Laws of Indices
Furthermore, 23 = 21 2 I understood!
21 u 22 = 2 u (2 u 2) 25 = 22 3 When the same bases are
27 = 23 4 multiplied, we should
22 u 23 = (2 u u (2 u 2 u 2) a5 = a2 3 add their indices!!
23 u 24 = (2 u u 2) u (2 u 2 u 2 u 2)
Similarly, a2 u a3 = (a u a) u (a u a u a)
Thus, if am and an are any two terms with the same base a and the powers m and n
respectively, then, am u an = am + n
2. Quotient law of indices
Let's study the following illustrations and try to investigate the idea of quotient law
of indices.
22 ÷ 2 = 22 = 2×2 = 2 = 22 – 1 I also understood!
2 2 When a base is divided
by another same base,
23 ÷ 2 = 23 = 2 × 2 × 2 = 22 = 23 – 1 powers should be
2 2 subtracted.
25 ÷ 22 = 25 = 2×2×2×2×2 = 23 = 25– 2
22 2×2
24 ÷ 26 = 24 = 2×2×2×2 = 1 = 1
26 2×2×2×2×2×2 22 26 – 4
Similarly, a5 y a2 = a5 = a×a×a×a×a = a3 =a5-2
a2 a×a
Thus, if am and an are any two terms with the same base a and the powers m and n
1
respectively, then, am ÷ an = am – n if m > n and am ÷ an = an – m if n > m
3. Power law of indices
(i) Let's study the following illustrations and try to investigate the idea of power
law of indices.
(22)3 = 22 u 22 u 22 = 22 2 2 = 26 = 22 u 3 When a base with some
(22)4 = 22 u 22 u 22 u 22 = 22 2 2 2 = 22 u 4 = 28 power has another power,
Similarly, the powers are multiplied.
(a3)2 = a3 × a3 = a3 3 = a2 u 3 = a6
Thus, if am is any term with the base a and the index m, then, (am) n = am u n
(ii) Let's study the following illustrations.
(2 u 3)2 = 22 u 32, (4 u 5)3 = 43 u 53
Similarly, (a u b)3 = a3 u b3
Thus, if a and b are any two terms, then, (a × b)m = am u bm Also, am bm = (ab)m
(iii) Let's study the following illustrations.
2 2 22 , 4 3 43 , Similarly, a 4 a4
3 32 5 53 b b4
= = =
Thus, if a and b are any two terms, then a m = am
b bm
149Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Laws of Indices
4. Law of zero index
Let's study the following illustrations and investigate the result when a base has
zero index. It’s interesting!
The value of a base with
20 = 21– 1 = 21 ÷ 21 = 21 =1 power 0, is always 1!
21
30 = 31 – 1 = 31 ÷ 31 = 31 =1
31
Thus, if a0 is any term with base a and power 0, then, a0 = 1 where a z 0.
Worked-out examples
Example 1: Which one is greater 25 or 52 ?
Solution:
Here, 25 = 2 × 2 × 2 × 2 × 2 = 32 and 52 = 5 × 5 = 25
Since 32 > 25, 25 is greater than 52.
Example 2: Express the numbers as a product of prime factor in exponential
form. a) 432 b) 675
Solution:
a) 2 432 b) 3 675
2 216 3 225
2 108 3 75
2 54 5 25
3 27 5
39
3 ? 675 = 3 × 3 × 3 × 5 × 5
?432 = 2 × 2 × 2 × 2 ×3 × 3 × 3 = 33 × 52
= 24 × 33
Example 3: Factorize the number and express in exponential form.
a) 216 b) 2744
Solution: b) 2 2744
a) 2 216
2 108 2 1372
2 54 2 686
3 27 7 343
3 9 7 49
37
? 216 = 2 × 2 × 2 × 3 × 3 × 3 ? 2744 = 2 × 2 × 2 × 7 × 7 × 7
= 23 u 33 = (2 u 3)3 = 63 = 23 u 73 = (2 u 7)3 = 143
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Example 4: Express the numbers as the products of powers of 10.
a) 600 b) 900000 c) 11000000
Solution: = 6 u 100 = 6 u 10 u 10 = 6 u 102
a) 600
b) 900000 = 9 u 100000 = 9 u 10 u 10 u 10 u 10 u 10 = 9 u 105
c) 11000000 = 11 u 1000000 = 11 u 10 u 10 u 10 u 10 u 10 u 10 = 11 u 106
Example 5: Find the product in their exponential forms.
a) 52 × 54 × 56 b) 3 3 × 34 4 × 34 5 c) 34 × 93 × 272
4
Solution:
a) 52 × 54 × 56 = 52+4+6 = 512
b) 3 3 × 34 2 × 43 5 = 3 3 – 2 5 = 3 6 = 36
4 4 4 4
c) 34 × 93 × 272 = 34 × (32)3 × (33)2 = 34 × 36 × 36 = 3 4+6+6 = 316
Example 6: Find the quotients in their exponential forms.
a) 35 ÷ 32 b) (2x)7 ÷ (2x) –2 c) 85 ÷ 44 d) 34 ÷ 93
Solution:
a) 35 ÷ 32 = 35–2 = 33
b) (2x)7 ÷ (2x) –2 = (2x)7–(–2) = (2x) 7+2 = (2x)9
c) 85 ÷ 44 = (23)5 ÷ (22)4 = 215 ÷ 28 = 215–8 = 27
1
d) 34 ÷ 93 = 34 ÷ (32)3 = 34 ÷ 36 = 1 = 32
36 – 4
Example 7: Find the value of a) 4 1 b) 8 2
2 3
Solution: 9 27
a) 4 1 22 1 = 2 2×12 = 21 = 2
9 32 2 3 3 3
2=
b) 8 2 = 23 2 = 2 3× 2 = 2 2 = 22 = 4
3 3 3 3 3 32 9
27 33
Example 8: Simplify a) 25 3 b) 33 × 35 × 37 c) 253 × 52
Solution: 23 32 × 95 54 × 1252
a) 25 3 = (25 – 3)3 = (22)3 = 22 u 3 = 26 = 64
23
b) 33 × 35 × 37 = 33 + 5 + 7 = 315 = 315 = 315 – 12 = 33 = 27
32 × 95 32 × (32)5 32 × 310 312
c) 253 × 52 = (52)3 × 52 = 56 × 52 = 56 + 2 = 58 = 1 = 1 = 1
54 × 1252 54 × (53)2 54 × 56 54 + 6 510 510 – 8 52 25
xa + b × xb + c × xc + a
Example 9: Simplify a) xb – c u xc – a u xa – b b) x2a × x2b × x2c
Solution:
151Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Laws of Indices
a) xb – c u xc – a u xa – b = xb – c + c – a + a – b = xq = 1.
b) xa + b × xb + c × xc + a a+b+b+c+c+a
x2a × x2b × x2c
= x x2a + 2b + 2c
= x2a + 2b + 2c = x2a + 2b + 2c – 2a –2b – 2c = x° = 1
x2a + 2b + 2c
EXERCISE 10.1
General Section - Classwork
1. Let's tell and write the answers as quickly as possible.
a) a × a2 = ......... b) a2 × a3 = ......... c) a5 × a4 = .........
d) am × an = ......... e) x2 ÷x = ......... f) x5 ÷ x2 = .........
g) x10 ÷ x7 = ......... h) xm ÷ xn = ......... i) (y2)3 = .........
j) (y3)4 = ......... k) (y6)3 = ......... l) (ya)b = .........
m) x° = ......... n) (2x)° = ......... o) p) xm–m = .........
2. Let's tell and write the products in their exponential forms.
a) 7 × 7 × 7 = ......... b) (–2)×(–2)×(–2)×(–2) = .........
c) (5p) × (5p) × (5p) × (5p) × (5p) = ......... d) 23 × 23 × 32 = .........
Creative Section - A
3. a) Define coefficient, base and index of an algebraic term with an example.
b) Define product law of indices taking two terms xa and xb.
c) Define quotient law of indices taking two terms px and py.
d) Define power law of indices taking (xa)b.
e) Write the law of zero index with an example.
4. Let's evaluate and identify which one is greater?
a) 23 and 32 b) 43 and 34 c) 25 and 52
d) 210 and 102 e) 54 and 45 f) 53 and 35
5. Let's factorize the following numbers and express in exponential form.
a) 16 b) 27 c) 128 d) 343 e) 625 f) 400 g) 864 h) 1125
6. Let's find the prime factors and express in exponential forms.
a) 36 b) 100 c) 216 d) 225 e) 196 f) 441 g) 1000 h) 1296
7. Let's express the following numbers as the products of power of 10.
a) 200, 2000, 20000, 200000 b) 5000, 50000, 500000, 5000000
8. Let's find the products in their exponential forms.
a) 2 ×22 × 23 b) 22 × 42 × 82 c) 273 × 92 × 3 45
d) 54 × 252 × 125 e) (3a)4 ×(3a)3 × (3a)–2 2 3 7
f) 4 × 4 ×
7 7
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9. Let's find the quotients in their exponential forms.
a) 27 ÷ 23 b) 39 ÷ 34 c) (2x)8 ÷ (2x)2 d) (3a)4 ÷ (3a)–3
e) 95 ÷ 36 f) 163 ÷ 82 g) 4 8 ÷ 43 h) 4 5 26
5 5 9 3
÷
10. Let's find the quotients in their exponential forms.
a) 34 ÷ 39 b) 53 ÷ 510 c) (5a) 4 ÷ (5a)7 d) (4p) –3 y (4p)2
e) 22 ÷ 43 f) 35 ÷ 94 g) 54 ÷ 253 h) 7 ÷ 492
11. Let's evaluate. 1 2 3 3
2 3 4 5
a) 4 1 b) 25 1 c) 4 d) 8 e) 16 f) 32
2 2 9 27 81 243
12. Let's simplify.
a) a5 u a7 b) p6 u p4 c) x7 ux 2 d) y10 u y 4 e) 4x5 u 3x
a9 p3 x3 y4 6x2
13. Let's simplify.
a) 22 3 b) 34 3 c) 57 2 d) 92 3 e) 43 4 f) 82 5 g) 272 2
54 32 23 29 310
Creative Section - B
14. Let's simplify.
a) 23 × 43 × 84 b) 32 × 93 × 274 c) 22 × 34 × 123
22 × 45 × 162 33 × 94 × 812 25 × 63
d) 64 × 92 × 253 e) 43 2 34 2 92 2 f) 53 2 u 82 3 54
32 × 42 × 156 25 32 272 252 42 2
u u u
15. Let's simplify.
a) xa b × xb – a b) xa – b × xb – c × xc – a
c) (xa)b – c × (xb)c – a × (xc)a – b d) (xp q)r × (xq r)p × (xr p)q
e) xa + b × xb + c × xc + a f) (x2)a + b × (x2)b + c × (x2)c + a
x2a × x2b × x2c (xa × xb × xc)4
16. a) If a = 1 and b = 2, find the value of (i) ab (ii) ba (iii) (a + b)a + b
b) If x = 5 and y = 3, find the value of (i) xy (ii) yx (iii) (x – y)x – y
c) If p = 10 and q = 1, find the value of p2 2pq q2
p q
d) If m = 15 and n = 5, find the value of = m2 n2 .
m n
17. a) If x = 5a and y = 5b, show that xy = 5a + b
b) If x = 7a – b and y = 7b a, show that xy = 1.
c) If x y = x, show that xy = 1.
d) If a = 2 and b = 1, show that (a + b)2 = a2 + 2ab + b2
e) If a = 5 and b = 3, show that (a – b)2 = a2 – 2ab + b2
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It's your time - Project work 14 = .........
18. a) Let's fill in the blanks with correct numbers. 19 = .........
1° = ......... 11 = ......... 12 = ......... 13 = ......... (–1)5 = ......,
15 = ......... 16 = ......... 17 = ......... 18 = ......... (–1)10 = ......,
i) What idea did you investigate ?
ii) What would be the value of: 150 , 11000 and 1x ?
b) Lets find the value of :
(–1)1 = ......, (–1)2 = ......, (–1)3 = ......, (–1)4 = ......,
(–1)6 = ......, (–1)7 = ......, (–1)8 = ......, (–1)9 = ......,
i) What idea did you investigate?
ii) What would be the value of (–1)20 , (–1)99 and (–1)500?
c) Write a short report about your investigation on a) and b), then present in
the class.
10.2 Multiplication of algebraic expressions
While multiplying algebraic expressions, the coefficients of the terms are multiplied
and the power of the same bases are added. For example:
Example 1: Multiply: 4x2 by 3x
Solution: 4 × 3 = 12 (Coefficients are multiplied.)
x2 × x = x2 + 1 = x3 (Power of the same bases are added.)
Here, 4x2 u 3x = 12x3
(i) Multiplication of polynomials by monomials
While multiplying a polynomial by a monomial, we multiply each term of a
polynomial separately by the monomial. For example:
Example 2: Multiply: (x + y) by a xy
Solution: ax ay a
Here, a u (x + y) = ax + ay x+y
Example 3: Multiply (3m2 – 2n2) by 5mn
Solution:
Here, 5mn u (3m2 – 2n2) = 5mn u 3m2 – 5mn u 2n2 Each term of 3m2 – 2n2 is separately
= 15m3n – 10mn3 multiplied by 5mn.
