 2px = q
3m + 2 2

8. Solve:

a. x + k + x = 3k 91
c. 2 x – 4x – 3 = k+x b. x + x + 13 = x + 13

1 d. x – 2 = 2( x – 1)
4x – 3 x +1 3

e. 3x –7 = 5 ( x – 1) f. a + 1 = 3 – a + 1
x +1 3 2

g. 3 x – 4 = 15 + 3 x h. 5 a – 3 = 5 a + 3
x +2 a +2
x + 40 a +5

i. x – 1 – x = 1 – x j. 6 x – 11 = 2 x + 1
3 x x +6

9. Solve:

a. xx + 1 – x–1 = 1 b. x+2 + x–2 =2
+ 1 + x–1 2 x+2 – x–2

c. x + a + x–a = 4 d. x + 2 + x–2 =6
x – a x+a x – 2 x+2

e. 11 + x² – 1 – x² = 3 f. x + 1–x = 4
+ x² + 1 – x² 5 1–x x 3

10. Solve:

a. x² + 11x + 22 – x² + 7x + 6 = 2 b. a² – 3a + 5 – 1 = a² – a + 1

c. 2x² + 5x – 3 = 2x² + 3x – 8 + 1 d. x² – 2x – 4 – x² – 3x – 3 = 1

e. 4 a – 1 – a + 20 = a + 4 f. x² – 3x + 5 – x² – 4x + 4 = 1

126 GREEN Mathematics Book-10

7

Simplification of Algebraic Expressions

Estimated Teaching Periods : 5

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

Contents

7.1 Introduction

Objectives

At the end of this unit, students will be able to:
introduce the rational expression and its generalization.
reduce the given algebraic expressions into the simplest form.
know the L.C.M. of given rational algebraic expressions.

Materials

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

GREEN Mathematics Book-10 127

7.1 Introduction

Rational expression

Those algebraic expressions which can be expressed in the form of p where q ≠ 0 is
called rational expressions. For example : ba, aa+-bb, x32–x1, etc.
q

If we reduce two or more algebraic rational expressions in the simple form, this

process is called simplification of rational expressions.

The following steps are necessary to simplify the algebraic fractions.

Steps :

i. Reduce both numerators and denominators to lowest term by factorization.

ii. Take L.C.M. of denominators of given algebraic expressions.

iii. Simplify the numerator until we get the numerator in simple form of fraction.

Look and learn the following worked out examples to be more clear.

Worked Out EXAMPLES

EXAMPLE 1 Simplify : x² + y² – x–y
x ² – y² 2(x + y)

Solution : x² + y² – x–y
x ² – y² 2(x + y)

= (x² + y²) – (x – y)
(x + y) (x – y) 2(x + y)

= (x² + y²) ×2– (x – y)²
2(x + y) (x – y)

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

= x² + 2xy + y²
2(x – y) (x + y)

= 2(x (x + y)² y)
– y) (x +

= x+y
2(x – y)

128 GREEN Mathematics Book-10

EXAMPLE 2 Simplify : a² + 1 – 2(a – 1)
a³ + 1 a+1 a² – a + 1

Solution : a² + 1 – 2(a – 1)
a³ + 1 a+1 a² – a + 1

= (a + 1) a² a + 1) + (a 1 1) – 2(a – 1)
(a² – + (a² – a + 1)

= a² + (a² – a + 1) – (2a2 – 2)
(a + 1) (a² – a + 1)

= a² + a2 – a + 1 – 2a² + 2
(a + 1) (a² – a + 1)

= 3–a
(a + 1) (a² – a + 1)

= 3–a
a³ + 1

EXAMPLE 3 Simplify : 2 – 2 + a 3 1 + 4
a–1 a–2 + a+2

Solution : 2 – 2 + a 3 1 + a 4 2
a–1 a–2 + +

= 2 + 3 – 2 + 4
a–1 a+1 a–2 a+2

= 2(a + 1) + 3(a – 1) – 2(a + 2) – 4(a – 2)
(a – 1) (a + 1) (a – 2) (a + 2)

= 2a + 2 + 3a – 3 – 2a + 4 – 4a + 8
(a + 1) (a –1) a2 – 4

= 5a – 1 – –2a + 12
a2 – 1 a2 – 4

= 5a – 1 + 2a – 12
a2 – 1 a2 – 4

= (5a – 1) (a2 – 4) + (2a – 12) (a2 – 1)
(a2 – 4) (a2 – 1)

