Algebraic Expressions
3. Let’s find the area of the shaded region using a2 – b2 = (a + b) (a – b).
a) 10 cm b) 15 cm c) 18 m
10 cm 3 cm 15 cm 18 m 5m
5m
2 cm
2 cm 3 cm f)
d) 21 m e) 25 ft
21 m 9 ft 25 ft 12 m 30 m
7m7m 9 ft 12 m
30 m
Creative section
4. Factorise: b) a2 – 10a + 16 - 6b – b2
d) x4 + 8x2 – 65 + 18y – y2
a) x2 + 6x + 5 – 4y – y2 f) 625y2 + 400y – 36 + 20z – z2
c) p2 – 12p – 28 + 16q – q2 h) 25x2 – 20xy – 21y2 + 10yz – z2
e) 9a2 – 30a + 24 – 8x – 16x2
g) 16p2 – 72pq + 80q2 – 6qr – 9r2
5. Resolve into factors. b) (x2 – 1) (y2 – 1) – 4xy
a) (a2 – b2) (c2 – d2) + 4abcd d) (9 – x2) (100 – y2) – 120xy
c) (p2 – 4) (9 – q2) + 24pq
Project work
6. a) Write any three expressions of your own in the form of a2 – b2. Then, factorise your
expressions.
b) Write any three expressions of your own in the form of a4 + a2b2 + b4. Then, factorise
your expressions.
c) Write any three pairs of trinomial expressions of your own and find the product of
each pair. Then, factorise each product. Check, are your answers correct?
[E.g. (a + b – c) (a – b + c) = a2 – b2 – c2 + 2bc, etc.]
(v) Expression of the form a3 + b3 or 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 – a2b + ab2 + a2b – ab2 + b3 = a3 + b3
Similarly, (a – b) (a2 + ab + b2) = 3 – b3
So, (a + b) and (a2 – ab + b2) are the factors of a3 + b3.
Also, (a – b) and (a2 + ab + b2) are the factors of a3 – b3.
Thus, to factorise the expressions of the forms a3 + b3 and a3 – b3, we should use the
following formulae.
a3 + b3 = (a + b) (a2 – ab + b2) a3 – b3 = (a – b) (a2 + ab + b2)
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Worked-out examples
Example 1: Factorise: a) 8a4 + 125a b) x6 – y6 c) a7 + 1
Solution: a5
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 2: 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)
(vi) 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.
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Example 3: Factorise: a) x2 + 7x + 12 b) 2x2 – 5x + 2 Algebraic Expressions
c) 2x2 – x – 6
Solution: 12 × 1 = 12
4 × 3 = 12
x2 x2 x x x x x + 3
a) x2 + 7x + 12 = x2 + (4 + 3) x + 12 x x
x
= x2 + 4x + 3x + 12 2x xx 12 34
x–2x2 x2 x 56 78
= x (x + 4) + 3 (x + 4) xx 9 101112
= (x + 4) (x + 3) x+4
2×2=4 2x – 1
4×1=4
x2 x2 x
b) 2x2 – 5x + 2 = 2x2 – (4 + 1) x + 2 2 xx xx 21
= 2x2 – 4x – x + 2
= 2x (x – 2) – 1 (x – 2) 2x – 1
= (x – 2) (2x – 1) x–2
2 × 6 = 12 2x + 3 2x + 3
4 × 3 = 12 x2 x2 x x x
x x(x–2) x(x–2)
c) 2x2 – x – 6 = 2x2 – (4 – 3) x – 6 x–2 2 xx xx
1 23
x–2 4 56
x–2
= 2x2 – 4x + 3x – 6 x–2
xx3
= 2x (x – 2) +3 (x – 2) 2x + 3
= (x – 2) (2x + 3) x(x–2) x(x–2) 3x–6
Example 4: Resolve into factors:
Solution: a) a2 – 3 + 2b2 b) 9 (x + y)2 + x + y – 8
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 )
b a a
=( – b ) ( a – 2b )
a b a
b) Let x + y = a
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)
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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 6.3
General section
1. In each of the following figures write the polynomial as the product of its factors.
a) x 3 b) x 3 c) x
x
xx
22 3
d) x x 1 e) x x 4 x
f) x
x x2 x2 x x
3 2
2 1
3 3
Creative section - A
2. Resolve into factors.
a) 8x3 + y3 b) 1 + 27a3 c) 128t3 – 2t d) x3y – 64y4
e) a6 + b3 f) 64x6y3 – 125 g) a6 – 64 h) 64x6 – y6
j) (x + 2)3 – 27 k) (x – y)3 – 8 (x + y)3
i) (a + b)3 + 1 l) p3 + 1
p3
m) a 3– b3 n) 27x3 – 30x2y + 40xy2 – 64y3 o) 8 – 6a – 9a2 + 27a3
b a
p) x4 + 1 q) p7 + 1
x2 p5
3. Factorise.
a) x2 + 4x + 3 b) x2 – 7x – 8 c) a2 – 27a + 180
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
n) 2x6 + 17x3 + 8 b2 b
x2 3y2
j) y2 – 2 – x2 l) 3 (x – y)2 – 10 (x – y) + 8
m) 8a4 – 14a2 – 9 o) 3x6 – 79x3 – 54
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Creative section - B
4. Simplify.
a) a2 – 4 b) x2 x2 – 9 15 c) 4a2 – 1
a2 – a – 6 + 2x – 2a2 – 3a – 2
d) a2 – 9 e) x3 + 3x2 – 4x f) 2x2 + 3x – 2
a3 + 27 x4 – x 2x2 – 5x + 2
g) a3 – a2b + ab2 – b3 h) (x2 – x – 6) (x2 + 4x + 3) i) (x2 – 16) (x + 2)
a4 – b4 (x2 – 9) (x + 1) (x2 – 2x – 8) (x2 + 5x + 4)
5. 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)
Project work
6. 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 + bx + c (ii) ax2 – bx – c (iii) ax2 + bx – c (iv) ax2 – bx + c
6.2 Highest Common Factor (H.C.F.) of algebraic expressions
To find the H.C.F. of monomial expressions, at first 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 monomial expressions.
To find the H.C.F. of the polynomial expressions, they are to be factorised and a
common factor or the product of common factors is obtained as their H.C.F.
Worked-out examples
Example 1: Find the H.C.F. of a2 – 9 and a2 + 4a – 21.
Solution:
1st expression = a2 – 9 a2 – 9 a2 + 4a – 21
2nd expression = a2 – 32 = (a + 3) (a – 3)
= a2 + 4a – 21 (a + 3) (a – 3) (a + 7)
? H.C.F. = a2 + 7a – 3a – 21
= a (a + 7) – 3 (a + 7)
= (a + 7) (a – 3)
= (a – 3)
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6.3 Lowest Common Multiple (L.C.M.) of algebraic expressions
The L.C.M. of monomial expressions is the common variable with the highest power.
To find the L.C.M. of the given polynomial expressions, they are to be factorised and
the product of common factors and the factors which are not common is taken as their
L.C.M.
Example 2: Find the L.C.M. of x2 – 25, x3 – 125 and x2 – 8x + 15
Solution:
1st expression = x2 – 25 = a2 – 52 = (a + 5) (a – 5)
2nd expression = x3 – 125 = x3 – 53 = (x – 5) (x2 + 5x + 25)
3rd expression = x2 – 8x + 15
= x2 – 5x – 3x + 15
= x (x – 5) – 3 (x – 5) = (x – 5) (x – 3)
? L.C.M. = (x – 5) (x + 5) (x – 3) (x2 + 5x + 25)
Example 3: Find the L.C.M. of a2 – 3a + 2, a2 – 5a + 6 and a2 – 4a + 3
Solution:
1st expression = a2 – 3a + 2 = a2 – 2a – a + 2 = a (a – 2) – 1 (a – 2) = (a – 2) (a – 1)
2nd expression = a2 – 5a + 6 = a2 – 3a – 2a + 6 = a (a – 3) – 2 (a – 3) = (a – 3) (a – 2)
3rd expression = a2 – 4 + 3 = a2 – 3a – a + 3 = a (a – 3) –1 (a – 3) = (a – 3) (a – 1)
? L.C.M. = (a – 1) (a – 2) (a – 3)
6.4 Simplification of rational expressions 3x , 25y , a + b,
7
Rational expressions can be expressed in the form of , where q ≠ 0.
x+y , etc. are a few examples of rational expressions.
x–y
When we multiply or divide rational expressions, the numerators and denominators
of each expression are factorised (if necessary). Then we simplify the expressions to
the simplest forms.
Worked-out examples
Example 1: Simplify: a) a2 + 3a +2 × a2 – 9 6 b) 4x2 – 81y2 ÷ 2x – 9y
Solution: a2 + a –6 a2 – a – 1 – 4a2 a – 2a2
a) a2 + 3a + 2 × a2 – 9 6 = a2 + 2a + a + 2 × a2 – a2 – 32 – 6
a2 + a – 6 a2 – a – a2 + 3a – 2a – 6 3a + 2a
= a (a + 2) + 1 (a + 2) × (a + 3) (a – 3)
a (a + 3) – 2 (a + 3) a (a – 3) + 2 (a – 3)
= (a + 2) (a + 1) × (a + 3) (a – 3)
(a + 3) (a – 2) (a – 3) (a + 2)
= a+1
a–2
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b) 4x2 – 81y2 ÷ 2x – 9y = 4x2 – 81y2 × a – 2a2 = (2x)2 – (9y)2 × a (1 – 2a)
1 – 4a2 a – 2a2 1 – 4a2 2x – 9y 12 – (2a)2 2x – 9y
= (2x + 9y) (2x – 9y) × a (1 – 2a)
(1 + 2a) (1 – 2a) 2x – 9y
= a (2x + 9y)
1 + 2a
When we add or subtract rational expressions with unlike denominators, we should find the
L.C.M. of denominators. Then, the simplification is performed as like the simplification of
fractions.