(ii) Multiplication of polynomials
While multiplying two polynomials, each term of a polynomial is separately
multiplied by each term of another polynomial. Then, the product is simplified.
For example: ab
Example 4: Multiply (a + b) by (x + y) x ax bx x+y
Solution: y ay by
Here, (x + y) u (a + b) = x (a + b) + y (a + b) a+b
= ax + bx + ay + by
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Example 5: Multiply (3x2 + 2x – 4) by (2x – 3) By vertical arrangement
Solution: 3x2 + 2x – 4
By horizontal arrangement
(2x – 3) (3x2 + 2x – 4) u 2x – 3
= 2x (3x2 + 2x – 4) – 3 (3x2 + 2x – 4) x3 4x2 – 8x
= 6x3 + 4x2 – 8x – 9x2 – 6x + 12
= 6x3 – 5x2 – 14x + 12 – 9x2 – 6x + 12
6x3 – 5x2 – 14x +12
EXERCISE 10.2
General Section - Classwork
Let's tell and write the products as quickly as possible.
1. a) 2x × 3x × 4x = ............. b) 4y × (– 5y3) = .............
c) ( 3ab × ( 4a2b) = ............. d) (– 2xy) × y × (– 3x2yz) = .............
e) p (p 5) = ............. f) 2x (3x 7) = .............
g) 5x2 (3x 2y) = ............. h) ab (a2 + b2) = .............
i) x(xy x y) = ............. j) ab (ab a 1) = .............
2. a) If a = x and b = 2x2, then 2ab = ..................
b) If x = 2p2 and y = 3p3, then 3xy = ..................
3. Let's investigate the tricky process of multiplication shown below.
1+ 2
x×x 1× 2
(x + 1)(x + 2)= x2... (x + 1)(x + 2) = x2 + 3x... (x + 1)(x + 2) = x2 + 3x + 2
a× a 2–3 2× –3
(a + 2 )(a–3) = a2 ... (a + 2)(a – 3) = a2 – 1.a ... (a + 2)(a – 3) = a2 – a – 6
Now, apply the tricky process shown above. Then, tell and write the products as
quickly as possible.
(a) (x + 2) (x + 1) = ......................... (b) (x + 2) (x + 3)= .........................
(c) (a + 3) (a + 4) = ......................... (d) (a + 3) (a – 2) = .........................
(e) (x – 3) (x + 5) = ......................... (f) (x 2) (x – 5) = .........................
(g) (x – 3) (x 2) = ......................... (h) (a 6) (a – 3) = .........................
Creative Section - A b) (–3x2) × 4x × (–x)
4. Let's simplify: d) (–3pq) × (–5qr) × pqr
f) (– 4x) × (–x2yz) × (–3y) × (–2z)
a) x × 2x × 3x2
c) (–2ab) × (–5a) × (–b)
e) mn × (–m2) × (–3n2)
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5. Let's find the products.
a) x (2x + 4) b) 2x (3x – 5) c) 3p (2p2 – 1)
d) – 3p (p2 – 2p + 3) e) – 3t2 (2t2 + t – 2) f) xyz (x2 – y2 – z2)
6. Let's simplify.
a) x (x + 2) + 3 (x + 2) b) 2x (x – 3) + 5 (x – 3)
c) 3x (x + y) + y (x + y) d) 2a (2a + b) – b (2a + b)
e) x (x2 – xy + y2) + y (x2 – xy + y2) f) a (a2 + ab + b2) – b (a2 + ab + b2)
7. Let's multiply.
a) (a + b) (a + ) b) (a – b) (a – ) c) (a + b) (a – )
d) (x + 2) (x +3) e) (y + 4) (y –5) f) (2a 5) (3a 4)
g) (4p – 3) (p 2) h) (3c 2d) (2c – 3d ) i) (x + y) (x – y)
j) (2x + 3y) (2x 3y) k) (7m 3n) (7m 3n) l) ( x + 2) (x2 2x + 3)
m) (x 3) (2x2 – 4x + 5) n) (a b) (a2 ab + b2) o) (x –y) (x2 xy + y2)
Creative Section - B c) x
y
8. Let's find the areas of the following rectangles. p
a) b)
nx
m ab
d) e) a f) 3x
2a b 8x y
2y
3a b 2a b
9. a) The length of a rectangular garden is (2x + 3y)m and its breadth is
(4x – y)m . Find the area of the garden.
b) A rectangular piece of carpet has length (3x + y)m and breadth (2x – y)m.
Find its area.
c) The floor of a bedroom is (5a –3)m long and (3b +2)m wide.
(i) Find the area of the floor.
(ii) Find its actual area if a = 2 and b = 1
d) The length of a play ground is (7p + 2q)m and width is (5p – 3q)m.
(i) Find the area of playground
(ii) If p = 5 and q = 2, find its actual area.
10. a) If x = (a + 7) and y = (a – 7), show that a2 = xy + 49
b) If p = (3x + 1) and q = (3x – 1), show that: pq 1 = x2
c) If a = (2x + 1) and b = (4x2 – 2x + 1) show 9
: ab – 1= x3
that 8
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Laws of Indices
It's your time - Project work!
11. a) Let's write any three pairs of monomial expressions and find the product
of each pair.
b) Let's write any three pairs of binomial expressions and find the product of
each pair.
c) Let's write any three pairs of monomial expressions such that the product
of each pair is 12x4y4.
10.3 Some special products and formulae
Da bC
ab a + b
(i) The product of (a + b) and (a + b) a a2
(square of binomials)
Let’s multiply (a + b) by (a + b) b ab b2 b
(a + b) u (a + b) = a (a + b) + b (a + b) A a+b b B
(a + b)2 = a2 + ab + ab + b2
= a2 + 2ab + b2 In the figure, a be the length of each side
Thus, (a + b)2 = a2 + 2ab + b2 of a smaller square. When the length is
Here, if (a + b)2 = a2 + 2ab + b2, then, increased by b, each side of the bigger
a2 + b2 = (a + b)2 – 2ab
square ABCD is (a + b).
Now, area of ABCD = a2 + ab + ab + b2
(a + b)u(a + b)=(a + b)2 = a2 + 2ab + b2
(ii) The product of (a – b) and (a – b) D a–b b C
Let’s multiply (a – b) by (a – b) (a – b)2 b(a – b) a–b
a
(a – b) u (a – b) = a (a – b) – b (a – b)
(a – b)2 = a2 – ab – ab + b2 b(a – b) b2 b
= a2 – 2ab + b2
A a–b b B
Thus, (a – b)2 = a2 – 2ab + b2
In the figure, the length and breadth of
Here, if (a – b)2 = a2 – 2ab + b2, then, the square ABCD are decreased by b.
a2 + b2 = (a – b)2 + 2ab Area of square
= (a – b)2 + b (a – b) + b (a – b) + b2
a2 = (a – b)2 + ab – b2 + ab – b2 + b2
a2 = (a – b)2 + 2ab – b2
? (a – b)2 = a2 – 2ab + b2
Now, let's recall the following important formulas,
1. (a + b)2 = a2 + 2ab + b2 = a2 – 2ab + b2 + 4ab = (a – b)2 + 4ab
2. (a – b)2 = a2 – 2ab + b2 = a2 + 2ab + b2 – 4ab = (a + b)2 – 4ab
Worked-out examples
Example 1: Find the squares of (x + 2) a) by geometrically
b) without using formula c) using formula
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Laws of Indices
Solution: Dx 1 1C
a) By geometrical process: x x2
1.x
The square of (x + 2) 1.x
(x + 2)2 = x2 + x + x + x + x + 1 + 1 + 1 + 1 x+2
= x2 + 4x + 4 1 1.x 1 1
b) Without using formula: 1 1.x 1 1
A x+2 B
The square of (x 2) = (x 2)2
= (x 2) (x 2)
= x (x 2) 2 (x 2)
= x2 + 2x + 2x + 4 = x2 + 4x + 4
c) By using formula: Consider, x = a and 2 = b
The square of (x 2) = (x 2)2 Now, (a + b)2 = a2 + 2ab + b2
? (x + 2)2 = (x)2 + 2.x.2 + (2)2
= (x)2 + 2. x . 2 + (2)2
= x2 + 4x + 4
Example 2: Expand: a) (2p2 3q2)2 b) 3x 12
Solution: 3x
Here,
a) (2p2 + 3q2)2 Let, 2p2 = a and 3q2 = b, then,
(a + b)2 = a2 + 2ab + b2
= (2p2)2 + 2. 2p2 . 3q2 + (3q2)2 ? (2p2 + 3q2)2 = (2p2)2 + 2. 2p2 . 3q2 + (3q2)2
= 4p4 + 12p2q2 + 9q4
b) 3x 12 = (3x)2 – 2 . 3x . 1 12 It is in the form of ( )2. So, using
3x 3x 3x the formula, (a – b)2 = a2 – 2ab + b2,
= 9x2 – 2 + 1 3x 1 2 = (3x)2 – 2 . 3x. 1 + 12
9x2 3x 3x 3x
Example 3: Simplify (2x 3y)2 – (2x 3y)2.
Solution:
(2x + 3y)2 – (2x – 3y)2 = (2x)2 2. 2x.3y + (3y)2 – [(2x)2 2. 2x.3y + (3y)2]
= 4x2 12xy + 9y2 – (4x2 12xy + 9y2)
= 4x2 12xy + 9y2 – 4x2 12xy – 9y2 = 24xy
Example 4: Express 9p2 + 24pq + 16q2 as a perfect square.
Solution: 9p2 24pq + 16q2
9p2 24pq + 16q2 (3p)2 ………. + (4q)2
= (3p)2 2. 3p. 4q + (4q)2 (3p)2 2. 3p. 4q. + (4q)2
= (3p + 4q)2 It is in the form a2 2ab + b2 which is equal to (a b)2.
Example 5: Find the squares of a) 99 b) 101
Solution:
a) Square of 99 = 992 = (100 – 1)2 = (100)2 – 2.100.1 + 12 = 10000 – 200+1 = 9801
b) Square of 101 = (101)2 = (100 + 1)2 = (100)2 + 2. 100.1 + 12
=10000 + 200 + 1 = 10201
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Example 6: If x + y = 5 and xy = 6, find the value of x2 + y2.
Solution:
Here, x + y = 5
On squaring both sides, we get
(x + y)2 = (5)2
or, x2 + 2xy + y2 = 25 Using (a + b)2 = a2 + 2ab + b2
or, x2 + 2 × 6 + y2 = 25 Putting xy = 6 [ xy = 6]
or, x2 + 12 + y2 = 25
or, x2 + y2 = 25 – 12
? x2 + y2 = 13 1
p2
Example 7: If p – 1 = 3, find the value of p2 – .
Solution: p
Here, p – 1 = 3
p
On squaring both sides, we get,
p 12 = (3)2
p
or, p2 – 2 × p × 1 + 1 =9 Using (a – b)2 = a2 – 2ab + b2
p p2
or, p2 + 1 =9+2
p2
? p2 +p12 = 11
Example 9: If m – 1 = 7, prove that : a) m2 1 = 51 b) m 1 2 = 53
Solution: m m2 m
Here, a)
m – 1 = 7
m
or, m 1 2 = 72 Squaring on the sides
m
or, m2 – 2 × m × 1 + 1 = 49
m m2
? m2 + 1 = 49 + 2 = 51 Proved.
m2
b) m 1 2 = m 1 2 + 4m × 1 (a + b)2 = (a – b)2 + 4ab
m m m
= 72 + 4
= 49 + 4
? m 1 2 = 53 Proved
m
159Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Laws of Indices
EXERCISE 10.3
General Section – Classwork
1. Let's tell and write the answers in the expanded forms.
a) (m + n)2 = .................................... b) (m – n)2 = ....................................
c) (p + q)2 = .................................... d) (p – q)2 = ....................................
e) (a + 1)2 = .................................... f) (a – 1)2 = ....................................
g) (x + 2)2 = .................................... h) (x – 2)2 = ....................................
2. Let's tell and write these expressions in their square forms.
a) x2 + 2xy + y2 = ......................... b) t2 – 2tr + r2 = .........................
c) x2 + 2.x.3 + 32 = ......................... d) y2 – 2.x.3 + 32 = .........................
e) a2 + 2.a.5 + 52 = ......................... f) b2 – 2.b.7 + 72 = .........................
g) p2 – 4p + 4 = ......................... h) r2 + 4r + 4 = .........................
Creative Section - A
3. Let’s study these diagrams and write the areas in algebraic expression forms
as shown in the example.