= 5a3 – 20a – a2 + 4 + 2a3 – 2a – 12a2 + 12
(a2 – 4) (a2 – 1)

= 7a3 – 13a2 – 22a + 16
(a2 – 1) (a2 – 4)

GREEN Mathematics Book-10 129

EXAMPLE 4 Simplify : x2 – 4 – 6 – x2 3 6 + 2 1
5x – 7x + x2 –

Solution : 4 – 3 +6 + 2 1
x2 – 5x – 6 x2 – 7x x2 –

= 4 – 3 + 2
x2 – 6x + x – 6 x2 – 6x – x + 6 x2 – 1

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

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

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

= 4x – 4 – 3x – 3 + 2x – 12
(x – 6) (x + 1) (x – 1)

= 3x – 19
(x – 6) (x + 1) (x – 1)

EXAMPLE 5 Simplify : 1+b – 1–b + 4b + 8b3
1–b 1+b 1 + b2 1 + b4

Solution : = 1+b – 1–b – 4b + 8b3
1–b 1+b 1 +b2 1 + b4

= (1 + b)2 – (1 – b)2 – 4b + 8b3
(1 – b) (1 + b) 1 + b2 1 + b4

= 1 + 2b + b2 – 1 + 2b – b2 – 4b + 8b3
1 – b2 1 + b2 1 + b4

= 4b – 4b + 8b3
1 – b2 1 + b2 1 + b4

= 4b + 4b3 – 4b + 4b3 + 8b3
(1 – b2) (1 + b2) 1 + b4

= 8b3 + 8b3
1 – b4 1 + b4

= 8b3 (1 + b4) + 8b3 (1 – b4)
(1 – b4) (1 + b4)

= 8b3 + 8b7+ 8b3 – 8b7
(1 – b4) (1 + b4)

= 16b3
1 – b8

130 GREEN Mathematics Book-10

EXAMPLE 6 Simplify : 2x + 3y + 2x – 3y – 16x4 + 54y3 + 81y4
4x2 + 6xy + 9y2 4x2 – 6xy + 9y2 36x2y2

Solution : 2x + 3y + 2x – 3y – 16x4 + 54y3 + 81y4
4x2 + 6xy + 9y2 4x2 – 6xy + 9y2 36x2y2

(2x+3y) (4x2–6xy+9y2) + (2x–3y) (4x2+6xy+9y2) 54y3
= – 16x4 + 36x2y2 + 81y4
(4x2 – 6xy + 9y2) (4x2 + 6xy + 9y2)

(2x)3 + (3y)3 + (2x)3 – (3y)3 54y3
= (6x4 + 36x2y2 + 81y4) – 16x4 + 36x2y2 + 81y4

= 16x4 + 16x3 – 16x4 + 54y3 + 81y4
36x2y2 + 81y4 36x2y2

= 16x3 – 54y3
16x4 + 36x2y2 + 81y4

= 2(8x3 – 27y3)
16x4 + 36x2y2 + 81y4

2(2x– 3y) (4x2 + 6xy + 9y2­)
=

(4x2 + 6xy + 9y2) (4x2 – 6xy + 9y2)

2(2x – 3y)
=

4x2 – 6xy + 9y2

EXAMPLE 7 Simplify : (2x – 3y)2 – 16z2 + (3x – 4z)2 – 4x2 + (4z – 2x)2 – 9y2
4x2 – (3y + 4z)2 9y2 – (4z + 2x)2 16z2 – (2x + 3y)2

Solution : (2x – 3y)2 – 16z2 + (3x – 4z)2 – 4x2 + (4z – 2x)2 – 9y2
4x2 – (3y + 4z)2 9y2 – (4z + 2x)2 16z2 – (2x + 3y)2

(2x – 3y)2 – (4z)2 (3y – 4z)2 – (2x)2 (4z – 2x)2 – (3y)2
= (2x)2 – (3y + 4z)2 + (3y)2 – (4z + 2x)2 + (4z)2 – (2x + 3y)2

(2x–3y+4z) (2x–3y–4z) (3y–4z+2x) (3y–4z–2x) (4z–2x+3y) (4z–2x–3y)
= (2x+3y+4z) (2x–3y–4z) + (3y+4z+2x) (3y– 4z –2x) + (4z+2x+3y) (4z –2x–3y)