Example 2: Simplify: a) x x 2 – x x 2 b) x2 + ax + a2 + x2 – ax + a2
Solution: + – x+a x–a
a) x x 2 – x = x (x – 2) – x (x + 2) = x2 – 2x – x2 – 2x = –4x
+ x–2 (x + 2) (x – 2) x2 – 4 x2 – 4
b) x2 + ax + a2 + x2 – ax + a2 = (x – a) (x2 + ax + a2) + (x + a) (x2 – ax + a2)
x+a x–a (x + a) (x – a)
= x3 – a3 + x3 + a3 = 2x3
x2 – a2 x2 – a2
Example 3: Simplify: x–1 + (x + 3 – 1) – (1 – 1 – 2x)
Solution: (2x – 1) (x + 2) 2) (x x) (1
x–1 + 3 1) – (1 – 1 – 2x)
(2x – 1) (x + 2) (x + 2) (x – x) (1
= x–1 + 3 – 1
(2x – 1) (x + 2) (x + 2) (x – 1) [– (x – 1)] [– (2x – 1)]
= (x – 1) (x – 1) + 3 (2x – 1) – (x + 2)
(2x – 1) (x + 2) (x – 1)
= x2 – 2x + 1 + 6x – 3 – x – 2
(2x – 1) (x + 2) (x – 1)
= x2 + 3x – 4
(2x – 1) (x + 2) (x – 1)
= (2x x2 + 4x – x – 4 1) = x (x + 4) – 1 (x + 4) = (2x (x + 4) (x – 1) 1) = (2x x+4 2)
– 1) (x + 2) (x – (2x – 1) (x + 2) (x – 1) – 1) (x + 2) (x – – 1) (x +
Example 4: Simplify: p–1 + p2 p–2 6 + p2 p– 5 15
Solution: p2 – 3p + 2 – 5p + – 8p +
p–1 + p2 p–2 6 + p2 p–5 15
p2 – 3p + 2 – 5p + – 8p +
= p2 – p–1 + 2 + p2 – p – 2 + 6 + p2 – p–5 + 15
2p – p 3p – 2p 5p – 3p
=p (p – p – 1 (p – 2) + p (p – p – 2 (p – 3) + p (p – p – 5 (p – 5)
2) – p 3) – 2 5) – 3
= (p p–1 + p–2 + (p – p–5 – 3)
– 2) (p – 1) (p – 3) (p – 2) 5) (p
= p 1 2 + p 1 3 + p 1 3 = p – 3+ p–2 +p – 2 = (p 3p –7 3)
– – – (p – 2) (p – 3) – 2) (p –
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Algebraic Expressions
Example 5: Simplify: m+n + m2 m–n n2 – m4 + 2n3 + n4
Solution: m2 + mn + n2 – mn + m2n2
m2 m+n n2 + m2 m–n n2 – m4 + 2n3 + n4
+ mn + – mn + m2n2
= (m + n) (m2 – mn + n2) + (m – n) (m2 + mn + n2) – (m2 + n2)2 2n3 + m2n2
(m2 + mn + n2) (m2 – mn + n2) – 2m2n2
= m3 + n3 + m3 – n3 – 2n3
(m2 + mn + n2) (m2 – mn + n2) (m2 + n2)2 – (mn)2
= (m2 + mn + 2m3 – mn + n2) – (m2 + mn + 2n3 – mn + n2)
n2) (m2 n2) (m2
= (m2 + mn 2m3 – 2n3 mn + n2)
+ n2) (m2 –
= 2 (m – n) (m2 + mn + n2)
(m2 + mn + n2) (m2 – mn + n2)
= 2( m– n)
m2 – mn + n2
Example 6: Simplify: (a – b)2 – c2 + (b – c)2 – a2 + (c – a)2 – b2
Solution: a2 – (b + c)2 b2 – (c + a)2 c2 – (a + b)2
(a – b)2 – c2 + (b – c)2 – a2 + (c – a)2 – b2
a2 – (b + c)2 b2 – (c + a)2 c2 – (a + b)2
= (a – b + c) (a – b – ) + (b – c + a) (b – c – a) + (c – a + b) (c – a – b)
(a + b + c) (a – b – c) (b + c + a) (b – c – a) (c + a + b) (c – a – b)
= a–b+c + b–c+a + c–a+b
a+b+c a+b+c a+b+c
= a – b + c +b–c+a+ c – a + b
a+b+c
= a + b + c = 1
a + b + c
EXERCISE 6.4
General section
1. Find the H.C.F. and the L.C.M. of the following factors:
a) (x + 2) (x – 1) and (x + 1) (x + 2) b) (a – 5) (a – 4) and (a + 4) (a – 5)
c) (x + y) (2x – y) and (2x – y) (x – y) d) (a + b) (a2 – ab + b2) and (a – b) (a + b)
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Creative section - A
2. Find the H.C.F.
a) ax2 + 2ax, ax3 – 4ax b) x2 – 1, x3 – 1
c) a3 – 8, a2 + 2a + 4 d) (x – 1)2, x2 – 3x + 2
e) x2 + 5x + 6, x2 – 4, x3 + 8 f) 2x2 + 5x + 2, 3x2 + 8x + 4, 2x2 + 3x – 2
g) m3 – 1, m4 + m2 + 1, m2 + m + 1 h) 2x3 – x2 – x, 4x3 – x, 8x4 + x
i) x3 – y3, x6 – y6, x4 + x2y2 + y4 j) a3 + b3, a4 + a2b2 + b4, a3 – a2b + ab2
3. Find the L.C.M.
a) (x + 1) (x + 2), (x + 2) (x – 2) b) (x – 1) (x – 2), (x – 2) (x – 3), (x – 3) (x – 1)
c) x2 – 1, x3 – 1 d) x2 – 9, 3x + 9
e) 4x2 – 25, 2x2 – x – 15 f) x4 + x2y2 + y4, x3 – y3
g) 2x2 – 3x – 9, 4x2 – 5x – 21, x3 – 9x h) 8x3 + y3, 8x3 – y3, 16x4 + 4x2y2 + y4
i) x2 – 5x + 6, x2 – 4x + 3, x2 – 3x + 2 j) x3 + 27, 2x3 – 6x2 + 18x, x2 – 3x + 9
4. Simplify:
a) x2 – a2 × 7a b) 4y2 – 9z2 × y– 2 c) 4a2 – 9b2 × x2y + xy2
ax + a2 x–a y2 – 4 2y – 3z x2 – y2 4a – 6b
d) a2 – b2 ÷ a2b – ab2 e) 2x + 6 ÷ 3x2 + 9x f) a2 – 4b2 ÷ a + 2b
a2b + ab2 a2b2 x2 – 9 2x2 – 6x a2 – 9x2 a – 3x
5. Simplify:
a) a2 b + b2 b) p 4 + 4 2 c) x2 x y2 + y2 y x2
a– b–a p2 – – p2 – –
d) 1 1 + 1 1 e) 1 – 1 f) y 1 3 – 2 1
2a + 2a – x–2 x–1 – 2y –
g) a 1 b – a+b h) p+q + q 1 p i) x3 + y3
– a2 – b2 p2 – q2 – x–y y–x
j) a–b + b–c + c–a k) x+2 – x–2 l) x2y – xy2
ab bc ca x–2 x+2 x–y x+y
m) y2 1 y) – x2 1 y) n) a2 + 2a – 1 o) 1 + 3
(x – (x – a–1 1–a y2 – 4 y2 + 5y + 6
p) a2 + ab + b2 + a2 – ab + b2 q) x2 + 2x + 4 + x2 – 2x + 4 r) (a 1 – a2 1 b2
a+b a–b x+2 x–2 – b)2 –
s) x+2 2 + 3 1 t) x–2 – x+ 1 u) x–2 – x2 x+1
x2 + x – x2 – x2 + 4x + 4 x2 – 4 x2 – 1 – 2x + 1
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Creative section - B
6. Simplify:
a) 1 – x 2 1 + x 1 2 b) x 1 1 – x 1 1 + 3 1
x + + + – x2 –
c) x x 2 + x – 4x d) 2xy – x–y + x+y
+ x–2 x2 – 4 x2 – y2 x+y x–y
e) x+1 + x–1 – 1 4 f) 2a 1 2x – 2a 1 2x – a2 x x2
2x3 – 4x2 2x3 + 4x2 x2 – – + +
7. Simplify:
a) 1 + 1 + 1
(x – 3) (x – 4) (x – 4) (x – 5) (x – 5) (x – 3)
b) ( 2 (a – 3) 5) + (3 – a–1 – 4) + (5 – a–2 – 3)
– 4) (a – a) (a a) (a
c) x2 – 2 + 6 – x2 – 2 + 3 + x2 – 1 + 2
5x 4x 3x
d) x2 x–1 2 + x2 x–2 6 + x2 x– 5 15
– 3x + – 5x + – 8x +
e) 2y + 5 + 11 – 16y
y2 + 6y + 9 y2 – 9 8y2 – 24y
8. Simplify:
a) x–y + x2 x + y y2 + 2y3
x2 – xy + y2 + xy + x4 + x2y2 + y4
b) a2 a–2 4 + a2 a +2 4 – a4 + 16 + 16
– 2a + + 2a + 4a2
c) x+3 9 + x–3 9 – 54 81
x2 + 3x + x2 – 3x + x4 + 9x2 +
d) 1 a +2 a2 – a–2 – 1 2a2 a4
+ a+ 1 – a + a2 + a2 +
9. Simplify:
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) a2 – (b – c)2 + b2 – (c – a)2 + c2 – (a – b)2
(c + a)2 – b2 (a + b)2 – c2 (b + c)2 – a2
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Unit Indices
7
7.1 Indices – review
An index is a number that tells 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 rthe 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.
7.2 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,
a am (2x)5 = 25 × x5, …
Negative index law b m = bm
Zero index law
Root laws of indices 1 1 5–3 = 1 or 53 = 1
a– m = am or am = a–m 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
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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 = am cor 4. a m = am
bm b b bm
cor 5. (ab)m = ambm cor 6. ambm = (ab)m
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(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 – 25
32
Solution: a) (16)43 b) (9–3)61 c) d) (160.5)32 e) 3 729 f) 3 64–1
a) (16)43 = (24) 34 = (2)4 × 34 = 23 = 8
b) (9–3)16 = (32)–3 × 61 = 3–6 × 16 = 3–1= 1
3
c) 243 – 25 = 35 – 52 = 25 52 = 2 =5 × 25 2 2= 4
32 25 35 3 3 9
d) (160.5)32 = (24)0.5 × 23 = 24 × 0.5 × 23 = 23 = 8
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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 – 12 125 31 ÷ 16 – 14 b) 127 × 286
16 64 81 217 × 166
1 1 1
Solution: c) (1 – 6–9)–1 + (1 – 69)–1 d) (a – b)–1 + (b – c)–1 + (c – a)–1
a) 25 – 21 125 31 ÷ 16 – 41
16 64 81
= 16 21 125 13 ÷ 81 14
25 64 16
= 4 2 × 12 5 3× 13 ÷ 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
c) LHS 5 4 × 5
5
= 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
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Example 4: Simplify:
Solution: a) 3 50ba5c–2 × 3 20ab5c8 b) 3 27x3y6 ÷ 4 81x4y8 c) 3 (a + b)–8 × (a + b)23
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)32 = (a + b)– 38 × (a + b)23 = (a + b)– 83 + 32 = (a + b) – 83+ 2
1
= (a + b)– 63 = (a + 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
xa2 xb2 xc2 = x a+b × x b+c × x c+a
a+b xb2 × b+c xc2 × c+a xa2
= 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)31 × 1 – n31
m–n m13
Solution:
m + (mn2)31 + (m2n)31 × 1 – n13 = m +m31 n23 + m23 n13 × m31 – n31
m– n m31 m–n m13
= m31 (m23 + n23 + m31 n31 ) × m13 – n31
m– n m31
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Indices
= (m31 – n13) [(m31)2 + m13.n31+ (n31)2] = (m31)3 – (n31)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 1 y= x x× x y= x x + y = RHS
y + y xy – 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 + 231
21
or, (x – 1)3 = (23 + 23 )3 [Cubing on both sides]
3 231 3 + 3. 223 . 231
or, x3 – 3x2 + 3x – 1 = 223 223 + 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.