Dx 1C
x x2 1.x x+1 Area of ABCD = x2 + 1.x + 1.x + 12
(x + 1)2 = x2 + 2x + 1
1 1.x 12
A x+1 B
D x 2C S x 3R M x yE
x2 2.x x2 x.y
a) x+2 b) c) x+y
x x x
2 2.x y x.y
A x+2 B
PQ P x+y R
Area of ABCD
Area of PQRS Area of PREM
4. Let's find the squares of the following expressions (i) by geometrical process
(ii) without using formula (iii) using formula.
a) (x + 1) b) (x + 2) c) (x + 3)
5. Let's find the squares of the following expressions.
a) (x + 3) b) (x – 4) c) (2y + 3) d) (x – 5y) e) (6m – 5n)
f) (4x + 3y ) g) (a2 + x2) 1 1
h) (x + 1 ) i) y y j) a 2a
x
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6. Let's expand.
a) (a + 3) 2 b) (2a – 5)2 c) (2x + 3y)2
d) 2t + 1 2 e) 3x + 1 2 f) c2 1 2
2t 3x c
7. Let's simplify.
a) (a + b)2 – 2ab b) (p – q)2 + 2pq c) (x + 3)2 – 3 (2x + 3)
d) (3m – 4)2 + 8(3m – 2) e) (a – t)2 + (a + t)2 f) (x + y)2 – (x – y)2
g) (2x – 3y)2 – (2x + 3y)2 h) (4b + 5c)2 – (4b – 5c)2 i) (3a – 5b)2 – (5a + 3b)2
8. Let's express the following expressions in the square forms.
a) x2 + 6x + 9 b) x2 – 8x + 16 c) 4a2 + 4ab + b2
d) p2 – 6pq + 9q2 e) 4x2 + 12xy + 9y2 f) 25x2 – 40xy + 16y2
g) 49a2 – 42ab + 9b2 h) x2 + 2 + 1 i) 4p2 – 2 + 1
x2 4p2
9. Find the squares of the following numbers by using the formula of (a + b)2 or
(a–b)2.
a) 49 b) 51 c) 99 d) 101 e) 999 f) 1001
Creative Section - B
10. a) If a + b = 3 and ab = 2, find the value of a2 + b2.
b) If p + q = 7 and pq = 12, find the value of p2 + q2.
c) If x – y = 1 and xy = 6, find the value of x2 + y2.
d) If m – n = 5 and mn = 14, find the value of m2 – n2.
11. a)
If p + 1 = 4, find the value of p2 + 1 and p– 1 2.
b) p p2 p
c) If m + 1 = 5, find the value of m2 + 1 and m– 1 2.
m m2 m
If x 1 = 3, find the value of x2 + 1 and x 1 2.
x x2 x
d) If y 1 = 6, find the value of y2 + 1 and y 1 2.
y y2 y
It's your time – Project Work!
12. a) Let's write the formulae you learnt in this lesson in a chart paper and
b) compare it to your friends' work.
c)
Let's write any three binomial expressions in the form of (a + b) and find
the squares of your expressions.
Let's write any three binomial expressions in the form of (a – b) and find
the square of your expressions.
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Laws of Indices
(iii) The product of (a + b) and (a – b)
Lets multiply (a – b) by (a + b).
(a + b) u (a – b) = a (a – b) + b (a – b) = a2 – ab + ab – b2 = a2 – b2
Thus, (a + b) (a – b) = a2 – b2
Da CD C Da bC
a–b (a + b) (a – b)
a–b
ab a2 – b2
b A a+b B
EF
AB Area of ABCD
Area of ABCD AB a2 – b2 = (a + b) (a – b)
= a2 Area of ABCDEF
= a2 – b2
(iv) The product of (a + b), (a + b) and (a + b) (cube of binomials)
Let’s find the product of (a + b)3.
(a + b)3 = (a + b) (a + b) (a + b)
= (a + b) (a + b)2
= (a + b) (a2 + 2ab + b2)
= a (a2 + 2ab + b2) + b (a2 + 2ab + b2)
= a3 + 2a2b + ab2 + a2b + 2ab2 + b3
= a3 + 3a2b + 3ab2 + b3
? (a + b)3 = a3 + 3a2b + 3ab2 + b3
Geometrical interpretation:
Let's take a cube of each side (a b). Then volume of cube = (a b)3. Which is
shown in the diagram.
a+b a a b
a bb
a+b a+b a a a
(a + b)3 = b bb
a3 + 3a2b + 3ab2 + b3
(v) The product of (a – b), (a – b) and (a – b)
Let’s find the product of (a – b)3.
(a – b)3 = (a – b) (a – b) (a – b)]
= (a – b) (a – b)2
= (a – b) (a2 – 2ab + b2)
= a (a2 – 2ab + b2) – b (a2 – 2ab + b2)
= a3 – 3a2b + ab2 – a2b + 2ab2 – b3
= a3 – 3a2b + 3ab2 – b3
? (a – 3)3 = a3 – 3a2b + 3ab2 – b3
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Worked-out examples
Example 1: Find the products of the following expressions.
a) (4x + 5) (4x – 5) b) (a2 + b2) (a2 – b2)
Solution:
a) (4x + 5) (4x – 5) Consider, 4x = a and 5 = b
= (4x)2 – 52 Now, (a + b) (a – b) = a2 – b2
= 16x2 – 25 ?(4x + 5) (4x – 5) = (4x)2 – 52.
b) (a2 + b2) (a2 – b2) Consider, a2 = a and b2 = b
= (a2)2 – (b2)2 Now, (a + b) (a – b) = a2 – b2
= a4 – b4
?(a2 + b2) (a2 – b2) = (a2)2 – (b2)2
Example 2: Find the product of (2x + 3y) (2x – 3y) (4x2 + 9y2).
Solution:
(2x + 3y) (2x – 3y) (4x2 + 9y2) = [(2x)2 – (3y2)] (4x2 + 9y2)
= (4x2 – 9y2) (4x2 + 9y2)
= (4x2)2 – (9y2)2 = 16x4 – 81y4
Example 3: Find the product of
a) 51 u 49 b) 102 u 98 by using the formula (a + b) (a – b) = a2 – b2.
Solution:
a) 51 u 49 = (50 + 1) (50 – 1) b) 102 u 98 = (100 + 2) (100 – 2)
= (50)2 – (1)2 = (100)2 – (2)2
= 2500 – 1 = 10000 – 4
= 2499 = 9996
Example 4: Simplify 2.6 × 2.6 1.4 × 1.4
Solution: 2.6 1.4
2.6 × 2.6 – 1.4 × 1.4 = (2.6)2 – (1.4)2 = (2.6 + 1.4) (2.6 – 1.4) = 2.6 + 1.4 = 4
2.6 1.4 2.6 1.4 2.6 – 1.4
Example 5: Find the cubes of a) (x + 2) b) (2m + n) c) 3p – 1
3p
Solution:
a) Cube of (x + 2)
= (x + 2)3
= x3 + 3 . x2 . 2 + 3 . x . 22 + 23 Using (a + b)3 = a3 + 3a2b + 3ab2 + b3
= x3 + 6x2 + 12x + 8
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b) Cube of (2m + n) Consider, 2m = a and n = b
= (2m + n)3 Now, (a + b)3 = a3 + 3a2b + 3ab2 + b3
? (2m + n)3 = (2m)3 + 3.(2m)2.n + 3.2m.n2 + n3
= (2m)3 + 3. (2m)2. n+ 3. 2m. n2 + n3
= 8m3 + 12m2n + 6mn2 + n3
c) Cube of 3p – 1 1
3p 3p
Consider, 3p = a and =b
= 3p – 13
3p Now, (a – b)3 = a3 – 3a2b + 3ab2 – b3
= (3p)3 – 3. (3p)2. 31p+ 3 . 3p 1 2 – 13 1 3 1
3p 3p 3p 3p
? 3p + = (3p)3 – 3. (3p)2.
= 27p3 – 3. 3p + 3. 3p. 1 – 1 + 3. 3p 1 2– 13
9p2 27p3 3p 3p
1 1
= 27p3 – 9p + p – 27p3
Example 6: Express x3 + 12x2 + 48x + 64 as a cube of a binomial.
Solution: x3 + 12x2 + 48x + 64
x3 + 12x2 + 48x + 64
= x3 + … + … + 43
= x3 + 3. x2. 4 + 3. x. 42 + 43
= (x + 4)3 = x3 + 3. x2. 4 + 3. x. 42 + 43.
It is in the form a3 + 3a2b + 3ab2 + b3 which is
equal to (a + b)3.
Example 7: If (a + b) = 5, find the value of a3 + b3 + 15ab.
Solution:
Here, (a + b) = 5
On cubing both sides, we get
? (a + b)3 = 53
or, a3 + 3a2b + 3ab2 + b3 = 125
or, a3 + b3 + 3ab (a + b) = 125
or, a3 + b3 + 3ab u 5 = 125
or, a3 + b3 + 15ab = 125
So, the required value of a3 + b3 + 15ab is 125.
Example 8: a) If x + y = 7 and xy = 6, find the value of x3 + y3.
Solution:
b) If a – 1 = 3, find the value of a3 – 1 .
a a3
Here, b) a– 1 =3
a
a) x + y = 7 and xy = 6
Now, x + y = 7 on cubing both side, we get,
on cubing both side, we get
Vedanta Excel in Mathematics - Book 7 164 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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(x + y)3 = 73 a– 1 3 = 33
or, x3 + 3x2y 3xy2 y3 = 343 a
1 1 1a3
or, x3 3xy (x y) + y3 = 343 or, a3 3a2 u a 3a u a2 = 27
or, x3 3 u 6 u 7 y3 = 343 or, a3 3 a – 1 1 = 27
or, x3 126 y3 = 343 a a3
? x3 y3 = 343 126 = 217
or, a3 3 u 3 1 = 27
a3
1
? a3 a3 = 27 9 = 36
EXERCISE 10.4
General Section - Classwork
1. Let's express as the product of two binomials.
a) x2 – y2 = ....................... b) p2 – q2 = .......................
= .......................
c) a2 – 4 = ....................... d) 9 – b2
2. Let's tell and write the products as quickly as possible.
a) (a + x) (a x) = ....................... b) (x + 1) (x – 1) = .......................
c) (a – 3) (a + 3) = ....................... d) (2 + p) (2 p) = .......................
3. Let's tell and write the expanded forms of the following cubes.
a) (a + x)3 = ........................................................................................
b) (a – x)3 = ........................................................................................
c) (m + n)3 = ........................................................................................
d) (m n) 3 = ........................................................................................
e) (x 1) 3 = ........................................................................................
f) (x 1) 3 = ........................................................................................
Creative Section - A
4. Let's find the area of the following rectangles.
a) b) c)
x–2 x+3 a–5
x+2 a+5
x–3
5. Let's find the products of the following expression by using formula.
a) (x + 2) (x –2) b) (a +3) (a –3)
c) (2p + 3) (2p – 3) d) (2a + 5b) (2a – 5b)
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e) x 1 x– 1 f) y 2 y– 2
2 2 3 3
g) (x2 + y2)(x2 –y2) h) (x2 + 9)(x2 – 9)
6. Let's simplify.
a) (a + 1) (a – 1) (a2 + 1) b) (x + 2) (x – 2) (x2 + 4)
c) (p + 3) (p –3) (p2 + 9) d) (x +y) (x –y) (x2 + y2)
e) (a + 2y) (a – 2y) (a2 + 4y2) f) (2x + 3y) (2x – 3y) (4x2 + 9y2)
7. Let's find the products by using the formula (a + b) (a – b) = a2 – b2
a) 11 × 9 b) 41 × 39 c) 52 × 48 d) 102 × 98
8. Let's simplify.
a) 1.8 × 1.8 – 1.2 × 1.2 b) 1.6 × 1.6 – 0.9 × 0.9
1.8 + 1.2 1.6 + 0.9
c) 2.4 × 2.4 – 1.6 × 1.6 d) 4.6 × 4.6 – 2.5 × 2.5
2.4 – 1.6 4.6 – 2.5
e) 0.7 × 0.7 – 0.3 × 0.3 f) 0.009 × 0.009 – 0.006 × 0.006
0.7 + 0.3 0.009 – 0.006
9. Let's find the cubes of the following expressions.
a) ( x +3) b) (2a –1) c) (2x + 3) d) (p – 4q)
e) (5 + 2b) f) (2 – 3y) 1 1
g) (a + a ) h) ( 2p – 2p )
10. Let's expand the following cubes. c) (2a + 3b)3
a) (2x + 1)3 b) (x 3y)3 f) 3x –31x 3
d) (4 m)3 e) (1 2z)3
11. Let's simplify.
a) (a + 2)3 – 6a (a + 2) b) (a – 2)3 + 6a (a – 2) c) (m + 3)3 – 9m (m 3)
d) (m 3)3 9m (m 3) e) (x y)3 x3 – y3 f) (x + y)3 (x y)3
12. Let's express the following expressions as cubes of binomials.
a) a3 + 3a2 + 3a + 1 b) x3 + 6x2 + 12x + 8
c) p3 – 9p2 + 27p – 27 d) m3 – 12m2 + 48m – 64
e) 8a3 – 12a2b + 6ab2 – b3 f) 27x3+54x2y + 36xy2 + 8y3
Creative Section - B
13. a) If x + y = 3 and xy = 2, find the value of x3 + y3.
b) If p + q = 4 and pq = 3, find the value of p3 + q3.
c) If a – b = 2 and ab = 3, find the value of a3 – b3.
d) If m – n = 3 and mn = 4, find the value of m3 – n3.