(2x–3y+4z) (3y–4z+2x) (4z–2x+3y)
= ++

(2x+3y+4z) (3y+4z+2x) (4z+2x+3y)

2x – 3y + 4z + 3y – 4z + 2x + 4z – 2x + 3y
= (2x + 3y + 4z)

(2x + 3y + 4z)
= (2x + 3y + 4z)
=1

GREEN Mathematics Book-10 131

EXAMPLE 8 Simplify : 1 + 2 1 + 4 + 8 1
a+1 a2 + a4 + 1 a8 –

Solution : 1 + 2 1 + 4 1 + 8
a+1 a2 + a4 + a8 – 1

= 1 1 + 2 + 4 1 + 8 – 1)
a+ a2 + 1 a4 + (a4 + 1) (a4

= a 1 1 + 2 1 + 4(a4 – 1) + 8
+ a2 + (a4 + 1) (a4 – 1)

= a 1 1 + 2 1 + 4a4 – 4+8
+ a2 + (a4 + 1) (a4 – 1)

= a 1 1 + 2 1 + (a4 4(a4 + 1) 1)
+ a2 + + 1) (a4 –

= a 1 1 + 2 1 + (a2 + 4 – 1)
+ a2 + 1) (a2

= a 1 1 + 2a2 – 2 + 4
+ (a2 + 1) (a2 – 1)

= a 1 1 + 2a2 + 2
+ (a2 + 1) (a2 – 1)

= a 1 1 + 2(a2 + 1)
+ (a2 + 1) (a2 – 1)

= a 1 1 + 2
+ a2 – 1

= a 1 1 + 2
+ (a + 1) (a – 1)

= a–1+2
(a + 1) (a – 1)

= (a + 1)
(a + 1) (a – 1)

= 1
a–1

132 GREEN Mathematics Book-10

EXERCISE 7.1

A. Very Short Questions

1. Simplify:

a. 1 – 1 b. a2 + b2 a c. x 1 y – x 1 y
3a 5a a–b b– + –

d. a 2 + a 3 b e. 6 1 + x 5 1 f. 3a – 3b
–b + x2 – + a–b a–b
1
a2 b2 b a x– y
a+b + a b
g. – b a h. (1 – ) ÷ (1 – ) i. 1
x
y–

B. Short Questions
2. Simplify the following:

a. a – a 1 b b. a(a + b) + a(a – b)
a2 – b2 + b2 – a2 a2 – b2

c. (b – 1 – a) – (c – 1 – b) d. 1 n2 – m2 1
c) (b b) (a m2 – – mn

e. 1 – (2a – 3b)2 f. 2a 1 3b – 2a
2a – 3b 8a3 – 27b3 – 4a2 – 9b2

g. a2 – 1 + 6 – a2 – 1 + 2 h. (1 – b ) – a + b
5a 3a a a – b

i. a – a2 a2 b2 j. x –x2a + x +x2a + x2 x2
a–b – – 4a2

C. Long Questions

3. Simplify the following algebraic expressions:

a. (x – 1 – 3) + (x – 2 –3) – x² – 3 + 2
2) (x 1) (x 3x

b. 1 – 2 – 3
a² – 5a + 6 a² – 4a + 3 a² – 3a + 2

c. 2a – 8 – a² a–4 12 – a–1
a² – 9a + 20 – 7a + a² – 4a + 3

d. a² + 2 + 6 + 1 + 3
5a a² + 7a + 12 a² + 6a + 8

e. 2 + 2 + 1
(x–2)(x–3) (x–1)(3–x) (1–x)(2–x)