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EXERCISE 7.1
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:
a) 29 × 2-6 b) 5–7 × 58 c) 11–5 ÷ 11–3 d) (25) 3
2
e) (64)– 2 f) 1 1 g) 8 – 4 h) 169 – 1
3 64 6 3 2
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 64–1 729 –1 3 1
64 4 100
n) 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 – 21 d) 25 – 12 125 31 ÷ 8 – 31
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
xa + 2b × x4a – 3b + ax – y + ay – x
e) 1 – 1 n + 1 – 1 m f) (1 – 3–5)–1 + (1 – 35)–1
xm – xn –
y–1 x–1 –1
g) (a + b)–1.(a–1 + b–1) h) x–1 + y–1
6. Show:
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 –1 b) 5x – 5x – 1 c) 3 5 × 2m – 4 × 2m – 2 1
3p + 1 + 4× 5x – 1 × 2m + 2 – 5 × 2m +
3p
d) 2n +2 × (2n – 1)n + 1 ÷ 4n e) 5–n × 625n –1 f) 9x × 3x – 1 – 3x
2n(n – 1) 53n – 2 × (5 × 32x + 1 × 3x – 2 – 3x
2)–1
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8. Simplify: b) a6b–2c4 ÷ 4 a4b–4c8
a) 25a2b2 × 3 27a3
d) 3 56p7q4
c) 4 16x8y4 ÷ 3 8x6y3 3 7p4q7
e) 4 216m7n5 ÷ 4 6–1m–1n f) 3 (a + b)–7 1
× (a + b) 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.
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Indices
7.3 Exponential equation
Let's take any two equations: 2x = 4 and 2x = 4.
In the equation 2x = 4, variable x is a base. It is a linear equation. However, in 2x = 4,
variable x is an exponent of the base 2. Such an equation in which variable appears
as an exponent of a base is known as exponential equation.
To solve an exponential equation, we need to have equations with the same base on
either side of the 'equal' sign. Then, we compare the powers of equal base and solve
the equation.
Let's study, the following axioms which are used in solving the exponential equations.
(i) If ax = ab, then x = b (ii) If ax = 1, then ax = a0 and x = 0
In this way, while solving an exponential equation, we should simplify the equation
till the equation is obtained in the form ax = ab or ax = 1.
Worked-out examples
Example 1: Solve: a) 2x = 32 b) 52x = 1 c) 7x – 2 = 1
Solution: 25
a) 2x = 32 b) 52x = 1 c) 7x – 2 = 1
or, 2x = 25 25 or, 7x – 2 = 70
? x=5 1 or, x – 2 = 0
or, 52x = 52 ? x=2
or, 52x = 5–2
or, 2x = – 2
or, x = – 1
Example 2: Solve: a) ( 2)x – 3 = ( 4 2)x + 1 b) 3x + 3x + 2 =10
Solution:
a) ( 2)x – 3 = ( 4 2)x + 1 b) 3x + 3x + 2 = 10
x–3 x+1 or, 3x + 3x × 32 = 10
or, 2 2 = 2 4
or, x – 3 = x + 1 or, 3x (1 + 9) = 10
2 4
or, 4x – 12 = 2x + 2 or, 3x = 10 =1
10
or, 2x = 14 or, 3x = 30
or, x = 7 or, x = 0
Example 3: Solve: a) 2x + 1 × 3x – 2 = 48 b) 9y + 1 = 32y + 1 + 54
Solution: b) 9y + 1 = 32y + 1 + 54
a) 2x + 1 × 3x – 2 = 48
or, 32y + 2 – 32y + 1 = 54
or, 2x × 21 × 3x × 3–2 = 48 or, 32y × 32 – 32y × 3 = 54
or, 32y (32 – 3) = 54
or, 2x × 3x × 2 = 48 or, 32y =9
32 or, 32y = 32
or, (2 × 3)x = 216 or, 2y =2
or, 6x = 63 ? y =1
? x =3
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Example 4: If ax = b, by = c and cz = a, prove that xyz = 1.
Solution: Another process:
by = c
Here, ax = b, by = c and cz = a
Now, ax = b or, axy = c [putting b = ax in by = c]
or, cxz = b [By putting a = cz in ax = b] or, cxyz = c [putting a = cz in axy = c]
or, bxyz = b [Putting c = by in cxz = b] or, xyz = 1
or, xyz = 1 proved.
Example 5: If 4p = 5q = 20–r, show that 1 = 1 + 1 = 0.
Solution: p q r
Another process:
Let, 4p = 5q = 20–r = k 4p = 5q, i.e. 4 = 5pq
Then, 4p = k ? 4 = kp1 20–r = 5q, i.e. 20 = 5– q
r
5q = k ? 5 = kq1 Now, 4 × 5 = 20
and 20–r = k 1 or, q q
or, = 5– r
? 20 = k r 5p × 5
Now, 4 × 5 = 20 5q +1 = 5– q
p r
11 1
or, = k– r q q
kp × kq p r
or, +1 = –
k1 + 1 k– 1
or, p q = r q q
p r
1 1 –1 or, +1 + =0
p q r
+ = or, q( 1 + 1 + 1 ) =0
p q r
? 1 + 1 + 1 = 0 Proved 1 1 1
p q r or, p + q + r =0
Example 6: Solve: 2x + 1 = 881 .
2x
Solution:
1 = 818
Here, 2x + 2x
or, 2x + 1 = 65
2x 8
1 65
Let 2x = a, then equation becomes a + a = 8
Now, a + 1 = 65
a 8
a2 + 1 65
or, a = 8
or, 8a2 + 8 = 65a
or, 8a2 – 65a + 8 = 0
or, 8a2 – 64a – a + 8 = 0
or, 8a(a – 8) –1(a – 8) = 0
or, (a – 8) (8a – 1) = 0
Either a – 8 = 0 i.e. a = 8 or, 2x = 23 ?x=3
1 1
or, 8a – 1 = 0 i.e. a = 8 or, 2x = 23 = 2–3 ? x = –3
Hence, x = ±3.
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Indices
EXERCISE 7.2
General section
1. a) If (ap × aq) ÷ ar = ax then express x in terms of p, q and r.
b) If (bm ÷ bn) × bp = by then express y in terms of m, n and p.
c) If (xm)n = xm × xn, express m in terms of n.
d) If ax = a , find the value of x.
e) If am = 1, is the value of m?
f) If xx = 4, what is the value of x?
2. Solve. b) 23x = 8 c) 32x = 1 d) 53x = 1
a) 2x = 4 9 125
e) mx – 2 = m2 f) 52x + 3 = 1 g) 1 = 64 h) 1 = 1
43x 7– 2x 49
i) 2 x 8 j) 4 2x= 16 –2 k) 22x + 1 = 23x – 1 l) 32x + 1 = 92x – 1
3 27 5 25
2=
m) 3 = 3x n) 2x – 2 = 2 × 82 o) 33x – 2 = 27 × 92 p) 53x–1 = 25 × 5x+1
3x 3
q) 42x – 1 – 2x + 1 = 0 r) 9x +1 = 1 s) 252x – 3 = 1 t) 49x +2 = 1
27x 625 343–2x
3. a) If px ÷ p4 = 1, find the value of x. b) If 3m ÷ 34 = 27, find the value of m.
c) If 5t = 0.04, find the value of t. d) If 103r = 1 , find the value of r.
0.001
Creative section - A
4. Solve.
a) 2x+5 = 16 b) (35)x – 1 = 25 c) ( 2)3x – 1 = ( 4)x – 2 d) ( 9)x – 3 = ( 3)x +2
f) (0.3)35x = 0.027
e) (0.5) x = 0.25 g) 2 1 2x = 2x h) 9 1 = 27–x
2 × × 32x
5. Solve.
a) 2x + 1 – 2x = 8 b) 3x + 1 – 3x = 54 c) 2x + 2x + 2 = 5
d) 7x + 7x + 1 = 56 e) 11x + 1 + 11x = 12 f) 3x + 2 + 3x + 1 = 1 1
3
g) 2x + 2x – 1 = 3 h) 3x – 3x – 2 = 8 i) 3x + 5 = 3x + 3 + 8
3
j) 2x + 2 + 2x + 3 = 1 k) 3x + 3 + 3x + 4 = 162 l) 5x + 5x + 1 + 5x + 2 = 155
2 3
m) 32x + 3 – 2.9x + 1 = 1 n) 9x – 2 + 2 × 32x – 3 = 63 o) 23x – 1 + 3 × 8x – 1 = 56
9
6. Solve.
a) 2x + 3 × 3x + 4 = 18 b) 2x + 3 × 3x + 2 = 432
c) 2x – 5 × 5x – 4 = 5
e) 23x – 5 × ax – 2 = 2x – 2 × a1 – x d) 72x + 1 × 52x – 1 = 7
5
f) 75x – 4 × a4x – 3 = 72x – 3 × ax – 2
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Indices
7. a) If a = bc, b = ca and c = ab, prove that abc = 1.
b) If xa = y, yb = z and zc = x, prove that abc = 1.
c) If a1x = b13 and ab = 1, prove that x + 3 = 0.
d) If a = 7x, b = 7y and aybx = 49, show that xy = 1.
e) If ax = by and ay = bx, show that x = y.
f) ax = by = cz and b2 = ac, show that y = 2xz
x+z
Creative section - B
8. a) If (a–1 + b–1) (a + b)–1 = ambn, prove that am – n = 1.
b) If m–1n2 7÷ m3n–5 –5 = mxny, prove that mx – 2y = 1.
m2n–4 m–2n3
9. Solve:
a) 3x + 1 = 9 1 b) 4x + 1 = 16 1
3x 9 4x 16
c) 4x – 6 × 2x + 1 + 32 = 0 d) 25x – 6 × 5x + 1 + 125 = 0
e) 2x + 21–x = 3 f) 3x – 1 + 32–x = 4
Project work
10. a) Take any base number and index number greater than 1 and less than 6. Then
verify the following laws of indices.