14. a) If a 1 = 2, find the value of a3 + 1 .
a a3
b) If p 1 = 5, find the value of p3 + 1 .
p p3
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c) If m – 1 = 3, find the value of m3 – 1 .
m m3
d) If x– 1 = 6, find the value of x3 – 1 .
x x3
It's your time - Project work!
15. a) Let’s take a few rectangular sheet of papers (photocopy paper). Then fold
them to get square sheet of papers as shown in the diagrams.
b) Again, let’s fold each square sheet of paper as shown in the diagram and complete
the sums.
(i) x x x x1
x x2 x2 x x2–12 x–1 x2–12 = ....x.2..–...1..2.... = ..........................
x
(ii) x 12 1 x x+1
1 xy
x
x x2 x2 x x2–y2 x–1 x2–y2 = ................ = ..........................
x
y2 y x+y
y
16. a) Let's find the products of any three pairs of binomials of the form (a + b)
(a – b). Then show that (a + b) (a – b) is always a2 – b2.
b) Let's find the product of any two sets of three binomials of the form (a + b)
(a + b) (a + b).
c) Let's find the products of any two sets of three binomials of the form (a – b)
(a – b) (a – b).
10.4 Division of algebraic expressions
(i) Division of monomials by monomials
While dividing a monomial by another monomial, we divide the coefficient of
dividend by the coefficient of divisor. Then, subtract the power of the base of
divisor from the power of the same base of dividend. For example:
Example 1: Divide a) 28x4y3 by 7x2y2 b) 42a5b4 by 6a4b4
Solution: =14278xx24yy23 = 4x4 – 2 y3 – 2 = 4x2y
a) 28x4y3 ÷ 7x2y2
7
b) 42a5b4 ÷ 6a4b4 = 42a5b4 = 7a5 – 4 b4 – 4 = 7abq = 7a
16a4b4
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(ii) Division of polynomials by monomials
While dividing a polynomial by a monomial, we divide each term of the
polynomial by the monomial separately. For example,
Example 2: Divide a) 8x4 – 6x3 by 2x2 b) 35a3b2 + 15a2b3 by 5ab
Solution:
a) (8x4 – 6x3) ÷ 2x2 Alternative process Checking:
2x2 ) 8x4 – 6x3 ( 4x2 – 3x 2x2 (4x2 3x)
= 8x4 – 6x3 = 2x2 u 4x2 2x2 u 3x
2x2 2x2 + 8x4 = 8x4 6x3
6x3
= 4x4 – 2 – 3x3 – 2 6x3
0
= 4x2 – 3x
? Quotient = (4x2 – 3x)
b) (35a3b2 + 15a2b3) ÷ 5ab 5ab ) 35a3b2 + 15a2b3 (7a2b + 3ab2
35a3b2 15a2b3
= 5ab + 5ab ±35a3b2 Checking:
5ab (7a2b 3ab2)
= 7a3 – 1 b2 – 1 + 3a2 – 1 b3 – 1 15a2b3 = 5ab u 7a2b 5ab u 3ab2
± 15a2b3 = 35a3b2 15a2b3
= 7a2b + 3ab2
0
? Quotient 7a2b + 3ab2
(iii) Division of polynomials by a binomial
While dividing a polynomial by a binomial at first we should arrange the
terms of divisor and dividend in descending (or ascending) order of power
of common bases. Then we should start the process of division dividing the
term of dividend with the highest power. For example,
Example 3: Divide (x2 + 2x – 15) by (x – 3).
Solution:
x – 3 ) x2 + 2x – 15 ( x + 5± Divide x2 by x o x2 ÷ x = x (quotient)
+x2 – 3x Multiply the divisor (x – 3) by the quotient x.
–+ x (x – 3) = x2 – 3x
Subtract the product from the dividend.
5x – 15
±5x 15 x2 + 2x – 15
0 ±x2 3x
5x – 15
Again, the remainder is the new dividend and repeat
the process till the remainder is not divisible by divisor.
Example 4: The length of a rectangular field is (x + 10) m and its area is
(2x2 + 17x – 30) sq.m.
(i) Find the breadth of the field.
(ii) If x = 10 m, find the actual length, breadth and area of the field.
Vedanta Excel in Mathematics - Book 7 168 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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Solution:
Here,
Length of a rectangular field (l) = (x + 10) m
Area of the field (A) = (2x2 + 17x – 30) sq.m
Breadth (b) = ?
Now, breadth = Area ÷ length
So, x + 10) 2x2 + 17x – 30 (2x – 3 (i) 2x2 ÷ x = 2x
– 2x2 +– 20x (ii) 2x(x + 10) = 2x2 + 20x
– 3x – 30 (iii) 2x2 + 17x – 2x2 – 20x = – 3x
+– 3x +– 30 (iv) – 3x ÷ x = – 3
(v) – 3(x + 10) = – 3x – 30
0 (vi) – 3x – 30 + 3x + 30 = 0
? Breadth (b) = (2x – 3)m
Again,
when x = 10 m, then actual length (l) = (x + 10)m = (10 + 10)m = 20m
breadth (b) = (2x – 3)m = (2 × 10 – 3) m = 17m
and area (A) = (2x2 + 17x – 30)m2
= (2 × 102 + 17 × 10 – 30) m2
= (200 + 170 – 30) m2 = 340m2 l × b = 20 m × 17 m = 340 m2
Example 5: Divide (x4 – y4) by (x y) Then, x4 ÷ x = x3
And, x3(x − y) = x4 − x3y
Solution: Again, x4 – y4 – (x4 – x3y) = +x3y – y4
Then,
x y ) x4 – y4 ( x3 x2y + xy2 y3 And, x3y ÷ x = x2y,
± x4 x3y Again, x2y(x – y) = x3y – x2y2
+ x3y – y4 Then, x3y – y4 – (x3y – x2y2) = x2y2 – y4
± x3y x2y2 And,
Again, x2y2 ÷ x = xy2
+ x2y2 – y4 Then, xy2 (x – y) = x2y2 – xy3
± x2y2 xy3 And, x2y2 – y4 – (x2y2 – xy3) = xy3 – y4
+ xy3 – y4 xy3 ÷ x = y3
± xy3 y4 y3(x – y) = xy3 – y4
xy3 – y4 – (xy3 – y4) = 0
0
EXERCISE 10.5
General Section A – Classwork
Let's tell and write the quotients as quickly as possible.
1. a) x2 ÷ x = ........ b) y2 ÷ y2 = ........ c) a4 ÷ a = ........
= ........
d) p4 ÷ p2 = ........ e) b3 ÷ b2 = ........ f) m4 ÷ m3
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2. a) x2y2 ÷ xy = ........ b) a3b3 ÷ ab = ........ c) p3q3 ÷ p2q2 = ........
d) p2q3 ÷ pq2 = ........ e) m4n4 ÷ mn2 = ........ f) x4y3 ÷ x2y3 = ........
3. a) 4x2 ÷ 2x = ........ b) 10y3 ÷ 5y = ........ c) 25a5 ÷ 7a3 = ........
d) 40a3b2 ÷ 4ab = ........ e) 20c2d3 ÷ 5c2d3= ........ f) 55x4y3 ÷ 11x2y2 = ........
Creative Section - A
4. Let's find the quotients.
a) 10x3 ÷ 5x b) 24x2 ÷ 6x2 c)18a2 ÷ 9a2
d) 32a3b2 ÷ 8ab e) 66x4y5 ÷ 11x3y2 f) – 20p4q5 ÷ 5p3q4
g) 45m7n6 ÷ ( – 9m2n4) h) –28a3t8 ÷ (– 7at5) i) 56a2b3c ÷ ( 8ab3c)
5. Let's divide and find the quotients.
a) (6x2 + 9x ) ÷ 3x b) (8a3 – 12a2) ÷ 4a
c) (25y6 – 30y5) ÷ 5y4 d) (2p + 4p2 – 6p3) ÷ 2p
e) (12x3 – 9x2 + 15x) ÷ 3x f) (14m4 – 21m3 – 28m2) ÷ 7m2
g) (10a4b3 – 15a3b4 + 20a2b2) ÷ 5ab
h) (36x5y4 + 45x4y5 – 63x3y3) ÷ (–9x3y3)
i) (44p3q4 – 55p4q3 + 66p5q2) ÷ ( – 11p3q2)
j) (6xyz – 9x2y2z2 – 12x3y3z3) ÷ (– 3xyz)
6. Let's find the quotients.
a) (a2 – 4) ÷ (a + 2) b) (a2 – 4) ÷ (a – 2)
c) (9x2 – 1) ÷ (3x + 1) d) (9x2 – 1) ÷ (3x – 1)
e) (x2 + 5x + 6) ÷ ( x + 2) f) (x2 + 6x + 8) ÷ (x + 4)
g) (y2 + y – 12) ÷ (y – 3) h) (y2 – 11y + 28) ÷ (y – 7)
i) (p2 – 3p – 10) ÷ (p + 2) j) (p2 – 5p – 24) ÷ (p + 3)
k) (4m2 + 5m – 6) ÷ (m + 2) l) (5m2 + 13m – 6) ÷ (m + 3)
m) (6x2 + 5x – 6) ÷ (3x – 2) n) (6x2 + x – 12) ÷ (2x + 3)
o) (4p2 + 12pq + 9q2 ) ÷ ( 2p + 3q) p) (16p2 – 40pq + 25q2 ) ÷ (4p – 5q)
7. a) The product of two algebraic expressions is 4x2 – 9. If one of the expressions is
(2x + 3), find the other expression.
b) The product of two algebraic expressions is (x2 – 4x – 32). If one of the
expressions is (x + 4), find the other expression.
c) The area of a rectangle is (x2 – 16) sq. cm. If its breadth is (x – 4) cm, find its
length.
d) The area of a rectangular field is (a2 – 25) sq. m. If its length is (a + 5) m, find
its breadth.
8. a) If a = 4x3y2, b = 3x2y3 and c = 6xy, find the value of ab .
c
21a4a4b64b,6s, hshowowththatatzxp+rq
b) If p = 7a3b3, q = 6a4b4 and r = = 3ab.
c) If x = 8a4b5, y = 4a5b4 and z =
y = 4b + 2a.
z
Vedanta Excel in Mathematics - Book 7 170 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Laws of Indices
Creative Section - B
9. Let's divide and find the quotients.
a) (2x3 – 5x2 – 24x – 18) ÷ (2x + 3) b) (3x2 – 10x2 + 7x + 10) ÷ (3x + 2)
c) (a4 – b4) ÷ (a + b) d) (a4 – b4) ÷ (a – b)
e) (a5 + b5) ÷ (a + b) f) (a5 – b5) ÷ (a – b)
g) (x6 – y6) ÷ (x2 – y2) h) (x6 – y6) ÷ (x3 – y3)
10.a) The length of a rectangular carpet is (x – 5)m and its area is (x2 – 12x + 35) sq. m.
(i) Find its breadth
(ii) If x 10m, find its actual length, breadth and area of the carpet.
b) The area of a rectangular garden is (x2 – 4x – 32) sq. m and its breadth is (x – 8)m.
(i) Find its length.
(ii) If x = 15m, find the actual length, breadth, and area of the garden.
11. a) Sunayana distributed (3x2 + 20x + 25) sweets equally among (3x + 5) friends
on her birthday.
(i) How many sweets did each of the friends receive?
(ii) If x = 5, find the total number of sweets distributed among friends, actual
number of friends and share of sweets.
b) On Children's Day Hari Narayan distributed (2x2 + 13x – 24) copies equally
among (2x – 3) number of children of a child care centre.
(i) Find the number of copies received by each child.
(ii) If x = 20, find the actual number of copies, students, and share of each
child.
It's your time - Project work!
12. a) Let's write any three pairs of monomial expressions with the same bases with
higher power of dividend. Divide the dividend by the divisor in each pair
and show that: Dividend = Quotient × Divisor + Remainder
b) Let's write any three pairs of binomial dividend and divisor in the forms of
(a2 – b2) and (a + b) or (a2 – b2) and (a – b).
Then, divide the dividend by the divisor and find quotient in each pair.
c) Let's find the products of any three pairs of binomial expressions. Then,
divide each product by one of the expressions in each pair and get another
expression.
E.g.: find o (x + 2) (x + 3) = x2 + 5x + 6, then find
(x2 + 5x + 6) ÷ (x + 2) = x + 3
10.5 Factors and factorisation – Introduction
We know that, in 2 × 5 = 10 o 10 is the product, 2, and 5 are the factors.
in 3 × 4 = 12 o 12 is the product, 3, and 4 are the factors.
Similarly, in x × x = x2 o x2 is the product, x, and x are the factors.
in 2a(a + b) = 2a2 + 2ab o 2a2 + 2ab is the product, 2a and (a + b)
are factors.
171Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Laws of Indices
Thus, when two or more algebraic expressions are multiplied, the result is called
the product and each expression is called the factor of the product. The process of
finding out factors of an algebraic expression is known as factorisation. It is also
known as resolution of the expression into its factors.