GREEN Mathematics Book-10 133

4. Simplify the following algebraic expressions:

a. a b + a b + 2ab + 4a³b
–b +b a² + b² a4 + b4

b. y – y + 2y2 + 4y4
x+y x–y x² + y² x4 – y4

c. a 1 b + a 1 b + 2a – 4a³
– + a² + b² a4 + b4

d. 1 a + 1 a + 2a + 16a³
– 2a + 2a 1 + 4a² 16a4 –1

5. Simplify the following algebraic expressions:

a. 1 +b – 1 –b – 4b + 8b³
1 –b 1 +b 1 + b² 1 + b4

b. a–b – a+b – 4ab – 8ab³
a+b a–b a2 + b² a4 – b4

c. 3 + x – 3 – x – 12x – 24x³
3 – x 3 + x x2 + 9 x4 + 81

6. Simplify the following algebraic expressions:

a. 3x – 1 1 + 3x + 1 1 – 81x4 2 + 1
9x² – 3x + 9x² + 3x + + 9x²

b. m–n + m+n – 2n³
m² – mn + n² m² + mn + n² m4 + m²n² + n4

c. a+c – a–c + 2a³
a² + ac + c² a² – ac + c² a4 + a²c² + c4

d. 2a – b – 2a + b + 16a³
4a² – 2ab + b² 4a² + 2ab + b² 16a4 + 4a²b² + b4

e. y+2 – y–2 – 2y³
y² + 2y + 4 y² – 2y + 4 y4 + 4y2 + 16

f. 1+x – 1–x + 2
1 + x + x2 1 – x + x2 1 + x2 + x4

7. Simplify the following algebraic expressions:

a. 1 – 2 a – 1 + 2
x–a 2x + x+a 2x – a

b. a 1 1 + 1 – 2 1 – 2
– a+1 2a + 2a – 1

c. a 1 3 – 1 + 1 – 1
– a–1 a+3 a+1

134 GREEN Mathematics Book-10

d. x 1 5 – 1 + 1 – 1
– x–3 x+5 x+3

e. x3 – x3 – 1 – 1
x–1 x+1 x–1 x+1

8. Simplify the following algebraic expressions:

a. (x – y)² – z² + (y – z)² – x² – (z – x)² – y²
x² – (y + z)² y² – (z + x)² z² – (x + y)²

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

c. a² (a – b)2 – c2 c2 + (b – c)2 – a2 + (c – a)2 – b2
– b2 – 2bc – b² – c2 – 2ca – a2 c² – a2 – 2ab – b2

d. (2b – 5c)² – 9a² + (5c – 3a)² – 4b² + (3a – 2b)² – 25c²
(3a + 5c)² – 4b² (2b + 3a)² – 25c2 (2b + 5c)2 – 9a2

9. Simplify the following algebraic expressions:

a. 8( 1+ 1+ 2x
x – 1) 8( x + 1) 8(x – 1)

b. 1 – 1 + 2b
1– b 1+ b 1+b

c. 3– 2– a –5 b
a+ b a– b a–b

d. 1+ 1 + 2
2 ( 2 – x) 2( 2+ x) 2+x

D. Very Long Questions

10. Simplify the following algebraic expressions:

a. 1 + 2 + 4 1 + 8 1
x+1 x² + 1 x4 + x8 –

b. 1 + 1 + 1 – 2a4
2a3 (a – x) 2a3 (a + x) a2 (a2 + x2) a8 – x8

c. 1 + 2a + 4a3 – 8a7
x+a x² + a2 x4 + a4 a8 – x8

d. (a + 1)2 ÷ 1 + a
a
1 – a + a2 + a + 1

GREEN Mathematics Book-10 135

11 11
e. + – –
y x y x2 y2 y2
1+ x y–x y2 – x2 1 + x2

f. 1 + x 1 1 + 2x – x x 1 + 4x3
– x2 + 1 + x4 + 1

g. 1 1 – a 1 – a2 – a4 1
a– a2 – a4 – 1 a8 –

h. 1 – (a 1 1)2 + 2 – 2
(a + 1)2 (a + 2)2 + (a + 1) a+2

i. 1 – 1 – 2x
1 + x + x2 1 – x + x2 1 + x2 + x4

j. a+b (a + b – c) + b+c (b + c – a) + c+a (c + a – b)
2ab 2bc 2ca

k. 2 – 2 + 1 – (y 1
y+1 y+2 (y + 1)2 (y + 2)2 + 1)2

l. ( 1 + 1 ) (a + b – c) + ( 1 + 1 ) (b + c – a) + ( 1 + 1 ) (c + a – b)
a b b c c a

m. 2 – 1 – 2 – 2
a–1 a2 + 1 a+1 a4 – 1

n. a+c + a– c + 2c3
a2 + ac + c2 a2 – ac + c2 a4 + a2c2 + c4

o. 1 + 2 + 4 + 8
x+1 x2 + 1 x4 + 1 x8 – 1

p. 2(x – 3) – (x2 x–1 12) – x–2
x2 – 9x + 20 – 7x + x2 – 8x + 15

q. 8a7 + 4a3 + 2a + 1
b8 – a8 b4 + a4 b2 + a2 a+b

Project Work:

a. Write the role of factorization, H.C.F. and L.C.M. in the simplification of algebraic
expressions. List them and discuss in your class.