(i) Product law (ii) Quotient law
(iii) Power law (iv) Law of negative index
(v) Root law of indices (vi) Law of zero index
b) Write any five exponential equations of your own in the form of ax = b, where a is
positive integer greater than 1 and b is the integer of power of a. Then, solve your
equations.
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Unit Simultaneous Linear Equations
8
8.1 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.
8.2 Method of solving simultaneous equations
There are various methods of solving simultaneous equations. Here, we discuss only
three methods:
(i) Graphical method (ii) Elimination method (iii) Substitution method
(i) Graphical method
In this method, we find a few pairs of solutions of each of two given equations.
The pairs of solutions of each equation are plotted in a graph and by joining
them two separate straight lines are obtained. The coordinates of the point of
intersection of two straight lines are the solutions to the given equations.
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Simultaneous Linear Equations
Worked-out examples
Example 1: From the given graph, find the solutions of the two equations represented
Solution: by the two straight lines.
In the graph, the coordinates of the point of intersection of the (3, 2)
two straight line l1 and l2 is (3, 2).
? (3, 2) is the common solution to the equations
represented by l1 and l2.
? x = 3 and y = 2 are the required solutions represented by
the two straight lines.
Example 2: Solve graphically x + y = –3 and 3x + y = 1.
Solution:
Here, x + y = –3 x0 1 2 (0, 1)
or, y = –x – 3 y –3 –4 –5
Again, 3x + y = 1 x0 1 2 (1, –2)
or, y = 1 – 3x y 1 –2 –5
The coordinates of the point of intersection of the two (0, –3) (2, –5)
(1, –4)
straight lines represented by the equations is (2, –5).
? (2, –5) is the common solution to the equations.
? x = 2 and y = –5.
EXERCISE 8.1
General section
1. Write down the solution to each pair of simultaneous equations represented by
straight-line graphs.
a) b) c)
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d) e) Simultaneous Linear Equations
f)
2. Copy and complete the table of values for x and y. Plot the coordinates separately
from each table in graph. Find the common solution to the equations from the graphs.
a) y = 5 – x y=x–1 b) y = 2x + 2 y = –x – 4
x0 6 x4 x1 –2 x2
y3 y0 –3 y8 y –4 –8
x–8 y = –2x – 4 x–1 –4 – x
c) y = 2 d) y = 3 y= 2
x2 4 x 3 –3 x –2 7 x2
y –1 y –6
y1 y –2 –1
Creative section
3. Solve the following simultaneous equations graphically:
a) y = 8 – x b) y = 2x + 2 c) y = x + 4 d) y = x + 3
y=x–2 y=x–1 y=2–x y = 3x – 1
e) y = x + 4 f) y = 2x – 4 g) x + y = 8 h) x – y = 1
y = 2x – 1 y=x–3 x–y=2 2x + 3y = 12
i) x + 2y = 14 j) x + 3y = 13 k) 5x – 3y = 11 l) 2x – 3y = 4
x–y=5 x+y=7 2x – 3y = –1 3x + y + 5 = 0
m) x + 3y = 7 n) x –2 = 2x – 6 o) x–8 = 3x – 3 p) x + y = 3
3x – y = 11 2 3 4 5 5 4 2
x – y = – 1
10 2 2
(ii) 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.
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Simultaneous 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,
we subtract one equation
x + 2y = 11 from the other to eliminate x.
+– x +– y =+– 7
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
7 = 14 multiplied by 5.
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
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Simultaneous 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+21 = 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 = 23+34 = 17 = RHS
? x = 6 and y = 4 12
Example 4: Solve x 10 y + 9 = 5 and x 15 y – 3 = 2.
Solution: + x–y + x–y
10 y + 9 = 5 ... (i) and x 15 y – 3 = 2 ... (ii)
x+ x–y + x–y
Let x + y = a and x – y = b. Then, equation (i) and (ii) reduce to
10 + 9 = 5 ... (iii) and 15 – 3 = 2 ... (iv) . Then, multiply equation (iv) by 3 and
a b a b adding (iii) and (iv),
10 + 9 + 45 – 9 =5+6
a b a b
or, 55 = 11 ? a=5
a
Now, substituting the value of a in equation (iii), we get,
10 + 9 =5 or, 9 = 3 ?b=3 Checking:
5 b b When x = 4 and y = 1,
From equation (i);
Again x + y = a, i.e. x + y = 5 ... (v)
and x – y = b, i.e. x – y = 3 ... (vi) LHS = 10 + 9 = 10 + 9
x+y x–y 4+1 4–1
Adding equations (v) and (vi), we get,
x+y+x–y=5+3 = 10 + 9 = 5 = RHS
or, x = 4 5 3
Substituting the value of x in equation (v), we get,
From equation (ii);
4+y=5 ?y=1 LHS 15 – 3 = 15 – 3
x+y x–y 4+1 4–1
15 3
Hence, x = 4 and y = 1. = 5 – 3 = 2 = RHS
(iii) 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.
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Simultaneous Linear 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
y+1 4 y–2 3
Solution:
x–1 3
y+1 = 4
or, 4x – 4 = 3y + 3
or, 4x – 3y = 7
or, 4x = 7 + 3y
? 7 + 3y
x = 4 ... (i)
Also,
x+2 = 4
y–2 3
or, 3x + 6 = 4y – 8
or, 3x – 4y = –14 ... (ii) Checking:
Now, substituting the value of x in equation (ii), we get, When x = 10 and y = 11,
3 7 + 3y – 4y = –14 From equation (i);
4
or, 21 + 9y – 16y = –14 LHS = yx+–11=1110+–11=192= 3 RHS
4 4
or, 21 – 7y = –56 From equation (ii);
or, 77 = 7y ? y = 11 LHS = xy+–22=1101+–22=192= 4 RHS
3
Again, substituting the value of y in equation (i), we get,
x= 7 + 3× 11 = 10
4
Hence, x = 10 and y = 11.
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Simultaneous Linear Equations
Example 7: The sum of two numbers is 20. If three times the smaller number is equal
Solution: to two times the bigger one, find the numbers.
Let the bigger number be x and the smaller be y.
From the first condition, From the second condition,
x + y = 20 3y = 2x .................... (ii)
or, y = 20 – x ............... (i)
Substituting the value of y from equation (i) in equation (ii), we get,
3 (20 – x) = 2x Checking:
or, 60 – 3x = 2x x = 12 and y = 8
or, x = 12 x + y = 12 + 8 = 20
Now, substituting the value of x in equation (i), we get,
2x = 2 × 12 = 24
y = 20 – 12 = 8 3y = 3 × 8 = 24
Hence, the required numbers are 12 and 8.
Example 8: The cost of four pencils and three pens is Rs 68. If two pencils and a pen
Solution: cost Rs 26, find the cost of each pencil and each pen.
Let, the cost of each pencil be Rs x and each pen be Rs y.
From the first condition, From the second condition,
4x + 3y = 68 ............. (i) 2x + y = 26
or, y = 26 – 2x ................. (ii)
Substituting the value of y from equation (ii) in equation (i), we get
4x + 3 (26 – 2x) = 68
or, 4x + 78 – 6x = 68
or, x = 5
Now, substituting the value of x in equation (ii), we get, y = 26 – 2 × 5 = 16
Hence, each pencil costs Rs 5 and each pen costs Rs 16.
Example 9: A father is three times as old as his daughter. Six years ago, he was four
Solution: times as old as his daughter was. Find their present age.
Let the present age of the father be x years and that of the daughter be y years.
From the first condition, From the second condition,
x = 3y .................. (i) x – 6 = 4 (y – 6) ............. (ii)
Substituting the value of x from equation (i) in equation (ii),
3y – 6 = 4 – 24
or, y = 18
Substituting the value of y in equation (i), we get, x = 3 × 18 = 54
Hence, the present age of the father is 54 years and that of the daughter is 18 years.
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Simultaneous Linear Equations
Example 10: The sum of the digits of a two-digit number is 7. When the places of the
digits are interchanged, the new number so formed will be 27 more than
the previous one. Find the number.
Solution:
Let the digit at tens place be x and at ones place be y.
Then, the number is 10x + y
When the places of the digits are interchanged the new number is 10y + x.
From the first condition, From the second condition,
x+y =7 10y + x = 10x + y + 27
or, y = 7 – x ................... (i) or, y – x = 3 ...................(ii)
Substituting the value of y from equation (i) in equation (ii),
7–x–x =3
or, x = 2
Substituting the value of x in equation (i), we get,
y =7–2=5
Hence, the required number is 10x + y = 10 × 2 + 5 = 25.
EXERCISE 8.2
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
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Simultaneous Linear Equations
d) 3x + 5y = 11 e) 2x + 3y = 3 f) 6x + 7y = 5
4x – y = 7 x + 2y = 1 7x + 8y = 6
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) 2 + 6 = 3 n) 9 – 4 = 5 o) x + 6 = 4
x y x y 5 y
10 9 3 8 x 9
x – 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
2x – y = 3 x = 2y + 1
3x + 4y = 2 2x + y = 6
i) 2x + y = 8 k) x + 3y = 7 j) 2x + y = 4 l) 2x – 3y = 6
2x – y = 8 x + 3y = 3
x–y=1 3x – y = 11
m) x–1 = 1 n) x–2 = 1 o) x+ y–2 =6
y+1 2 y–2 2 3
x–2 = 1 x + 2 = 3 y + x + 1 = 8
y+2 3 y + 2 4 2
Creative section - B
5. a) The sum of two numbers is 30. If three times the smaller number is equal to two
times the bigger one, find the numbers.
b) The total cost of a cap and a sunglasses is Rs 1,250. If the cost of the sunglasses is
Rs 250 more than the cost of the cap, find the cost of each of these two items.
c) The total cost of a bag and an umbrella is Rs 2,000. If the cost of the bag is Rs 1,000
less five times the cost of the umbrella, find the cost of each of these two items.