Of course, the process of factorisation is the reverse process of multiplication. For
example,
Multiplication of 2x(x – 3) = 2x2 – 6x and factorisation of 2x2 – 6x = 2x (x – 3)
Multiplication of (a + b) (a – b) = a2 – b2 and factorisation a2 – b2 = (a + b) (a – b)
Now, let’s learn the process of factorisation of different types of expressions.
(i) Factorisation of expressions which have a common factor in each of its
term
In this type of expression, the factor which is present in each term is taken as
common and each term of the expression is divided by the common factor. The
product form of the common factor and the quotient represents the factorisation
of the expression.
Worked-out examples
Example 1: Factorise a) ax + bx b) 2px2 – 6p2x c) 4ax2 + 6a2x – 8ax
Solution:
a) ax + bx x is present in each term. So, it is the common factor.
= x (a + b) x is taken as common, then ax ÷ x = a and bx ÷ x = b
b) 2px2 – 6p2x 2px2 – 6p2x Direct process
= 2px u x – 2 u 3p u px = 2px (x – 3p) 2px2 ÷ 2px = x and 6p2x ÷ 2px = 3p
= 2px (x – 3p)
c) 4ax2 + 6a2x – 8ax Remember!
= 2ax (2x + 3a – 4) Among the factors which are present in each term, the
factor with the least power is taken as a common factor.
Example 2: Resolve into factors 3a (x + y) – 4b (x + y)
Solution: (x + y) is the common factor.
3a (x + y) – 4b (x + y)
= (x + y) (3a – 4b) 3a(x + y) ÷ (x + y) = 3a and 4b(x + y) ÷ (x + y) = 4b
(ii) Factorisation of expressions having a common factor in the groups of
terms
In this case, terms of the given expression are arranged in groups in such a way
that each group has a common factor. For example:
Vedanta Excel in Mathematics - Book 7 172 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Laws of Indices
Example 3: Factorise a) a2 + ab + ca + bc b) x2 – 3a + 3x – ax
Solution: Terms are arranged in groups.
a) a2 + ab + ca + bc
= a (a + b) + c (a + b) a is common in a2 + ab and c is common in ca + bc
= (a + b) (a + c) (a + b) is common.
b) x2 – 3a + 3x – ax Arranging the terms in suitable groups
= x2 + 3x – ax – 3a
= x (x + 3) – a (x + 3) x is common in x2 + 3x and a is common in ax – 3a
= (x + 3) (x – a)
(iii) Factorisation of expressions having the difference of two squared terms
The algebraic expression a2 – b2 is the difference of two squared terms. Here,
a2 and b2 are the squared terms and a and b are their square roots respectively.
We have learnt that a2 – b2 is the product of (a + b) and (a – b)
? a2 – b2 = (a + b) (a – b) Square root of a2
a2 – b2 = (a + b) (a – b)
Thus, (a + b) and (a – b) are the factors of a2 – b2.
Square root of b2
To factorise such expressions, we should re-write
the given terms in the form of a2 – b2.
Then, a2 – b2 = (a + b) (a – b) represents the factorisation of the expression.
Example 4: Factorise a) 4x2 – 9y2 b) 81x4 – 16y4.
Solution: Square root of 4x2 = 2x and square root of 9y2 = 3y
a) 4x2 – 9y2
= (2x)2 – (3y)2
= (2x + 3y) (2x – 3y) Using a2 – b2 = (a + b) (a – b)
b) 81x4 – 16y4 Square root of 81x4 and 16y4 are 9x2 and 4y2
= (9x2)2 – (4y2)2
= (9x2 + 4y2) (9x2 – 4y2) Using a2 – b2 = (a + b) (a – b)
= (9x2 + 4y2) [(3x)2 – (2y)2] 9x2 – 4y2 is still in the form a2 – b2.
So, it is further factorised.
= (9x2 + 4y2) (3x + 2y) (3x – 2y)
173Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Laws of Indices
Example 5: Simplify by factorisation process.
a) 24 u 40 – 20 u 24 b) 652 – 552 c) 49 u 51
Solution:
a) 24 u 40 – 20 u 24 b) 652 – 552 c) 49 u 51
= 24 (40 – 20) = (65 + 55) (65 – 55) = (50 – 1) (50 + 1)
= 24 u 20 = 120 u 10 = 502 – 12
= 480 = 1200 = 2500 – 1 = 2499
(iv) Factorisation of expressions of the form x2 + px + q
While factorising a trinomial expression of the form x2 + px + q, we should
search any two numbers a and b such that a + b = p and ab = q. Clearly, a
and b must be the factors of q. Then px is expanded in the form ax + bx and
factorisation is performed by grouping.
Example 6: Factorise a) x2 + 5x + 6 b) a2 – 3a – 28
Solution: b) a2 – 3a – 28 = a2 – (7 – 4)a – 28
a) x2 + 5x + 6 = x2 + (2 + 3)x + 6 = a2 – 7a + 4a – 28
= a(a – 7) + 4(a – 7)
= x2 + 2x + 3x + 6 = (a – 7) (a + 4)
= x(x + 2) + 3(x + 2)
= (x + 2) (x + 3)
EXERCISE 10.6
General Section - Classwork
1. Let's tell and write the factors and products of these expressions as quickly as
possible.
a) a × a2 factors are .............................................. product is .........................
b) 2x2 × x factors are ............................................ product is .........................
c) 3p (p + 1) factors are ........................................... product is .........................
d) y2 (y – 2 ) factors are ............................................ product is .........................
2. Let's factorise, tell and write the factors as quickly as possible.
a) ax + ay = ............................. b) ax – ay = .............................
(c) ak + bk = ............................. (d) ak – bk = .............................
(e) a2 + a = ............................. (f) x2 – x = .............................
(g) x2 + xy = ............................. (h) p3 – p2q = .............................
Vedanta Excel in Mathematics - Book 7 174 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Laws of Indices
3. a) a2 – b2 = ............................. b) b2 – c2 = .............................
c) x2 – 22 = ............................. d) y2 – 32 = .............................
e) p2 – 16 = ............................. f) r2 – 25 = .............................
4. Let's study the following tricky ways of factorisation carefully.
3+6 3×6 –3–6 –3×(–6)
x2 + 9x + 18 = (x + 3) (x + 6) x2 – 9x + 18 = (x – 3) (x – 6)
6–3 6×(-3) 3–6 3×(–6)
x2 + 3x – 18 = (x + 6) (x – 3) x2 – 3x – 18 = (x + 3 ) (x – 6)
Could you investigate the idea of tricky ways of factorisation. Apply your
investigation, tell and write the factors as quickly as possible.
a) x2 + 5x + 6 = ........................ b) x2 – 5x + 6 = ........................
c) x2 + x – 12 = ........................ d) x2 – x – 6 = ........................
Creative Section
5. Let's factorise.
a) 2ax + 2bx b) 2x2y – 2xy2 c) 2abx – 4aby
d) 3mx2 – 6mx e) ax2 + a2x – ax f) 4x5y4 – 2x4y5 – 8x3y3
g) a(x – y) + b(x – y) h) x (a + b) – y(a + b) i) 2x (x – y) + y(x – y)
6. Let's resolve into the factors.
a) ax + bx + ay + by b) ma + na + mb + nb c) xy + xz + y2 + yz
f) x2 – zx + xy – yz
d) 3ax – bx – 3ay + by e) 4ax – 3bx – 8ay + 6by i) x2 + 4a + 4x + ax
g) a2b + ca – ab2c – bc2 h) a2bc + c2a – ab2 – bc
7. Let's factorise. b) x2 – 9 c) m2 – 16 d) p2 – 25
a) a2 – 4 f) 25x2 – 36y2 g) 49p2 – 81q2 h) 64b2 – 9c2
e) 4a2 – 9b2 j) 8x2y2 – 18 k) 75a2b2 – 27x2y2 l) 80a3 – 5ab2
i) 100m2 – 49n2
8. Let's resolve into the factors.
a) a4 – 16 b) x4 – 81 c) y4 – 625 d) 16p4 – q4
g) a4 – b4 h) 16p4 – 81q4
e) 81m4 – n4 f) x4 – 16y4
9. Let's factorise. b) a2 + 7a + 12 c) p2 + 6p + 8 d) x2 + 7x + 10
a) x2 + 3x + 2 f) x2 – 9x + 20 g) m2 – 9m + 14 h) b2 – 11b + 30
e) y2 – 8y + 15 j) x2 – 3x - 28 k) a2 + 4a – 12 l) x2 + 5x – 24
i) x2 – 2x - 15
175Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Laws of Indices
10. Let's simplify by factorisation process.
a)13 × 50 – 40 × 13 b) 28 × 60 – 30 × 28 c)36 × 24 + 26 × 36
f)752 – 552
d)152 – 52 e)272 – 172 i)101 × 99
g)21 × 19 h)79 × 81
It's your time - Project work!
11. a) Let's write any five pairs of difference of square numbers in the form
a2 – b2, where a2 > b2. Then, factorise and simplify the each pair.
[E.g. 52 – 32 = (5 + 3) (5 – 3) = 8 × 2 = 16]
b) Let's write any five binomial expressions of the form x2 – a2, where a2 is a
square number. Then factorise your expressions.
12. a) Let's find the products of any four pairs of binomials of the forms
(i) (x + a) (x + b) (ii) (x + a) (x – b) (iii) (x – a) (x + b) (iv) (x – a) (x – b),
where a and b are any natural numbers and a > b.
b) Again, factorise each product to get the factors.
[E.g. (x + 1) (x + 2) = x2 + 3x + 2 and
x2 + 3x + 2 = x2 + (1 + 2)x + 2 = x2 + x + 2x + 2 = x(x + 1) + 2(x + 1)
= (x + 1) (x + 2)
10.6 Simplification of rational expressions
We know that 1, 2, 1 ,– 3 , etc. are the rational numbers. Similarly,
2 5
x x2 x + 2
3 , y2 , x – 7 , etc. are called rational expressions. Here, we shall discuss about
addition, subtraction, multiplication, and division of rational expressions.
(i) Multiplication and division of rational expressions
In multiplication, we simplify the numerical coefficients as in case of
multiplication of fraction. In the case of variables we apply the product and
quotient rules of indices. In division, we should multiply the dividend by the
reciprocal of divisor.
Worked-out examples
Example 1: Multiply 4a3b2 × 10x4y3 .
5x3y2 12a4b3
Solution:
4a3b2 10x4y3 = 14a3b2 × 120 x4y3 = 2x4 – 3 × y3 – 2 = 2xy
5x3y2 u 12a4b3 5x3y2 × 12a4b3 3a4 – 3 b3 – 2 3ab
13
Vedanta Excel in Mathematics - Book 7 176 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Laws of Indices
Example 2: Divide: 5x2y ÷ 15a3y3
Solution: 8ax3 16a2x2
5x2y ÷ 15a3y3 = 15x2y × 126a2x2 = 2a2x4y = 2x4 – 3 = 2x
8ax3 16a2x2 81ax3 135a3y3 3a4x3y3 3a4 – 2 y3 – 1 3a2y2
Example 3: Simplify: 6y2 × 2x3 ÷ 5xy
Solution: 4x2 3y3 6ab
6y2 × 2x3 ÷ 5xy = 261y2 × 12x3 × 6ab = 6abx3y2 = 6ab = 6ab
4x2 3y3 6ab 4x2 3y3 5xy 5x3y4 5y4 – 2 5y2
21 1
EXERCISE 10.7
General Section - Classwork
1. Let's tell and write the answers as quickly as possible.
a) x2 × a = ................................... b) x3 × a = ...................................
a2 x a3 x
c) x2 × y = ................................... d) x3 × xy22= ...................................
y2 x y3
2. a) a2 ÷ a = ................................... b) a3 ÷ a2 = ...................................
x2 x x3 x2
c) x2 ÷ x = ................................... d) x4 ÷ y2 = ...................................
y3 y y4 x2
Creative Section
3. Simplify.
a) 2x3 × 9a2 b) 4y4 × 10z4 c) x4y3 × 15ab2
3a3 8x2 5z3 12y3 5a3b3 9x2y
d) 4a4 ÷ 8a3 e) 6y5 ÷ 18y3 f) x3y2 ÷ x2y
3b3 9b2 5x4 10x6 a3b2 a2b
4. Simplify.
a) 2x2 × 10y4 × 9x4 b) 4a3 × 9b6 × 6a c) x2y3 × a2b3 ÷ ab
3y3 6x3 4y4 3b2 8a5 15b3 a3b2 x3y2 xy
d) 3a4b3 ÷ 6a2b2 × 8x2y e) x2y2 × abxy ÷ xyz f) a3 × c3x ÷ a2c2
5x3y4 15x2 9ab2 a3b3 x3y3 a2b2 x3y2 ab2 b2xyz
ii) Addition and subtraction of rational expressions with the same
denominators.
In this case, we should add or subtract the numerators and the common
denominator is written as the denominator of the new rational expression.
For example:
177Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 7
Laws of Indices
Example 1: Simplify a) 3x + y + 3x – y b) a2 – b2
Solution: 4a 4a a+b a+b
a) 3x + y 3x – y = 3x + y + 3x – y =2346ax 3x It’s easier, it is same as the
4a + 4a 4a = 2a addition of like fractions.
such as 2 + 1 = 2+1 !