136 GREEN Mathematics Book-10

8

Equations

Estimated Teaching Periods : 11

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

Contents

8.1 Word problems on two variables equations
8.2 Word problems on quadratic equations

Objectives

At the end of this unit, students will be able to:
define quadratic equations and its types.
solve the quadratic equations by factorization method, completing square method and
by using formula.
solve the word problems involving simultaneous equations.
solve the word problems involving quadratic equations.

Materials

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

GREEN Mathematics Book-10 137

8.1 Word problems on two variables equations

We can solve some veriable problems with the help of simultaneous equations. We make
two separate simultaneous equations and solve by any method.

Worked Out EXAMPLES

EXAMPLE 1 The sum of two numbers is 16 and their difference is 2. Find the
numbers.
Solution : Let the greater number be x and the smaller number be y
Case I
x + y = 16 ......................... (i)
Case II
x – y = 2 ............................ (ii)
Adding equation (i) and (ii)
x + y = 16
x – y = 2
2x = 18
or, x = 18

2
∴ x = 9
Putting the value of x in equation (i),
x + y = 16
or, 9 + y = 16
or, y = 16 – 9
∴ y = 7
∴ Required two numbers are 9 and 7.

138 GREEN Mathematics Book-10

EXAMPLE 2 A two digit number is 24 more than 3 times the sum of its digits.
When we change the position of digits, the new number is 18 less
Solution : than it. Find the number.

Let the digit in ten's place be x and the digit in unit's place be y

∴ The number is 10x + y

Case I

10x + y = 3(x + y) + 24

or, 10x + y = 3x + 3y + 24

or, 10x – 3x + y – 3y = 24

7x – 2y = 24 ............................. (i)

Case II 10y + x = (10x + y) – 18


or, 10y + x = 10x + y – 18

or, 9x – 9y = 18

or, 9(x – y) = 18
or, x – y = 18

9
x – y = 2 ................................ (ii)

Multiplying eqn. (ii) by 2 then subtracting eqn. (ii) from eqn. (i)

7x – 2y = 24

–2x +– 2y = –4
5x = 20

or, x = 20
5

∴ x = 4

Putting the value of x in equation (ii)

x – y = 2

or, 4 – y = 2

or, – y = 2 – 4

or, – y = – 2

∴ y = 2

∴ The required number = 10x + y

= 10 × 4 + 2

= 40 + 2 = 42

GREEN Mathematics Book-10 139

EXAMPLE 3 A father said to his son, "5 years ago I was 4 times as old as you
were and 9 years hence I shall be only twice as old as you." Find
Solution : their present ages.

Let, present age of father be x years and present age of son be y years.

Then,

5 years ago father's age = (x – 5)

5 years ago son's age = (y – 5)

9 years hence father's age = x + 9

9 years hence son's age = y + 9

Case I
x – 5 = 4(y – 5)
or, x – 4y = – 15 ............. (i)

Case II

x + 9 = 2(y + 9)
or, x + 9 = 2y + 18
or, x – 2y = 9 .................. (ii)
Multiplying equation (ii) by 2 then subtracting from (i)
x – 4y = – 15
–2x –+ 4y = – 18
– x = – 33
∴ x = 33
Putting the value of x in equation (i)
x – 4y = – 15
or, 33 – 4y = – 15
or, – 4y = – 15 – 33
or, – 4y = – 48
y = 48

4
∴ y = 12
∴ Father's present age is 33 years.
Son's present age is 12 years.

140 GREEN Mathematics Book-10

EXAMPLE 4 In a fraction, the numerator is 1 less than the denominator. If 7
is subtracted from the numerator and 5 from the denominator the
Solution : fraction becomes 4 . Find the fraction.

5

Let the numerator and denominator of the fraction be x and y
respectively i.e. fraction = x .

y

Now,

Case I

x = y – 1 ................ (i)

Case II
x–7 = 4

y–5 5
or, 5(x – 7) = 4(y – 5)
or, 5x – 4y = –20 + 35
or, 5(y – 1) – 4y = 15 [from equation (i) x = y + 1]
or, 5y – 5 – 4y = 15
∴ y = 20
Again, putting the value of y in equation (i)
x=y–1
or, x = 20 – 1

∴ x = 19

Now, required fraction is
x = 19

y 20

EXAMPLE 5 The total cost of 4kg of basmati rice and 5kg of sugar is Rs. 860. If
the cost of 10kg of basmati rice is the same as 9kg of sugar, find the
Solution : rate of cost of these goods.