6. a) The cost of a pen is three times and Rs 4 more than the cost of a pencil. If three
pencils and two pens cost Rs 71, find the cost of each item.
b) The cost of tickets of a comedy show of 'Gaijatra' is Rs 700 for an adult and
Rs 500 for a child. If a family paid Rs 3,100 for 5 tickets, how many tickets were
purchased in each category?
c) The cost of 4 kg of chicken and 5 kg of mutton is Rs 7,200. If the cost of 4 kg of
chicken is the same as the cost of 1 kg of mutton, find the rate of cost of chicken
and mutton.
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Simultaneous Linear Equations
7. a) Mother is three times as old as her daughter. Three years ago she was four times as
old as her daughter was. Find their present ages.
b) The sum of the present ages of a father and his son is 40 years. Four years hence
father will be three times as old as his son will be. Find their present ages.
c) The differences of the present ages of a father and his daughter is 30 years. Four
years ago father was seven times as old as his daughter was. Find their present
ages.
d) Three years hence a father will be four times as old as his son will be. Before two
years he was seven times as old as his son was. Find their present ages.
8. a) The sum of the digits of a two-digit number is 7. When the places of the digits are
interchanged the new number so formed will be 9 more than the previous one.
Find the number.
b) The sum of the digits of a two-digit number is 10. If 18 is subtracted from the
number, the places of the digits are reversed. Find the number.
c) A certain number of two-digit is three times the sum of its digits. If 45 is added to
it, the digits are reversed. Find the number.
Project work!
9. 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
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) + = .......
2x – y = .......
x – y = ....... x +2y = ....... 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
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Unit Quadratic Equation
9
9.1 Quadratic equation – review
Let's consider two equations x + 3 = 0 and x + y = 7.
In x + 3 = 0, the highest power of the variable x is 1. So, it is called a first degree
equation of one variable. A first degree equation is also called a liner equation.
In x + y = 7, the highest power of each variable is 1. Therefore, it is also a first degree
equation with two variables x and y.
On the other hand, let's consider another equation x2 + 2x + 3 = 0. In this equation,
the highest power of the variable x is 2. So, it is called a second-degree equation. A
second degree equation of one variable is also called a quadratic equation.
Every quadratic equation can be expressed in the form ax2 + bx + c = 0, where a,
b, and c are real numbers and a ≠ 0. Therefore, ax2 + bx + c = 0 is known as the
standard form of a quadratic equation.
The table given below shoes the various forms of quadratic equations.
Standard form of Every quadratic equation can be expressed in the form
quadratic equation ax2 + bx + c = 0, is known as the standard form of a quadratic
equation.
Adfected quadratic The standard form of quadratic equation ax2 + bx + xc = 0
equation which has the variables both in degree 2 and 1 is known as
adfected quadratic equal. For example,
x2 + 5x + 6 = 0, 2x2 + 3x – 2 = 0, 3x2 – 4x = 0, etc. are
adfected quadratic equations.
Pure quadratic The quadratic equation of the form ax2 + c = 0, where the
equation middle term with variable of degree 1 is missing is known as
pure quadratic equation. For example:
x2 – 4 = 0, x2 = 9, 2x2 – 6 = 0, etc. are pure quadratic equations.
9.2 Solution of quadratic equations
A quadratic equation is a second degree equation. Therefore, we obtain two solutions
(or roots) of the variable from a quadratic equation. In other words, a quadratic
equation has two roots. There are various methods of solving quadratic equations.
Here, we discuss three methods.
(i) Factorisation method (ii) Completing square method (iii) By using formula
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Quadratic Equations
(i) Solving a quadratic equation by factorisation method
In this method, we factorise the quadratic expressions ax2 + bx + c or ax2 – b and
express as the product of two linear factors. Then, each linear factor is separately
solved to get the required solutions of the equation by applying zero factor property.
In zero factor property, if p . q = 0, , either p = 0, or q = 0. In other words, if the
product of two numbers is 0, one or both of the numbers must be 0.
Worked-out examples
Example 1: Solve: x2 – 9 = 0.
Solution:
Alternative process Checking the solutions
When x = – 3
x2 – 9 = 0 x2 – 9 = 0 x2 – 9 = 0
or, x2 – 32 = 0 or, x2 = 9 or, (–3)2 – 9 = 0
or, (x + 3) (x – 3) = 0 or, x2 = (±3)2 or, 9 – 9 = 0
or, x = ± 3
0=0
Either, x + 3 = 0, i.e. x = – 3 When x = 3, x2 – 9 = 0
or, 32 – 9 = 0
or, x – 3 = 0, i.e. x = 3 or, 9 – 9 = 0
? x = ±3 0=0
So, x = –3 and 3 are
Remember, (+3)2 = 9 and (–3)2 = 9 correct solutions.
Therefore, 9 = (±3)2
Example 2: Solve: 2x2 + 3x –2 = 0. Checking the solutions
When x = –2
Solution: 2 × (–2)2 + 3 × (–2) – 2 = 0
or, 8 – 6 – 2 = 0
2x2 + 3x – 2 = 0
or, 2x2 + 4x – x – 2 = 0 or, 0 = 0
or, 2x (x + 2) – 1 (x + 2) = 0 When x = 1
2
or, (x + 2) (2x – 1) = 0 1 3
2 2
Either, x + 2 = 0, i.e. x = –2 2 × ( )2 + – 2 =0
or, 2x – 1 = 0, i.e. x = 1 or, 1 + 3 – 2 =0
2 2 2
? x = –2 and 1 =0
2 or, 4 – 2 =0
or, 2 0
Hence, x = –2 and 1 are the correct
solutions. 2
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Example 3: Solve: 11 – x = 5 Example 4: Solve: 3x – 2 = 3x – 8
Solution: x+3 x+1 Solution: 2x – 3 x+4
11 – x = x 5 1 3x – 2 = 3x – 8
x+3 + 2x – 3 x+ 4
or, (11 – x) (x + 1) = 5(x + 3) or, (3x – 2) (x + 4) = (2x – 3) (3x – 8)
or, 11x + 11 – x2 – x = 5x + 15 or, 3x2 + 12x – 2x – 8 = 6x2 – 16x – 9x + 24
or, –x2 + 10x – 5x + 11 – 15 = 0 or, 3x2 + 10x – 8 = 6x2 – 25x + 24
or, –x2 + 5x – 4 = 0 or, 3x2 – 35x + 32 = 0
or, x2 – 5x + 4 = 0 or, 3x2 – 32x – 3x + 32 = 0
or, x2 – 4x – x + 4 = 0 or,x(3x – 32) – 1(3x – 32) = 0
or, x(x – 4) – 1(x – 4) = 0 or, (3x – 32) (x – 1) = 0
or, (x – 4) (x – 1) = 0 Either, 3x – 32 = 0, i.e., x = 32 = 1032
3
Either, x – 4 = 0, i.e., x = 4 or, x – 1 = 0, i.e., x = 1
or, x – 1 = 0, i.e., x = 1 Hence, 1032 and 1 are the required solutions of
the given equation.
Hence, 4 and 1 are the required
solutions of the given equation.
Remember that we have to simplify the equation in the form of rational expression till we
get ax2 + bx + c = 0 form. Then, we should factorise it and get the required solutions of
the given equation.
Example 5: Solve: x+3 – 2x – 3 = x–3 .
Solution: x+2 x–1 2–x
x+3 – 2x – 3 = x–3
x+2 x–1 2–x
or, x + 3 – x–3 = 2x – 3
x + 2 2–x x–1
or, x+3 + x–3 = 2x – 3
x+2 x–2 x–1
or, (x – 2) (x + 3) + (x + 2) (x – 3) = 2x – 3
(x + 2) (x – 2) x–1
or, (x – 1) (2x2 – 12) = (2x – 3) (x2 – 4)
or, 2x3 – 12x – 2x2 + 12 = 2x3 – 8x – 3x2 + 12
or, 2x3 – 12x – 2x2 + 12 – 2x3 + 8x + 3x2 – 12 =0
or, x2 – 4x =0
or, x (x – 4) =0
Either, x = 0
or, x – 4 = 0, i.e. x = 4
? x = 0 and 4
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Quadratic Equations
Example 6: Solve: x + 1–x = 2 1
1–x x 6
Solution:
Let x = a, then 1–x = 1
1–x x a
Now, a + 1 = 13
or, a 6
or, a2 + 1 13
= 6 When, a = 3 , x = 3
a 2 1–x 2
6a2 + 6 = 13a
or, x = 9
or, 1–x 4
or, 6a2 – 13a + 6 = 0 or,
4x = 9 – 9x
9
or, 6a2 – 9a – 4a + 6 = 0 x = 13
or, 3a(2a – 3) –2(2a – 3) = 0 2 x 2
3 1–x 3
or, (2a – 3) (3a – 2) =0 When, a = , =
Either, 2a – 3 = 0, i.e. a
or, 3a – 2 = 0, i.e. a = 3 or, x = 4
2 or, 1–x 9
2 or,
= 3 9x = 4 – 4x
4
9 4 x = 13
13 13
Hence, and are the required solutions.
Example 7: Find the quadratic equation whose roots are –2 and 3.
Solution:
Since the roots of the required quadratic equation are –2 and 3,
x = –2 and x = 3
x + 2 = 0 and x – 3 = 0
? (x + 2) (x – 3) = 0
or, x2 – 3x + 2x – 6 = 0
or, x2 – x – 6 = 0
Hence, x2 – x – 6 = 0 is the required quadratic equation.