5 5 5
b) a2 – b2 = a2 – b2 = (a + b) (a – b) = (a – b)
a+b a+b a+b (a + b)
EXERCISE 10.8
General Section - Classwork
Tell and write the answers as quickly as possible.
1. a) 1 + 1 = ........................ b) 2 + 3 = ........................
x x y y
c) 3a + 4a = ........................ d) a + b = ........................
b b x x
e) 3 + 5 = ........................ f) x + y = ........................
a–b a–b a+ b a+b
2. a) 2 – 1 = ........................ b) 5 – 3 = ........................
x x = ........................ a a
= ........................
c) 5x – 2x d) a – b = ........................
a a 2y 2y
e) x – y f) 2y + z = ........................
a+ b a+b p–q p–q
Creative Section
3. Let's simplify.
(a) 5x + 3x (b) 2a + 3a + 5a (c) 3y + 5y – 9y
2a 4a 3x 2x 6x 4b 2b 8b
(d) x – 2x + 5x (e) ax + 3ax – 7ax (f) 4by + 3by – 5by
6y 3y 9y 2b 5b 10b 7ax 14ax 28ax
4. Let's simplify.
(a) 2x + y + x + 2y (b) 4a + 3b + 2a + b (c) 2x + y + 2x – y
3 3 2 2 4a 4a
(d) x+y + x–y (e) x+y – x–y (f) x2 + y2
2xy 2xy 2xy 2xy x+y x+y
(g) a2 + 2a – b2 + 2a (h) x + y (i) x – y
a2 + b2 a2 + b2 x2 – y2 x2 – y2 x2 – y2 x2 – y2
Vedanta Excel in Mathematics - Book 7 178 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit Equation, Inequality and Graph
11
11. 1 Open statement and equation - Looking back
Classroom - Exercise
1. Let's tell and write whether the following statements are 'true', 'false' or open
statements.
a) The sum of 2 and 5 is 7. It is ................................. statement.
b) The difference of 9 and 4 is 6. It is ................................. statement.
c) The sum of x and 4 is 8. It is ................................. statement.
d) The difference of y and 3 is 9. It is ................................. statement.
e) The product of 5 and p is 10. It is ................................. statement.
2. Let's tell and write the equations and find the values of the variables.
a) The sum of x and 2 is 5, equation is ........................ and x = .........
b) The difference of y and 1 is 6, equation is ........................ and y = .........
c) The product of 3 and p is 6, equation is ........................ and p = .........
d) The quotient of x divided by 2 is 4, equation is ........................ and x = .........
3. Let's tell and write the values of variables that make the following open
statements (or equations) true.
a) x + 2 = 7, x = ...................... b) x – 2 = 7, x = ......................
c) 4y = 8, y = ...................... d) m = 3, m = ......................
2
Let's suppose, the sum of x and 5 is 9. i.e. x + 5 = 9. It is a mathematical statement.
Here, unless x is replaced by any number, we cannot say whether x + 5 = 9
is a true or a false statement. For example:
When x is replaced by 1, then 1 + 5 = 6, which is false.
When x is replaced by 2, then 2 + 5 = 7, which is false.
When x is replaced by 3, then 3 5 = 8 which is false.
When x is replaced by 4, then 4 5 = 9 which is true.
Thus, the mathematical statements which cannot be predicted as true or false
statements until the variable is replaced by any number are known as Open
Statements.
x > 3, x + 2 < 7, x + 3 = 7, 2x = 10, etc. are also the examples of open statements.
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Equation, Inequality and Graph
11.2 Linear equations in one variable
Let’s consider any two open statements: x + 3 > 7 and x + 3 = 7.
The open statement x + 3 > 7 can be true for many values of x. However, the
other open statement x + 3 = 7 can be true only for a fixed value of x. Such open
statement containing ‘equal to’ (=) sign and can be true only for a fixed value of
variable is called an equation.
Furthermore, in x + 3 = 7, the equation has only one variable which is x. So, it is
the equation in one variable. Also, the variable x has power (or exponent) 1. So, it is
called a linear equation. Thus, x + 3 = 7 is a linear equation in one variable.
11.3 Solution to equations
Let’s consider an equation, x + 3 = 7.
This equation can be true only for a fixed value of x which is 4. So, 4 is called the
solution (or root) of the equation. The process of getting a solution to an equation
is called solving equation.
Worked-out examples
Example 1: Solve a) x + 2 = 8 b) 7a = 21 c) t =5
Solution: 2
a) x + 2 = 8 Direct process
x+2 =8
or, x + 2 – 2 = 8 – 2 Subtracting 2 from both sides
or, x = 8 – 2
or, x = 6
or, x = 6
b) 7a = 21 Direct process
or, 7a = 21 Dividing both sides by 7 7a = 21
7 7 21
or, a = 3 or, a = 37
or, a =
c) t =5 Direct process
2 t
2 =5
or, t u 2 = 5 u 2 Multiplying both sides by 2
2 or, t = 5 × 2
or, t = 10
or, t = 10
Example 2: Solve the equations and check the solutions. 2 3
9x + 2y −
a) 7 (x – 10) = 10 – 3x b) 11 + 7 = 43 c) y 1 = 1
Solution: Checking:
a) 7 (x – 10) = 10 – 3x Putting x = 8 in the given equation:
or, 7x – 70 = 10 – 3x 7(x – 10) = 10 – 3x
or. 7 (8 – 10) = 10 – 3 × 8
or, 7x + 3x = 10 + 70
or, 7 ( – 2) = 10 – 24
or, 10x = 80 or, – 14 = – 14
LHS = RHS
or, x = 80 =8
10 ? x = 8 is the required solution.
Vedanta Excel in Mathematics - Book 7 180 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Equation, Inequality and Graph
b) 9x + 7 = 43 Checking
11
9x Substituting x = 44 in the given equation:
11
or, = 43 7 9x 7 = 43
11
9x u 444
or, 11 = 36 or, 9 11 + 7 = 43
or, 9x = 11 u 36 or, 9 × 4 + 7 = 43
or, x = 11 × 36 4 or, 36 + 7 = 43
91
43 = 43 ? LHS = RHS
or, x = 11 × 4 Hence, x = 44 is the required solution.
or, x = 44 Checking
Substituting y = 5 in the given equation:
2 3
c) 2 = 3 y+1 = 2y – 1
y+1 −
2y 1 or, 2 = 2 3 – 1
5+1 ×5
or, 2(2y 1) = 3(y 1) 2 3
or, 6 = 10 –
or, 4y 2 = 3y 3 1
1 3
or, 4y 3y = 3 2 or, 3 = 9
or, y = 5 1 = 1 ? LHS = RHS
3 3
Hence, y = 5 is the required solution.
Example 3: Solve a) 40% of x = Rs 80 b) x + 25 % of x = Rs 555.
Solution: b) x + 25 % of x =Rs 555
a) 40 % of x =Rs 80 or, x + 25 u x = Rs 555
100
40 x
or, 100 u x = Rs 80 or, x+ 4 = Rs 555
or, 2x = Rs 80 or, 4x + x = Rs 555
5 4
or, 2x = 5 × Rs 80 or, 5x = Rs 555
4
or, x = 5 × Rs 80 = Rs 200
2 or, 5x = 4 u Rs 555
? x = Rs 200. or, x = 4 × Rs 555
5
? x= 4 × Rs 111 = Rs 444.
EXERCISE 11.1
General Section - Classwork
1. Let's tell and write 'true', 'false', or 'open' statement in the blank spaces.
a) The sum of 4 and 5 is 10. It is ............................ statement.
b) The difference of x and 5 is 3. It is ............................ statement.
c) The product of 6 and 7 is 42. It is ............................ statement.
d) The quotient of x divided by 3 is 2. It is ............................ statement.
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2. Let's make equations. Then tell and write solutions of the equations.
a) The sum of x and 6 is 9. Equation is ...................... solution is ........
b) The product of y and 3 is 15. Equation is ...................... solution is ........
c) The difference if a and 2 is 10. Equation is ...................... solution is ........
d) The quotient of x divided by 4 is 3. Equation is ...................... solution is ........
Let's tell and write the values of letters as quickly as possible.
3. a) x + 2 = 7, x = ............... b) x + 5 = 9, x = ...............
c) x – 4 = 10, x = ............... d) x – 2 = 5, x = ...............
e) 2x = 8, x = ............... f) 3x = 15, x = ...............
g) x = 5, x = ............... h) x = 4, x = ...............
2 3
4. a) 2x – 1 = 5, x = ............... b) 3x + 2 = 14, x = ...............
c) 5x – 3 = 7, x = ............... d) 3x = 6, x = ...............
5
e) x + 1 = 3, x = ............... f) x–2 = 5, x = ...............
2 3
Creative Section - A
5. a) Define open mathematical statements with examples.
b) Write the difference in the meaning of x + 2 > 5 and x + 2 = 5.
c) What is an equation? Write with an example.
d) Define solution of an equation with an example.
6. Let's solve the equations and check the solutions.
a) 3x + 1 = 7 b) 6x – 13 = 5
c) 8x + 13 = 6x + 25 d) 5x – 18 = 2x + 3
e) 2(3x + 4) = 5(2 + x) f) 9(7x – 2) = 5 (10x – 1)
g) 3x + 2(x + 2) = 22 – 3 (2x – 5) h) 6(y – 3) – 5(y – 8) = 48 – 3(y – 2)
i) 5.8x – 7.3 = 3.8 x + 2.7 j) 1.4x – 9.8 = 10.2 – 3.6x
7. Solve each of the following equations and check the answer in each case.
a) x =2 b) 5x = 15 c) 7x = – 14
3 6 8
d) 3x = – 6 e) 2x = 8 f) 3x = 9
5 9 3 4 2
g) x−1 =3 h) 4x − 5 = 5 i) x−1 = x 2
2 3 2 3
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j) 3x − 1 = 2x 3 k) 4 = 5 l) 3 = 7
5 7 x−1 x 1 x 1 3x 1
m) x 3 = 4 n) 2x – 1 = 1
3x − 5 5 x+4 2
Creative Section - A
8. Let's make equations and solve the equations.
a) 5% of x = Rs 10 b) 10% of x = Rs 50 c) 20% of x = 60 kg
f) 90% of x = 360 students
d) 30% of x = 90 km e) 40% of x = Rs 200
9. Let's make equations and solve the equations.
a) x + 10% of x = 55 kg b) x + 20% of x = Rs 96 c) x + 25% of x = Rs 125
d) x – 15% of x = 34 km e) x – 30% of x = Rs 70 f) x – 75% of x = Rs 75
g) x = 160 + 20% of x h) x = Rs 300 + 50% of x i) x = 600 − 20% of x
10. Let's solve.
a) 3x – 1 = 1 b) 5x + 1 = 2 c) x – x =2
8 4 2 12 4 3 2 3
d) x + x =7 e) x 1 + x 6 =5 f) 2x 1 + x 3 =2
3 4 2 3 3 4
g) 2 + 3 =5 h) 3 + 2 =7 i) 5 – 1 = 1
x x 2x x 4n 2n 8
It's your time - Project work!
11. a) Let's write any integer to the right hand side of each of the following equation
such that the solution of each equation will be an integer. Also, check the
solution.
(i) x + 7 = ........ (ii) x – 5 = ........ (iii) 3x = ........ (iv) x = ........
(v) 2x + 1 = ........ (vi) 3x – 8 = ....... (vii) x 4
2 x–3
2 = ....... (viii) 5 = .......
b) Let's write any integer in the blank spaces such that the solution must be
an integer. Then solve the equations and find the solutions. Also, check the
solution.
(i) x + ........ = 9 (ii) x – ........ = –2
(iii) 2x + ........... = 5 (iv) 3x – ........... = 2
c) Let's make linear equations of your own and solve each equation to get the
given solutions.
(i) x = 1 (ii) x = 3 (iii) x = –2 (iv) x = 4 (v) x = –1
11.4 Applications of equations
We use equations to find the unknown value of any quantity. For this, we should
consider the unknown value of the given verbal problems as the variables like
x, y, z, a, b, c, etc. Then, the verbal problems should be translated into mathematical
sentences in the form of equations. By solving the equations, we obtain the required
values.
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Worked-out examples
Example 1: The sum of two numbers is 50. If one of the numbers is 35, find the
other number.
Solution: Answer checking:
Let the other number be x. If the required number is 15,
Now, x + 35 = 50 then, 35 + 15 = 50
or, x = 50 – 35 = 15 which is given in the question.
Hence, the required number is 15.
Example 2: There are 555 students in a school. If the number of boys is 55
more than that of girls, find the number of boys and girls.
Solution:
Let the number of girls be x.
Then, the number of boys = (x + 55)
Now, number of girls number of boys = total number of students
or, x + (x + 55) = 555 Answer checking:
Number of boys = 305 and number of girls = 250
or, 2x 55 = 555 Then, 305 + 250 = 555
which is given in the question.
or, 2x = 555 55
or, x = 500 = 250
2
? Number of girls = x = 250
Also number of boys = (x 55) = (250 55) = 305
Example 3: If the sum of three consecutive even numbers is 36, find the numbers.
Solution:
Let the smallest even number be x.