Let the cost of basmati rice be Rs. x per kg and the cost of sugar be
Rs. y per kg.

Case I
4x + 5y = Rs. 860 ................ (i)

GREEN Mathematics Book-10 141

Case II

10x = 9y

or, x = 9y .................. (ii)
10

Putting the value of x in equation (i), we get,

4x + 5y = Rs. 860

or, 4 9y y + 5y = Rs. 860
10

or, 36y + 5 y = Rs. 860
10

or, 86y = Rs. 860 × 10

or, y = Rs. 8600
86

∴ y = Rs. 100

Put y = 100 in equation (ii), we get,
x = 9 × 100

10

= 90
Therefore, the cost of basmati rice (x) is = Rs. 90 per kg and the cost
of sugar (y) = Rs. 100 per kg.

EXAMPLE 6 Avipsa takes 3 hours more than Anusuya to cover a distance of

30km. But if Avipsa increases her speed double, she can travel
30km in 2 hours less than Anusuya. Find the speed of each of them.

Solution :

Let the speed of Avipsa = x km/hr

and the speed of Anusuya = y km/hr

Case I 30 = 30 + 3
x y

30 – 3y0 = 3 ............. (i)
x

Case II 30 = 30 – 2
2x y
30 – 30 = – 2 ........... (ii)
or, 2x y

142 GREEN Mathematics Book-10

Subtracting equation (ii) from (i)
30 – 30 = –2

2x y

30 – 3y0 = 3
x
–+ –

30 – 30 = – 5
2x x

or, 60 – 30 = 5
2x

or, 10x = 30
or, x = 30

10

∴ x = 3

Putting the value of x in equation (i)

30 – 3y0 = 3
x

or, 30 – 30 = 3
3y

or, 10 – 30 = 3
y

or, 3y0 = 7

or, 7y = 30

y = 30 km/hr.
7
∴ S peed of Avipsa = 3km/hr

Speed of Anusuya = 370 km/hr.

EXERCISE 8.1

A. Very Short Questions

1. a. What do you mean by simultaneous equations?

b. Solve for 'y' :

i. x = 2 and x + y = 0

ii. x + y = 10 and x =4
iii. x +y=4 and 2 0
x=
21

GREEN Mathematics Book-10 143

c. The sum of 2 numbers is 10. If the second number is 6 then first number is
.....................

d. The difference of 2 numbers is 20. If the first number is 10 then the second
number is ...........................

e. The product of 2 numbers is 16. If the first number is ................... then the
second number is 4.

B. Short Questions

2. The difference between two numbers is 7. Twice of the smaller one needs 1 to be
equal to the greater one. Find the number

3. Four years ago, a father was 10 times as old as his son. 12 years hence, he will be
twice as old as his son. Find their present ages.

4. The ratio of two numbers is 4:7. When we add 5 to each number we get another
ratio 5:8. Find the numbers

5. 3 boys and 4 girls can do a project in 7 days. 2 boys and 3 girls finish the same
project in 10 days. If the project is given to single boy or single girl, how many days
does a boy or girl take to finish it?

C. Long Questions

6. a. When a number is added to two times of itself, the sum is 21. Find the number.

b. The ratio of two numbers is 7:2. The difference between numbers is 40. Find
them.

c. If the sum of two numbers is 20 and their difference is 4, find the numbers.

d. If 39 is subtracted from the square of a number, the result is 130. Find the
positive number.

e. 1 of the age of a mother is the age of her daughter. If the difference of their
6
ages be 35 years, find the age of the mother.

f. Divide 11 into two parts so that their product will be 24.

g. The sum of two positive number is 29. If one of them is 5 more than that of the
other, find the numbers.

7. a. The total cost of 5 pens and 3 pencils is Rs. 365. If the cost of 11 pens is same as
the cost of 8 pencils, find the cost of a pen and a pencil separately.

b. One pen and 3 copies cost Rs. 62 while 2 pens and 4 copies cost Rs. 100. What
is the cost of a pen and a copy?

c. A father was 3 times as old as his son in 1980 A.D. and 5 times as old in 2008
3
A.D. In which year was the son born?

d. The present ages of a father and his son are 40 years and 8 years respectively.
How many years ago the product of their ages was 105? Find it.

144 GREEN Mathematics Book-10