EXERCISE 9.1
General section
1. Use the zero factor property to solve each equation.
a) (x + 1) (x – 1) = 0 b) (x + 2) (x – 3) = 0 c) (x – 3) (2x + 1) = 0
f) 3x (4 – 3x) = 0
d) (2x – 5) (5x – 2) = 0 e) x (2x – 1) = 0
2. Solve each equation:
a) x2 = 9 b) 2x2 = 8 c) 9y2 = 16 d) 4 = 25
y2
e) x2 – 36 = 0 f) x2 – 81 = 0 g) 25x2 – 64 = 0 h) 49y2 – 100 = 0
i) 2x = 6 j) 3x = 4 k) 5 – 36 = 0 l) 64x – 7 =0
3 x 4 3x x2 5 7 x
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Quadratic Equations
3. Find the quadratic equations which have the following roots.
a) 3 and 5 b) –2 and 3 c) –4 and –2 d) 4 and –3
Creative section - A
4. Solve each equation by factorisation method:
a) x2 – 4x = 0 b) 2x2 – 10x = 0 c) 3x2 – 12x = 0
f) x2 – 3x + 2 = 0
d) x2 + 3x + 2 = 0 e) x2 + 7x + 12 = 0 i) 2x2 – 3x + 1 = 0
g) x2 – 8x + 15 = 0 h) x2 – x – 12 = 0 l) 2x2 + 21 = 17x
j) 3x2 – 10x – 8 = 0 k) 5x2 + 8x – 21 = 0
Creative section - B
5. Solve these equations by factorisation method:
a) x– 1 = 6 b) x 3 = x 2 1 c) 1 1 = x 1
x– 2 x 2x – – 2x + 4x –
d) x 2 1 = 3x 1 e) 3x – 8 = 5x – 2 f) 1 – 4x = 1 4
+ 5x – x–2 x+5 2x + 1 – 5x
g) 3 + x 4 1 = 2 h) x 3 + 2 = 2 i) x+4 + x–4 = 3 1
x + –1 x–3 x x–4 x+4 3
j) x+5 + x–5 = 2 1 k) x–2 + x+2 = 2x + 6 l) x–2 + 3x – 11 = 4x + 13
x–5 x+5 2 x+2 x–2 x–3 x–3 x – 4 x+1
6. Solve these equations by factorisation method:
a) x + x–3 = 5 b) x + x–2 = 3 1
x–3 x 2 x–2 x 3
c) 2x – 3 – 4 x–1 =3 d) 3x + 1 + x+1 = 2 1
x–1 2x – 3 x+1 3x + 1 2
7. a) The hypotenuse of a right-angled triangle is 10 cm and the difference between the
other two sides is 2 cm, find two unknown sides of the triangle.
b) If the sides of a right-angled triangle are (3x + 1) cm, 3x cm and (x – 1) cm, find
the length of it's sides.
8. a) If the sum of the squares of two consecutive natural numbers is 25, find the
numbers.
b) Find two natural numbers which differ by 3 and the sum of whose squares is 117.
c) If the sum of the squares of two consecutive natural even numbers is 52, find the
numbers.
(ii) Solving a quadratic equation by completing the square
In this method, we transpose the constant term (c) to R.H.S. and L.H.S. is expressed
as a perfect square expression. Study the following example.
Let’s take the standard form of quadratic equation.
ax2 + bx + c = 0
or, ax2 + bx = – c (Transposing c to R.H.S.)
or, ax2 + b x = – c (To make the coefficient of x2 unity, dividing both sides by a)
a a a
or, x2 + b x = – c
a a
or, x2 + b x + b2 = b 2 – c (Adding the square of half of the coefficient of x
a 2a 2a a to both sides)
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Quadratic Equations
or. x2 + 2. b x + b2 = b2 – c
2a 2a 4a2 a
or, x + b 2 = b2 – 4ac
2a 4a2
or, x + b =± b2 – 4ac
2a 2a
or, x = – b ± b2 – 4ac = – b ± b2 – 4ac
2a 2a 2a
Thus, the required roots of x are – b + b2 – 4ac and – b – b2 – 4ac .
2a 2a
Worked-out examples
Example 1: Solve: x2 – 11x + 24 = 0
Solution:
x2 – 11x + 24 = 0
or, x2 – 11x = – 24
or, x2 – 11x + 11 2 = 11 2 – 24
22
11 121
or,x2 – 2.x. 2 x + 11 2 = 4 – 24
2
Taking +ve sign, Taking –ve sign,
or, x – 11 2 = 25 x – 11 = 5 x – 11 = –5
2 4 2 2 2 2
or, x – 11 2 = ± 5 2 or, x = 5 + 11 or, x = –5 + 11
2 2 2 2 2 2
or, x – 11 =± 5 = 16 = 8 = 6 =3
2 2 2 2
Hence, 8 and 3 are the required solutions.
Example 2: Solve: 4x2 – x – 3 = 0 by completing the square.
Solution: 4x2 + x – 3 = 0 Taking +ve sign,
or, 4x2 + x = 3 x + 1 = 7
8 8
or, 4x2 + x = 3 7 1 6 3
4 4 4 or, x = 8 – 8 = 8 = 4
or, x2 + x + 12 = 1 2 + 3
4 8 8 4
Taking –ve sign,
or, x2 + 2.x.81 + 12 = 1 + 3 x + 1 = – 7
8 64 4 8 8
or, x + 1 2 = 49 or, x = – 7 – 1
8 64 8 8
1 = ±87 8
or, x + 8 = – 8 =–1
Hence, –1 and 3 are the required solutions.
4
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Quadratic Equations
Example 3: Solve x 2 1 = x 3 1 + x 4 2 by completing the square.
Solution: – + +
x 2 1 = x 3 1 + 4 2
– + x+
or, 2 = 3 (x + 2) + 4 (x + 1)
x–1 (x + 1) (x + 2)
or, 2 = 7x + 10 2
x–1 x2 + 3x +
or, (x – 1) (7x + 10) = 2(x2 + 3x + 2)
or, 5x2 – 3x – 14 = 0
or, 5x2 – 3x = 14 [Dividing each term by 5]
5 5 5
3 14
or, x2 – 5 x = 5 Taking +ve sign,
or, x2 – 3 x+ 32 = 3 2+ 14 x – 3 = 17
5 10 10 5 10 10
17 3
or, x2 – 2.x.130 + 32 = 9 + 14 or, x = 10 + 10 = 2
10 100 5
Taking –ve sign,
or, x – 3 2 = ±1170 2
10
3 x – 3 = – 17
10 = ±1107 10 10
or, x – 17 3 –7
Hence, or, x = – 10 + 10 = 5
7
2 and – 5 are the required solution.
Example 4: Express x2 – 8x + 21 in the form of (x – a)2 + b, where a and b are whole
numbers. Also, find the roots of x2 + ax – b = 0.
Solution:
x2 – 8x + 21 = x2 – 8x + 42 – 42 + 21
= x2 – 2.x.4 + 42 – 16 + 21
= (x – 4)2 + 5, which is in the form (x – a)2 + b. where a = 4 and b = 5.
Again, x2 + ax – b = 0
or, x2 + 4x – 5 = 0
or, x2 + 5x – x – 5 = 0
or, x(x + 5) –1 (x + 5) = 0
or, (x + 5) (x – 1) = 0
Either x + 5 = 0, i.e. x = –5
or, x – 1 = 0, i.e. x = 1
Hence, the required roots of x2 + ax – b = 0 are –5 and 1 when a = 4 and b = 5.
EXERCISE 9.2
General section
1. Express both sides of these equations in square forms. Then, solve:
a) x2 + 2.x. 1 + 12 = 4 b) x2 + 2.x.3 + 32 = 16
c) x2 – 2. x. 4 + 42 = 25 d) x2 – 2.x.5 + 52 = 81
e) x2 + 2.x.12 + 1 2 = 4 f) x2 – 2.x.53 + 3 2 = 4
2 9 5 25
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Quadratic Equations
2. Express both sides of these equations in square forms. Then, solve.
a) x2 + x + 1 2= 1 2+6 b) x2 – 7x + 7 2 = 7 2 – 12
2 2 22
c) x2 – 3x + 3 2 = 3 2 + 10 d) x2 – 10x + 52 = 52 – 21
22
e) x2 – x + 1 2 = 1 2 + 20 f) x2 + 5x + 5 2 = 5 2 + 24
22 22
Creative section - A
3. Solve each equation by completing the square:
a) x2 – 3x = –2 b) x2 + 2x = 8 c) x2 + 3x = 10
d) x2 – 2x – 3 = 0 e) x2 – 7x + 12 = 0 f) x2 + 3x – 28 = 0
g) 2x2 – 7x + 3 = 0 h) 3x2 + 5x – 2 = 0 i) 6x2 – 5x – 6 = 0
j) 3x2 = 10x – 3 k) 5x2 = 8x + 21 l) 15x2 – 2ax = a2
m) x2 + 2 = 3x n) 9x + 1 = 6 o) x + 3 = 3x – 7
4 2 x x + 2 2x – 3
p) 3x – 8 = 5x – 2 q) 2x + 2x – 5 = 8 1 r) 2x – 1 – 2x + 1 = –2 2
x–2 x+5 x–4 x–3 3 2x + 1 2x – 1 3
Creative section - B
4. a) Express x2 + 4x + 7 in the form of (x + a)2 + b, where a and b are whole numbers.
Hence, solve the equation x2 – bx + a = 0
b) Express x2 – 10x + 27 in the form of (x – p)2 + q, where p and q are whole numbers.
Hence, solve the equation x2 – px + q2 = 0
c) Express 2x2 – 4x + 5 in the form of a(x – h)2 + k, where a, h, and k are whole
numbers. Hence, find the roots of x2 + hx – ak = 0.
(iii) Solving a quadratic equation by using formula
We obtained the following two roots of x while solving the quadratic equation of
the form ax2 + bx + c = 0 by completing the square.
x= – b + b2 – 4ac and x = – b – b2 – 4ac
2a 2a
These roots can be written as x = – b ± b2 – 4ac
2a
where, a is the coefficient of x2, b is the coefficient of x and c is the constant
term. We use this formula to find the required solutions of the given quadratic
equation.
For the quadratic equation ax2 + bx + c = 0, a ≠ 0, the expression b2 – 4ac is
called discriminant. It is generally denoted by 'D'.
? D = b2 - 4ac.
Worked-out examples
Example 1: Solve 16x2 – 9 = 0 by using the formula.
Solution: 16x2 – 9 = 0
or, 16x2 + 0.x + (–9) = 0
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Quadratic Equations
Here, a = 16, b = 0 and c = 9
? x = – b ± b2 – 4ac = –0± 02 – 4 × 16 × (–9) = ± 24 =± 3
2a 2 × 16 32 4
Example 2: Solve 2x2 + 5x – 12 = 0 by using formula.
Solution:
2x2 + 5x – 12 = 0
or, 2x2 + 5x + (–12) = 0
Here, a = 2, b = 5 and c = –12
? x = – b ± b2 – 4ac = –5± 52 – 4 × 2 × (–12) = – 5 ± 11
2a 2×2 4
Taking +ve sign, Taking –ve sign,
x = – 5 + 11 = 6 = 3 x = – 5– 11 = – 16 = –4
4 4 2 4 4
? x = 3 and – 4
2
Example 3: Solve x–2 + x+2 = 2x + 6 by using the formula.