Then, the second consecutive even number = x + 2
And, the third consecutive even number = x + 4
Now, x + (x + 2) + (x + 4) = 36 Answer checking:
or, 3x + 6 = 36 If the required consecutive even
or, 3x = 36 – 6 numbers are 10, 12 and 14, then
or, 10 + 12 + 14 = 36
x = 30 = 10 which is given in the question.
3
? The first even number = x = 10
The second even number = x + 2 = 10 + 2 = 12
The third even number = x + 4 = 10 + 4 = 14
Hence, the required consecutive even numbers are 10, 12 and 14.
Example 4: Mother divides Rs 1,30,000 between her son and daughter in the
ratio 6 : 7. Find the shares of the son and the daughter.
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Solution: Answer checking:
Let the share of the son = Rs 6x Rs 60,000 + Rs 70,000 = Rs 1,30,000
And, 60,000 : 70,000 = 6 : 7
The share of the daughter = Rs 7x which are given in the question.
Now, 6x + 7x = Rs 1,30,000
or, 13x = Rs 1,30,000
or, x = Rs 1,30,000 = Rs 10,000
13
? The share of the son = Rs 6x = Rs (6 × 10,000) = Rs 60,000
The share of the daughter = Rs 7x = Rs (7 × 10,000) = Rs 70,000
Hence, the shares of the son and the daughter are Rs 60,000 and Rs 70,000 respectively.
Example 5 : The length of a rectangular field is 10m longer than its breadth
and its perimeter is 100m. Find the length and the breadth of the
field.
Solution:
Let, the breadth (b) of the rectangular field be x m.
Then, the length (l) of the field = (x + 10) m
Now, the perimeter of the field = 100 m Answer checking:
or, 2 (l + b) = 100 m l = 30 and b = 20 m
or, 2 (x + 10 + x) = 100 m Thus, l is 10 m longer than b
or, 2 (2x + 10) = 100 m Also, perimeter = 2(l + b)
or, 4x + 20 = 100 m
= 2(30 m + 20m)
= 2 × 50 m
or, 4x = (100 – 20) m = 100 m
or, x = 80 m = 20 m which is given in the question.
4
? The breadth of the field (b) = x m = 20 m
The length of the field (l) = (x + 10) m = (20 + 10) m = 30 m
Example 6: 1 th part of a pole is inside the mud, 2 th part is inside the water and
Solution: 4 5
the remaining length of 7 m is above the surface of the water. Find
the length of the pole.
Let the length of the pole be x m.
The length of the pole inside the mud = 1 of x m = x m.
4 4
2 2x
The length of the pole inside the water = 5 of x m = 5 m.
The length of the pole above the surface of the water = 7 m
Now, x = x + 2x + 7
or, 4 5
x = 5x + 8x + 140
20
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or, x = 13x + 140
20
or, 20x = 13x + 140 Answer checking:
or, 20x – 13x = 140 x = 20 = 5 m and 2x = 2 × 20 = 8 m
4 4 5 5
or, 7x = 140 Then, 5 m + 8 m = 13 m and 20 m – 13 m = 7 m
140 which is given in the question.
7
or, x = = 20
Hence, the required length of the pole is 20 m.
Example 7: A father is five times as old as his son. Six years hence he will be
only three times as old as his son. Find their present age.
Solution:
Let, the present age of the son be x years.
Then, the present age of the father = 5x years.
6 years hence the age of the son will be (x 6) years.
6 years hence the age of the father will be (5x 6) years.
According to the question, Answer checking:
5x 6 = 3 (x 6) Present age of the father = 30 years
Present age of the son = 6 years
or, 5x 6 = 3x 18 After 6 years:
or, 5x 3x = 18 6 The age of father = 30 + 6 = 36 years.
The age of son = 6 + 6 = 12 years
or, 2x = 12 Here, 30 = 5 times 6
and 36 = 3 times 12
or, x= 12 = 6 years which are given in the question
2
? The present age of the son = x years = 6 years
The present age of the father = 5x years = 5 u 6 years = 30 years.
Hence, the present age of the father and the son are 30 years and 6 years respectively.
EXERCISE 11.2
General Section - Classwork
1. Let's tell and write the equation corresponding to the following mathematical
statements where x is the unknown number. Also, find the value of x.
Mathematical Statement Corresponding Equation Value of x
a) The sum of two numbers is 10
and one of them is 3.
b) The difference of two numbers
is 6 and the smaller one is 9.
c) The difference of two numbers
is 7 and the greater one is 12.
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d) The product of two numbers is
18 and one of them is 9.
e) Six times a number is 42.
f) Two times a number increased
by 7 is 17.
g) 4 less than three times a number
is 11.
h) A number exceeds the other by 5
and their sum is 11.
i) The sum of two consecutive odd
numbers is 12.
j) The quotient of dividing 56 by x
is 8.
Creative Section - A
Let's solve these problems by making linear equations.
2. a) The sum of two numbers is 25. If one of the numbers is 10, find the other
number.
b) The difference of two numbers is 16. If the greater number is 40, find the
smaller one.
c) The difference of two numbers is 28. If the smaller number is 10, find the
greater one.
3. a) Two complementary angles differ by 10q. Find the angels.
b) There are 32 students in a class and 18 of them are girls. Find the number of
boys.
c) There are 3 more boys than girls in a class and the total number of students
is 33, find the number of girls and boys.
d) There are 2 less boys than girls in a class and the total number of students is
34, find the number of boys and girls.
4. a) If the sum of two consecutive numbers is 23, find them.
b) If the sum of two consecutive even numbers is 34, find them.
c) If the sum of two consecutive odd numbers is 12, find them.
d) If the sum of three consecutive odd numbers is 33, find them.
5. a) Find two numbers whose sum is 35 and the greater number exceeds the
smaller one by 9.
b) The sum of two numbers is 20. If the smaller number is 4 less than the bigger
one, find the numbers.
c) Find two numbers whose sum is 15 and difference is 3.
d) A sum of Rs 100 is divided into two parts. If the greater part exceeds the
smaller one by Rs 20, find the parts of the sum.
e) A sum of Rs 180 is divided into two parts. If the smaller part is Rs 50 less
than the greater part, find the parts of the sum.
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f) Sunayana and Anamol donated some money in a Coronavirus relief fund.
The amount donated by Anamol is Rs 250 more than that of Sunayana. If
they donated Rs 1000 in total, find the amount donated by each of them.
6. a) If 7 books cost Rs. 300 more than 5 same books, find the cost of a book.
b) Six times a number is 21 less than nine times the same number, find the
number.
c) Shashwat's weekly earning is earns Rs. 4,500 more than the earning of
4 days. What is his daily earning?
d) After 42 years, Akanksha will be four times as old as she is now. Find her
present age.
e) If a number is as many greater than 79 as it is less than 97, find the number.
7. a) If 2 part of the distance between two places is 34 km, find the distance
3
between the places.
b) 3 part of the total number of students of a school are girls. If there are 240
4
boys, find the number of girls.
c) If 10% discount on the price of an item amounts to Rs 100, find the price of
the item.
d) If 13% VAT on the selling price of an article amounts to Rs 130, find the
selling price of the article.
e) If 20% of profit on the cost price of an item amounts to Rs 400, find the cost
price of the item.
8. a) Divide Rs 650 in the ratio of 2 : 3.
b) A father divides Rs 54,000 between his son and daughter in the ratio of 4 : 5.
Find the shares of each of them.
c) Pratik, Devashish and Bishwant invest a sum of Rs 1,50,000 on a business in
the ratio of 2 : 3 : 5. Find the share of each of their investment.
d) If the angles of a triangle are in the ratio 1 : 2 : 3, find the size of each angle.
e) If two complementary angles are in the ratio of 2 : 3, find the angles.
Creative Section - B
9. a) 1 part of a pole is inside the mud, 2 part is inside the water, and the remaining
5 3
length of 8 m is above the surface of the water. Find the length of the pole.
s15pepnat r34t
b) Bipin had some money. He part of his money to buy goods for his
birthday, he gave his sister of the money, and he gave the rest of
Rs 120 to his mother. How much money did he have in the beginning?
10. a) The length of a rectangular ground exceeds its breadth by 5m. If its
perimeter is 110m, find its length and breadth.
b) The breadth of a rectangular handkerchief is 10cm shorter than its length. If
the perimeter of the handkerchief is 100cm, find its length and breadth.
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c) The length of a rectangular lawn is two times of its breadth. If its perimeter is
120 m, find its length and breadth.
d) The length and breadth of a rectangular park are in the ratio 5 : 2. If its
perimeter is 168 m, find its length and breadth.
11. a) A father is 25 years older than his son. Five years ago, he was six times as old
as his son was. Find their present ages.
b) A mother is 24 years older than her daughter. Eight years ago, she was seven
times of her daughter’s age. Find their present ages.
c) A son is 30 years younger than his father. After four years the father will be
three times as old as his son. Find their present ages.
d) The present age of a father is three times than that of his son. Five years ago
he was five times as old as his son was. Find their present ages.
e) A woman is thrice as old as her daughter. Twelve years hence, she will be
twice as old as her daughter. Find their present ages.
It's your time - Project work!
12. a) How many students are there in your class and how many girls are there?
Make an equation and find the number of boys.
b) How many students are there in your class and how many boys are there?
Make an equation and find the number of girls.
c) How many teachers are there in your school and how many are male
teachers? Make an equation and find the number of lady teachers.
Let's have a fun!
13. Let's ask your friends to think of any number and multiply the
number by 4. Then, tell them to add 6 to the product and divide the sum by 2.
At last, tell them to subtract two times the original number from the quotient
and ask their answer. Are they surprised? Solve it by making an equation.
11.5 Trichotomy – Review
Let’s consider any two whole numbers 3 and 7.
There is only one way to compare these two numbers.
Either, 3 < 7 (3 is less than 7) or 7 > 3 (7 is greater than 3).
But, 3 = 7 or 3 > 7 or, 7 < 3 are not true comparisons.
Thus, if a and b are any two whole numbers, only one comparison from the following
can be true comparison between them.
Either, a = b or, a < b or, a > b.
Such a property of whole numbers is known as Trichotomy property. The sign
‘=’ (equal to), ‘<’ (less than) and ‘>’ (greater than) are the trichotomy signs.
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Negation of trichotomy <
<
Let’s consider any two numbers 4 and 9.
<
Here, 4 < 9 or 9 > 4 are true comparisons.
But, 4 </ 9 (4 is not less than 9) or 9 / 4 (9 is not greater than 4) are false comparisons.
Here, 4 </ 9 is the negation of 4 < 9 and 9 / 4 is the negation of 9 > 4.
Thus, ‘</ ‘ (is not less than) is the negation of ‘<’ (is less than), / (is not greater than)
is the negation of ‘>’ (is greater than) and ‘z' (is not equal to) is the negation of '='
(is equal to).
Trichotomy rules
1. Let's suppose that a, b, and c are any three whole numbers, where a > b, then
(i) a + c > b + c [Addition axiom] eg, 9 > 4, then 9 + 3 > 4 + 3
(ii) a – c > b – c [Subtraction axiom] eg, 8 > 5, then 8 – 2 > 5 – 2
(iii) a u c > b u c [Multiplication axiom] eg, 4 > 3, then 4 u 2 > 3 u 2
(iv) a > b [Division axiom] eg, 9 > 6, then 9 > 6
c c 3 3
Thus, when an equal positive number is added to or subtracted from or
multiplied or divided by both sides of trichotomy sign the sign remains the
same.
2. Let's suppose that, a and b are any two whole numbers and c is a negative
integer.
(i) If a > b, then a u (– c) < b u (– c) eg, 4 > 2, then 4 u (– 2) < 2 u (– 2)
(ii) If a > b, then a < b eg, 10 > 8, then 10 < 8
–c –c –2 –2
Thus, when both sides of trichotomy sign are multiplied or divided by an equal
negative number, the sign ‘<’ is changed to ‘>’ and the sign ‘>’ is changed to ‘<’.
EXERCISE 11.3
General Section - Classwork
1. Let's tell and write ‘true’ or ‘false’ in the blank spaces.
a) –2 < – 7 ............... b) 0 > – 5 ............... c) 6 < – 6 ...............
d) 2 – 4 = 8 – 6 ............... e) 10 < -5 ............... f) –6 > –9 ...............
–5 5 3 3
2. Let's insert the appropriate trichotomy sign ( <, > or = ) in the boxes.
a) 9 –9 b) –9 9 c) 7 – 11 4–8
d) 4 × (–2) –20 –18
–4 × (–2) f) 10 g) 15
–4 8 –3 3
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Creative Section
3. a) Define trichotomy property of whole numbers with examples.
b) Define negation of trichotomy with examples.
c) State the addition, subtraction, multiplication, and division axioms of the
trichotomy rules with examples.
x y
d) If x > y, then (i) x × (–a) < y (–a) and (ii) –a < –a . Justify these facts with
one example of each.
4. Let's rewrite the following trichotomy statements using negation signs.
a) x > 5 b) x < 8 c) – a = – 7 d) p – 4 < 10
e) 12 = y + 4 f) – 9 > a – 4 g) x + 3 > – 6 h) y – 5 < 2
5. Write the following statements using trichotomy signs. Also, write the
negation of each statement. Rewrite each negation by using signs.
a) x is less than 3 b) y is greater than – 5 c) – a is equal to – 2
d) x + 4 is less than 6 e) 10 is greater than p – 9 f) p + r is less than q + r
11.6 Inequalities
'The sum of x and 5 is 9' is an open mathematical statement.
Here, x + 5 = 9 is the open mathematical statement containing 'equal to' (=) sign.
Such type of open mathematical statement is called an equation.
On the other hand, if the open mathematical statement contains trichotomy sign
such as < (less than), > (greater than), ≤ (less than and equal to), ≥ (greater than
and equal to), it is an inequality. Inequality is also called inequation. For example:
x > 5 (x is greater than 5), y < 7 (y is less than 7), x ≤ 9 (x is less than and equal to 9),
p ≥ – 2 (p is greater than and equal to – 2), etc. are a few examples of inequalities.
11.7 Replacement set and solution set
Let’s consider a set of natural number less than 5. Then, N = {1, 2, 3, 4}.
Consider an inequality x < 3.
Now, let’s replace x by the natural numbers of the set N.
When x = 1, 1 < 3 It is true. When x = 2, 2 < 3 It is true.
When x = 3, 3 < 3 It is false. When x = 4, 4 < 3 It is false.
Thus, the inequality x < 3 is true only for a certain values of x taken from the set of
natural numbers N, less than 5.
Here, the set of values of x that makes the inequality true is {1, 2} and it is known
as Solution set. Similarly, the set of natural numbers, N = {1, 2, 3, 4} from which
numbers are used to replace x in the inequality is known as the replacement set.
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11.8 Graphical representation of solution sets
We use number lines to show the solution sets of the given inequalities graphically.
Let's study the following illustrations.
x>3 x≤3
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
The circle at 3 shows that 3 is not The solid circle shows that 3 is
included in the solution set. included in the solution set.
x>–6 x≤ –2
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
–5< x <5 –7≤ x ≤ 7
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Worked-out examples
Example 1: If R = {0, 1, 2, 3, 4, 5} be the replacement set of the inequality
Solution: 2x + 3 < 9, find its solution set and illustrate graphically.
Here, the replacement set, R = {0, 1, 2, 3, 4, 5}
The given inequality is, In x < 3
when x = 0, 0 < 3 is true
2x + 3 < 9 when x = 1, 1 < 3 is true
when x = 2, 2 < 3 is true
or, 2x + 3 – 3 < 9 – 3 when x = 3, 3 < 3 is false
? The solution set = {0, 1, 2}
or, 2x < 6
or, 2x < 6
2 2
or, x < 3
0≤x≤2
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Example 2: Find the solution sets of the following inequalities and represent
them graphically. (i) 3x – 1 > x + 9 (ii) 4 – 5x ≤ 2x + 18
Solution:
(i) 3x – 1 >x+9
or, 3x – 1 – x > 9
or, 2x – 1 >9 1 is added to both sides.
or, 2x – 1 + 1 > 9 + 1
or, 2x > 10
or, 2x > 10 Both sides are divided by 2.
2 2
x>5
or, x > 5
? Solution set = {6, 7, 8, …} -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
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(ii) 4 – 5x d 2x + 18
or, 4 – 5x – 2x d 18
or, 4 – 7x d 18
or, 4 – 4 – 7x d 18 – 4 4 is subtracted from both sides.
or, – 7x d 14
or, – 7x t 14 Both sides are divided by – 7. So, the sign d is changed into t
or, – x7 t ––27 x ≥ -2
? Solution set = {– 2, – 1, 0, 1, 2,…} -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Example 3: Solve 4 < x + 2 d 9 and represent the solution set graphically.
Solution: 2 is subtracted from all sides.
4<x+2d9
2<x≤7
or, 4 – 2 < x + 2 – 2 d 9 – 2
or, 2 < x d 7
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Example 4: The difference of two times a number and 5 is less than and equal to –9.
(i) Express it using trichotomy sign.
(ii) Solve the inequality and draw a graph to represent it.
Solution:
(i) Let the number be x.
According to the given statement, 2x – 5 d – 9
(ii) or, 2x – 5 + 5 d – 9 + 5 5 is added to both sides.
or, 2x d – 4 Both sides are divided by 3.
or, 2x d –2
2 2
or, x d – 2 x ≤ –2
? Solution set = {– 2, – 3, – 4, …}
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Example 5: When one-third of a number is subtracted from 2, the difference is
greater than and equal to 4.
(i) Express it using trichotomy sign.
(ii) Solve the inequality and represent it graphically.
Solution:
(i) Let the number be x. x
3
According to the given statement, 2– t4
(ii) 2– x t4
3
or, 2–2– x t4–2 2 is subtracted from both sides.
3
or, – x t2
3
u (– x ( Both sides are multiplied by –3.
or, – 3 3 d 2 u (– 3) So, the sign t is changed into d
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 193 Vedanta Excel in Mathematics - Book 7
Equation, Inequality and Graph
or, x d –6
?Solution set = {–6, –7, –8, ...}
x ≤– 6
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
EXERCISE 11.4
General Section - Classwork
1. Let's tell and write the inequalities represented by each of the following graphs.
a) b)
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
c) d)
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
e) f)
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
g) h)
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
2. Let's represent these inequalities in the number lines.
a) x > – 4 b) x d 2
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
c) – 4 < x < 4 d) – 6 d x d
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Creative Section
3. a) What is an inequality? Write with an example.
b) In what way x + 2 < 7 is different from x + 2 = 7?
c) Define replacement set and solution set of an inequality with examples.
4. Let's draw graphs and represent the following inequalities.
a) x > 3 b) x < 4 c) x < – 2 d) x t – 3
e) 2 < x < 7 f) – 6 d x d 6 g) – 5 < x d3 h) – 2 d x < 6
Vedanta Excel in Mathematics - Book 7 194 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Equation, Inequality and Graph
5. If the replacement set is W = {0, 1, 2, 3, 4, 5}, let's find the solution set of the
following inequalities and represent them graphically.
a) x < 4 b) x t 1 c) x d 5 d) x + 1 < 5
e) x – 2 t 1 f) x < 5 g) 2x > 4 h) x – 3 d 1
3 3 5 5 4 4
6. Let's find the solution sets of the following inequalities and represent them
graphically.
a) x + 1 < 5 b) x + 2 > 5 c) x – 1 d 2 d) x – 4 t – 2
e) 3 < x + 9 f) 10 > x + 7 g) 7 t x + 10 h) 3 d x – 3
i) 3 – x d – 2 j) 2x > 6 k) 3x – 4 < 2 l) 2x – 5 >, 10 x
m) x < – 1 n) 2x > – 4 o) x – 1 d – 3 p) 2x –3 t 1
2 3 2 – 6 2
7. Let's solve the following inequalities and show the solutions graphically.
a) 5 < x + 1 < 9 b) 3 < x + 2 < 10 c) – 1 < x + 3 < 7
d) – 2 < x – 1 < 5 e) – 4 d 2x d 6 f) – 9 d 3x d 15
8. a) The sum of two times a number and 2 is less than 12.
(i) Express the statement by using trichotomy sign.
(ii) Solve the inequality and draw a graph to show the solution.
b) The sum of three times a number and 7 is greater than and equal to – 8.
(i) Express the statement by using trichotomy sign.
(ii) Solve the inequality and represent the solution graphically.
c) The difference of four times a number and 5 is less than and equal to 3.
(i) Express the statement by using trichotomy sign.
(ii) Solve the inequality and show the solution graphically.
d) When 5 times a number is subtracted from 7 the result is greater than and
equal to –3.
(i) Express the statement in a mathematical sentence.
(ii) Solve it and show the solution in a number line.
e) The difference of 5 and one-third of a number is less than and equal to 4.
(i) Express the statement in a mathematical sentence.
(ii) Solve it and show the solution in a number line.
9. a) The length a line segment is always greater than its x 1 x 3
BC
each part. On the basis of this fact, write down two A
inequalities from the figure along side. Solve them
and represent the solution graphically. A
b) The sum of the length of two sides of any triangle is always
greater than the length of third side. On the basis of this 3x 2x
fact, write down three inequalities from the triangle given B 3x + 4 C
below. Solve them and represent the solution graphically.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 195 Vedanta Excel in Mathematics - Book 7
Equation, Inequality and Graph
11.9 Function machine and relation between the variables x and y
Let's study the adjoining diagram of a function 1
machine. When different numbers are inserted 2 Inputs
3
inside the machine, it does certain process to bring
out results under the certain given rule. Here, the 1+4 5
inserted numbers are called inputs and the result 2+4 6 Outputs
coming out from the machine are called outputs. The 3+4 7
rule to give outputs is 'add 4 to each input'.
In the given function machine,
When input = 1, output = 1 + 4 = 5
When input = 2, output = 2 + 4 = 6
When input = 3, output = 3 + 4 = 7
Now, let’s represent every number of input by x and every result of output
by y. Then, the input and output can be shown in the table as well as in the
arrow-diagram.
xy
Input (x) 12 3 15
Output (y) 56 7
26
37
In the above function machine, its rule of producing outputs is ‘add 4 to each input.’
So, when input is x, the output y is x + 4. It is written as:
y=x+4
Thus, y = x + 4 is an equation with two variables Y
x and y. Here, when we put a certain value of y,
we obtain the corresponding value of y. So, the (3, 7)
value of y depends on the value of y. In this way, (2, 6)
x is known as an independent variable and y is (1, 5)
known as a dependent variable.
The input values of x and the corresponding O X
output values of y can also be shown in the form X' Y'
of ordered pairs such as: (1, 5), (2, 6), (3, 7), etc. If
we plot these ordered pairs in a graph paper and
join them, we obtain a straight line.
Therefore, y = x + 4 is a type of equation that always gives a straight line and it is
also called a linear equation.
Vedanta Excel in Mathematics - Book 7 196 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Equation, Inequality and Graph
Worked-out examples
Example 1: In the given function machine input Input (x)
the numbers from 1 to 5 and show the
outputs in a table and by an arrow- RULE Output (y)
diagram. ×3
Solution:
Here, the rule of getting outputs (y) is ‘multiply input x y
(x) by 3’. 1 3
? y = 3x 2 6
3 9
Input (x) 1234 5 4 12
Output (y) 3 6 9 12 15 5 15
arrow-diagram
Example 2: Discover the relation between input (x) and output (y) from the
given table. Then, find the missing numbers.
Solution: Input (x) 1 2 3 4 5 67
Output (y) 3 4 5 … … ……
Here, when input is 1, output = 3 o 1 + 2 o (input + 2)
when input is 2, output = 4 o 2 + 2 o (input + 2)
when input is 3, output = 5 o 3 + 2 o (input + 2)
Similarly, when input is x, output = x + 2
?The required relation between input and output is y = x + 2.
Now, by using the relation y = x + 2
when x = 4, y = 4 + 2 = 6 when x = 5, y = 5 + 2 = 7
when x = 6, y = 6 + 2 = 8 when x = 7, y = 7 + 2 = 9
Hence, the required missing numbers in the given table are 6, 7, 8, and 9.
Example 3: The relation between input (x) and output (y) is given by y = 2x 1.
Input the numbers from 0 to 5 in the relation and show the output in
the table.
Solution:
Here, the relation is y = 2x 1
When x = 0, y = 2 u 0 1 = 1; when x = 1, y = 2 u 1 1 = 3
When x = 2, y = 2 u 2 1 = 5; when x = 3, y = 2 u 3 1 = 7
When x = 4, y = 2 × 4 + 1 = 9
When x = 5, y = 2 × 5 + 1 = 11
Input (x) 012345
Output (y) 1 3 5 7 9 11
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 197 Vedanta Excel in Mathematics - Book 7
Equation, Inequality and Graph
EXERCISE 11.5
General Section - Classwork
1. Let's insert the inputs and tell and write the outputs by using the given rules.
012 –1–2 –3 24 6
Rule is + 1 Rule is – 1 Rule is × 3
2. The relation between the input (x) and the output (y) is y = x + 3. Let's tell
and write the missing inputs and outputs as quickly as possible.
Input (x) 0 1 4 – 1 ...... ...... – 5
Output (y) 3
...... ...... ...... 5 0 ......
3. Let's tell and write the relation between input (x) and output (y) as quickly as
possible.
a) Input = 1, output = 1 1 = 2 b) Input = 1, output = 1 × 2 = 2
Input = 2, output = 2 1 = 3 Input = 2, output = 2 × 2 = 4
Input = 3, output = 3 1 = 4 Input = 3, output = 3 × 2 = 6
? The relation is .......................... ? The relation is ..........................
x y x y
c) 1 1=0 d) 2u1 1 3
2 1=1 2u2 1 5
1 3 1=2 1 2u3 1 7
2 2 2u4 1 7
3 3
4
? The relation is .......................... ? The relation is ..........................
Creative Section
4. a) Let's input the numbers from 0 to 5 and show the outputs in the table from
the given function machine.
Vedanta Excel in Mathematics - Book 7 198 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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