Solution: x+2 x–2 x–3
x–2 + x+2 = 2x + 6
x+2 x–2 x–3
or, (x – 2)2 + (x + 2)2 = 2x + 6
(x + 2) (x – 2) x–3
or, x2 – 4x + 4 + x2 + 4x + 4 = 2x + 6
x2 – 4 x–3
or, 2x2 + 8 = 2x + 6
x2 – 4 x–3
or, (x2 – 4) (2x + 6) = (x – 3) (2x2 + 8)
or, 2x3 + 6x2 – 8x – 24 = 2x3 + 8x – 6x2 – 24
or, 12x2 – 16x = 0
or, 4 (3x2 – 4x) = 0
or, 3x2 + (–4) x + 0 = 0
Here, a = 3, b = – 4 and c = 0
? x = – b ± b2 – 4ac = – (–4) ± (–4)2 – 4 × 3 × 0 = 4 ± 4
2a 2×3 6
Taking +ve sign, Taking –ve sign,
x = 4 + 4 = 4 x = 4 – 4 = 0 = 0
6 3 6 6
? x = 4 and 0.
3
Example 4: The sum of two numbers is 9 and their product is 20. Find the numbers.
Solution:
Let one of the numbers be x and another be y.
From the first condition, From the second condition,
x+y=9
or, y = 9 – x ................... (i) xy = 20 ....................... (ii)
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Quadratic Equations
Substituting the value of y from equation (i) in equation (ii),
x (9 – x) = 20
or, 9x – x2 – 20 = 0
or, x2 – 9x + 20 = 0
or, x2 – 5x – 4x + 20 = 0
or, x (x – 5) – 4 (x – 5) = 0
or, (x – 5) (x – 4) = 0
Either, x – 5 = 0, i.e. x = 5
or, x – 4 = 0, i.e. x = 4
Substituting x = 5 in equation (i), y = 9 – 5 = 4 or
Substituting x = 4 in equation (i) y = 9 – 4 = 5
Hence, the required numbers are 4 and 5.
Example 5: The perimeter of a rectangle is 36 cm and its area is 80 cm2. Find its
Solution: length and breadth.
Let the length of the rectangle be x cm and breadth be y cm.
From the first condition, From the second condition,
2 (x + y) = 36 xy = 80 ....... (ii)
or, x + y = 18
or, y = 18 – x ....... (i)
Substituting the value of y from equation (i) in equation (ii),
x (18 – x) = 80
or, 18x – x2 – 80 = 0
or, x2 – 18x + 80 = 0
or, x2 – 10x – 8x + 80 = 0
or, x (x – 10) – 8 (x – 10) = 0
or, (x – 10) (x – 8) = 0
Either, x – 10 = 0, i.e. x = 10
or, x – 8 = 0, i.e. x = 8
Substituting x = 10 in equation (i), y = 18 – 10 = 8
Substituting x = 8 in equation (i), y = 18 – 8 = 10
Hence, the length of the rectangle is 10 cm and breadth is 8 cm.
EXERCISE 9.3
General section
1. a) Define quadratic equation with an example.
b) What are the roots of equation ax2 + bx + c = 0, a ≠ 0?
c) In the quadratic equation x2 – 5x + 6 = 0, what is the product of roots?
d) What are the roots of the quadratic equation x2 – 25 = 0?
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Quadratic Equations
2. Replace a, b, and c by their given values and evaluate x = – b ± b2 – 4ac .
2a
a) a = 1, b = 3, c = 2 b) a = 1, b = 5, c = 6
c) a = 1, b = 1, c = –6 d) a = 2, b = –5, c = 2
e) a = 3, b = – 10, c = 8 f) a = 2, b = – 13, c = 15
Creative section - A
3. Solve each equation by using the formula:
a) x2 – 4 = 0 b) x2 = 9 c) 2x2 – 32 = 0 d) 50x2 = 8
h) 15x2 = – 10x
e) x2 – x = 0 f) 2x2 – 6x = 0 g) 6x2 = –3x
4. Solve these equations by using the formula:
a) x2 – 3x + 2 = 0 b) x2 – 5x + 6 = 0 c) x2 + 2x – 3 = 0
d) x2 + 2x – 8 = 0 e) 3x2 – 5x + 6 = 2(x2 + 3) f) 4x(x + 7) + 10 = 2(x2 + 5)
g) 2x2 + 3x + 1 = 0 h) 3x2 – 2x – 21 = 0 i) 3x2 + 10 = –11x
j) (x + 1) (2x + 3) = 4x2 – 22 k) x + 2 = 7 l) 4x2 = 4 x + 3
Creative section - B 3 x 3 15
5. Solve:
a) x+3 = 2x – 1 b) 1 = 2 (x – 3) c) x + x+1 = 13
2x – 7 x–3 x–2 x (x – 1) x+1 x 6
d) x+3 + x–3 = 221 e) x–1 + x–3 = 331 f) 1 + 2 = 6
x–3 x+3 x–2 x–4 x–2 x–1 x
g) x+3 + x–3 = 2x – 3 h) x–1 + x+2 = x–3 + x+4
x+2 x–2 x–1 x+1 x–2 x+3 x–4
6. Solve the following problems by using any method:
a) The sum of two numbers is 10 and their product is 21. Find the numbers.
b) The difference of two numbers is 3 and their product is 108. Find the numbers.
c) The sum of the ages of a father and his son is 36 years and the product of their ages
is 180. Find their ages.
d) Find the length and the breadth of a rectangular plot whose area is 660 sq.m and
its perimeter is 104 m.
e) The length of a room is 2 m longer than its breadth. If the area of its floor is
63 sq.m, calculate the length and the breadth of the room.
Project work
7. a) Make the quadratic equations in x and solve to get the following values of x.
(i) x = ±1 (ii) x = ±2 (iii) x = ±7 (iv) x = ±9
(v) x = 2, 3 (vi) x = 3, –2 (viii) x = –3, 2 (viii) x = –3, –2
b) Write one quadratic equation of each of the following forms and solve them.
(i) x2 + ax + b = 0 (ii) x2 + ax – b = 0 (iii) x2 – ax – b = 0 (iv) x2 – ax + b = 0
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Unit Ratio and Proportion
10
10.1 Ratio – review
Suppose Ram has Rs 100 and Hari has Rs 50. One way to compare their money is by
subtraction. Since Rs 100 – Rs 50 = Rs 50, Ram has Rs 50 more than Hari.
Rs 100
We can also compare their money by division. Since Rs 50 = 2, Ram has two times
(or double) as much money as Hari has.
A comparison of quantities of the same kind by division is called a ratio.
Thus, a ratio tells how many times a quantity is greater or smaller than another
quantity of the same kind.
If a and b are any two quantities of the same kind and expressed in the same unit, the
a b
ratio of a to b is a : b or b and that of b to a is b : a or a .
10.2 Different types of ratios
(i) Compounded ratio
Suppose a : b and c : d be two ratios. Then, a : b × c : d i.e. ac : bd is called a
compounded ratio of these two given ratios. Thus, a compounded ratio is the
product of two or more ratios. For example, 9 5 3
10 3 2
the compounded ratio of 9 : 10 and 5 : 3 = × = = 3 : 2
(ii) Duplicate and sub-duplicate ratios
Let's take a ratio a : b. When it is compounded with itself a new ratio of square
terms so formed is called a duplicate ratio of a : b, and a : b is called the
sub-duplicate ratio of the duplicate ratio.
Duplicate ratio of a : b = a × a = a2 = a2 : b2
b b b2
Sub-duplicate ratio of a2 : b2 = a : b
Thus, the duplicate ratio of 2 : 3 = 22 : 32 = 4 : 9
The sub-duplicate ratio of 25 : 16 = 25 : 16 = 5 : 4
(iii) Triplicate and sub-triplicate ratios
Let's take a ratio a : b. When it is multiplied by itself three times, a compounded
ratio of cube terms so formed is called a triplicate ratio of a : b, and a : b is
called the sub-triplicate ratio of the triplicate ratio.
a a a a3
Triplicate ratio of a : b = b × b × b = b3 = a3 : b3
Sub-triplicate ratio of a3 : b3 = a : b
Thus, triplicate ratio of 2 : 3 = 23 : 33 = 8 : 27
The sub-triplicate ratio of 125 : 64 = 3125 : 364 = 5 : 4
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Ratio and Proportion
(iv) Inverse ratio (Reciprocal ratio)
Let's take a ratio a : b. Here, a is called the antecedent and b is called the
consequent. If we interchange the antecedent and consequent the new ratio
b : a so formed is called the inverse (or reciprocal) ratio of a : b.
1
Thus, the inverse or reciprocal ratio of 2 : 5 = 2 =5:2
5
Worked-out examples
Example 1: Find the ratio of 80 cm to 1.2 m.
Solution:
Here, 1.2 m = 120 cm
? The ratio of 80 cm to 120 cm = 80 = 2 = 2 : 3
120 3
Example 2: Compare the ratios 2 : 3 and 4 : 5.
Solution:
Here, 2 : 3 = 2 and 4 : 5 = 4
3 5
L.C.M. of the denominators 3 and 5 is 15.
? 23 = 2 × 5 = 10 and 4 = 4 × 3 = 12
3 × 5 15 5 5 × 3 15
Since, 10 < 12 , 2 : 3 < 4 : 5
15 15
Example 3: If a : b = 2 : 5 and b : c = 3 : 4, find a : c and a : b : c.
Solution: 2 3
5 4
Here, a : b = 2 : 5 = and b : c = 3 : 4 =
Now, a : b × b : c = a × b = a = a : c = 2 × 3 = = 3 : 10
b c c 5 4
Again, the numbers represented by b in a : b and b : c are 5 and 3 respectively.
? L.C.M. of 5 and 3 is 15
Now, a = 2 = 2×3 = 6 and b = 3 = 3×5 = 15
b 5 5×3 15 c 4 4×5 20
? a : b : c = 6 : 15 : 20
Example 4: If x + 3y : 3x + 2y = 3 : 4, find x : y and x2 + y2 : x2 – y2.
Solution:
Here, x + 3y : 3x + 2y = 3 : 4
i.e. x + 3y = 3
3x + 2y 4
or, 9x + 6y = 4x + 12y
or, 5x = 6y
x 6
or, y = 5
? x:y = 6
5
Again, x2 = 62 = 36 and y2 = 52 = 25
? x2 + y2 : : x2 – y2 = 36 + 25 : 36 – 25 = 61 : 11
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 145 Vedanta Excel in Mathematics - Book 9
Ratio and Proportion
Example 5: If a : b = 2 : 5, find the value of 5a – b : 5a + b.
Solution: Alternative process
Here, a : b = 2 : 5 5a – b 5a – b 5× a – 1
Let, a = 2x and b = 5x 5a + b b b
= =
Now, 5a – b : 5a + b 5a + b 5× a
b b +1
= 5a – b
5a + b
2
= 5 × 2x – 5x = 5x =1:3 = 5 × 5 –1 = 1 = 1 : 3
5 × 2x + 5x 15x 5 × 2 +1 3
5
Example 6: Divide Rs 840 among three persons Anamol, Shadikshya and Shadeep in the
ratio 3 : 4 : 5.
Solution:
Let, the shares of Anamol, Shadikshya and Shadeep be Rs 3x, Rs 4x and Rs 5x respectively.
Now, 3x + 4x + 5x = Rs 840
or, 12x = Rs 840
or, x = Rs 70
? The share of Anamol = Rs 3x = Rs 3 × 70 = Rs 210
The share of Shadikshya = Rs 4x = Rs 4 × 70 = Rs 280
The share of Shadeep = Rs 5x = Rs 5 × 70 = Rs 350
Example 7: The ratio of copper and zinc in a certain mass of brass is 3:2. If the mass
Solution: of copper is 240 g, calculate the mass of zinc.
Let the mass of zinc be x g.
Now, 240 = 3
x 2
or, x = 160
Hence, the required mass of zinc is 160 g.
Example 8: Two numbers are in the ratio 5 : 7. If their LCM is 140, find the numbers.
Solution:
Let the required number be 5x and 7x.
According to the question; LCM of 5x and 7x = 140
or, x × 5 ×7 = 140
? x =4
Hence, the first number = 5x = 5 × 4 = 20
and the second number = 7x = 7 × 4 = 28.
Example 9: Two numbers are in the ratio 4 : 5. When 4 is added to each term, the ratio
Solution: becomes 5 : 6. Find the numbers.
Vedanta Excel in Mathematics - Book 9 146 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Ratio and Proportion
Let the required numbers be 4x and 5x respectively.
According to the question, 4x + 4 = 5
5x + 4 6
or, 24x + 24 = 25x + 20
or, x = 4
? The first number = 4x = 4 × 4 = 16
The second number = 5x = 5 × 4 = 20
Example 10: In 40 l of milk, the ratio of pure milk and water is 3 : 5. How much pure
milk should be added to the mixture, so that the ratio of milk and water
will be 4 : 5?
Solution:
Let the volume of pure milk in 40 l of milk = 3x l.
The volume of water in 40 l of milk = 5x l
Now, 3x + 5x = 40 l
or, x = 5 l
? Volume of pure milk = 3x = 3 × 5 l = 15 l
Volume of water = 5x = 5 × 5 l = 25 l
Again, the let the volume of pure milk to be added be y.
Then, according to question; 15 + y = 4
or, 25 y = 55
Hence, the required volume of pure milk to be added is 5 l.
Example 11: The ratio of the present ages of Bhurashi and her mother is 1 : 4. After 6
years, the ratio of their ages will be 5 : 14. Find their present ages.
Solution:
Let the present ages of Bhurashi and her mother be x year and 4x year respectively.
Now, according to the question,
x+6 = 5
4x + 6 14
or, 20x + 30 = 14x + 84
or, x = 9
? The present age of Bhurashi = x = 9 years
The present age of her mother = 4x = 4 × 9 = 36 years
EXERCISE 10.1
General section
1. Rewrite and complete the following by filling the blank spaces with correct answers.
a) In the ratio 2 : 3, ............... is the antecedent and ............... is consequent.
b) The duplicate ratio of 4 : 5 is ...............
c) The sub-duplicate ratio of 81:100 is ...............
d) The triplicate ratio of 1 : 2 is ...............
e) The sub-triplicate ratio of 27:64 is ...............
f) The inverse ratio of 8 : 9 is ...............
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 147 Vedanta Excel in Mathematics - Book 9
Ratio and Proportion
2. Find the ratios of the following:
a) 60 cm and 1.5 m b) 1.5 kg and 750 g c) Rs 2.50 and 75 paisa
d) 5 pairs and a dozen
e) 2.5 years and 9 months
3. Find the compounded ratios of the following ratios:
a) 3 : 5 and 5 : 6 b) 2 : 5 and 10 : 7 c) 7 : 4 and 8 : 21
4. Find the duplicate ratios of the following ratios:
a) 3 : 4 b) 5 : 2 c) 6 : 7
c) 49 : 36
5. Find the sub-duplicate ratios of the following: c) 7 : 6
a) 4 : 25 b) 16 : 9
6. Find the triplicate ratios of the following ratios:
a) 2 : 3 b) 5 : 4
7. Find the sub-triplicate ratios of the following ratios:
a) 27 : 8 b) 64 : 125 c) 216 : 343
8. Find the inverse ratios of the following ratios:
a) 3 : 8 b) 9 : 5 c) 4 : 11
9. Compare the ratios: b) 3 : 5 and 4 : 7 c) 3 : 10 and 4 : 5
a) 1 : 2 and 2 : 3
10. a) If a : b = 3 : 4 and b : c = 2 : 5, find (i) a : c (ii) a : b : c
b) If x : y = 2 : 3 and y : z = 6 : 7, find (i) x : z (ii) x : y : z
11. a) If 6x = 9y, find the value of x : y. b) If 5x – 4y = x + y, find x : y.
c) If 3a – 5b = 41, find a : b. d) If 4m + n = 54, find m : n.
3a + 5b 8m – n
e) If 4x + 3y : 2x + y = 5 : 2, find y : x.
f) If a2 – 2ab + b2 : a2 + 2ab + b2 = 1:4, find a : b.
12. a) If x – 2y = 2y – x find: (i) x : y (ii) (x2 – y2 : x2 + y2)
b) If a : b = 3 : 2, find the value of (i) b – 2a (ii) a2 – ab .
3 ab + b2
c) If x : y : z = 2 : 3 : 4, find the value of x2 + yzzx.
y2 +
d) If 5x – 4y = x + y, find the value of 4x – y .
4x + y
Creative section - A
13. a) Divide Rs 5,250 in the ratio of 2 : 5.
b) An alloy contains nickel and chromium in 3 : 2 ratio, find the mass of each metal
in 540 g of alloy.
c) If the angles of a triangle are in the ratio 1 : 3 : 5, find them.
d) If the angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4, find them.
e) Two numbers are in the ratio 7 : 5 and their difference is 22. Find the numbers.
Vedanta Excel in Mathematics - Book 9 148 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Ratio and Proportion
14. a) The ratio of two numbers is 3 : 5. If the antecedent is 24, find the consequent.
b) The ratio of two numbers is 4 : 7. If the greater number is 49, find the smaller one.
c) The ratio of number of boys and girls in a class is 4 : 3. If there are 20 boys, find
the number of girls.
d) Sterling silver is an alloy of copper and silver in the ratio of 1 : 9. If a certain mass
of alloy contains 85 g of copper, find the mass of silver.
Creative section - B
15. a) If (x + 4) : (3x + 1) is the duplicate ratio of 3 : 4, find the value of x.
b)
If the duplicate ratio of 2 is x–6 , find the value of x.
c) 5 3x – 5
If (x + 1) : (4x – 1) is the triplicate ratio of 2 : 3, find the value of x.
d) If (3a – 7) : (4a + 3) is the sub-triplicate ratio of 8 : 27, find the value of a.
16. a) Two numbers are in the ratio 4 : 3. If 4 is added to each number, the ratio becomes
5 : 4. Find the numbers.
b)
Two numbers are in the ratio 2 : 5. If 8 is subtracted from each number, the ratio
c) becomes 4 : 13. Find the numbers.
d) What is the common number to be added to each term of the ratio 5 : 8 to make the
ratio 4 : 5?
17. a)
What is the common number to be subtracted from each term of the ratio 11 : 8 to
b) get the new ratio 2 : 1?
18. a) The ratio of the present ages of Bishnu and Shiva is 2 : 3. After 4 years, if the ratio
b) of their ages will be 5 : 7, find their present ages.
19. a) The present ages of Ram and Hari are in the ratio 6 : 5. Before 6 years, if the ratio
of their age was 9 : 7, find their present ages.
b)
The ratio of two numbers is 2 : 3 and their LCM is 30, find the numbers.
The ratio of two numbers is 6 : 7 and their LCM is 210, find the numbers.
In 50 l of milk, the ratio of pure milk and water is 2 : 3. How much pure milk should
be added to the mixture so that the pure milk and water will be in 5 : 6 ratio?
In 35 l of an acid, the ratio of acid and water is 3 : 4. How much acid be added to
the solution so that the ratio of acid and water will be 5 : 4?
10.3 Proportion
Let’s take any two ratios 6 : 10 and 15 : 25.
Here, 6 : 10 = 6 = 3 = 3 : 5 and 15 : 25 = 15 = 3 = 3 : 5
10 5 25 5
Thus, the ratios 6 : 10 and 15 : 25 are equal ratios.
The equality of ratios is called a proportion.
So, 6 : 10 and 15 : 25 are said to be in proportion. Here, the terms 6, 10, 15 and 25 are
called proportional.
Suppose , the terms a, b, c and d are in proportion. We can write it as a : b = c : d or
a : b :: c : d and we read it as ‘a is to b as c is to d.’
Here, the terms a, b, c, and d are the first, second, third and the fourth proportional
respectively.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 149 Vedanta Excel in Mathematics - Book 9
Ratio and Proportion
In a proportion, the first and the fourth proportional are called extremes. The second
and the third proportional are called means.
Extremes
Means
a:b c:d
1st 2nd 3rd 4th
Proportionals Proportionals
In a:b = c:d i.e. a = c , so, a × d = b × c
b d
We can use this fact to test whether the given ratios are in proportion or not.
Continued proportion
If a : b = b : c = c : d = ... i.e. a = b = c = ..., the terms a, b, c, d, ... are said to be in
b c d
continuous proportion. In a continued proportion, the ratio of the first to the second
term is equal to the ratio of the second to the third term and so on. In a continued
proportion.
1st proportional 3rd proportional
a:b = b:c
In a = b , b2 = ac Mean proportional
b c
? b = ac , i.e. the mean proportional = 1st proportional × 3rd proportional
10.4 Properties of proportion
Suppose, a, b, c, and d are in proportion. Then, we can verify the following properties
shown by these proportionals.
(i) Invertendo b d
a c a c
If b = d , then =
Proof a c a c
b d b d
Here, = , then 1 ÷ = 1 ÷
or, 1 × b = 1 × d
a c
or, b = d Proved
a c which if
in a c b d
Thus, the property of the proportion b = d , then a = c is known as
invertendo.
Example: For 2 : 3 = 6 : 9 using invertendo property 3 : 2 = 3 × 3 : 2 × 3 = 9 :6
? 2 : 3 = 6 : 9 implies 3 : 2 = 9 : 6
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