Page 97 - Start Up Mathematics_8 (Non CCE)

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Example 12: Find the products: (a) a ¥ (2ab – 5c) (NCERT) (b) (7a b ) ¥ (a + b)
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Solution: (a) a ¥ (2ab – 5c) (b) (7a b ) ¥ (a + b)
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2 2
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= (a ) ¥ (2ab) – (a ) ¥ (5c) = (7a b ) ¥ a + (7a b ) ¥ b
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= 2 ¥ (a ¥ a) ¥ b – a ¥ 5 ¥ c = 7 ¥ (a ¥ a) ¥ b + 7 ¥ a ¥ (b ¥ b)
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= 2a b – 5a c = 7a b + 7a b
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3 2
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Example 13: Multiply 10x and (–x – y ). Verify the result for x = 1 and y = 1 .
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Solution: (10x ) ¥ (–x – y ) = (10x ) ¥ (–x ) – (10x ) ¥ (y )
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= –10 ¥ (x ¥ x ) – 10x y = –10x – 10x y
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2 2
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LHS = 10x ¥ (–x – y ) RHS = –10x – 10x y
Ï Ê ˆ 1 2 ¸ 2 1ˆ 2
Ê
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Ì
= 10(1) ¥ -()1 3 - Á ˜ ˝ = –10(1) – 10(1) Á ˜
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Ó Ë ¯ ˛ Ë 3¯
{ } Ê 91 1
-- ˆ
1
= 10 ¥ --1 9 = 10 ¥ Á 9 ˜ = –10 ¥ 1 – 10 ¥ 1 ¥ 9
Ë
¯
Ê -10 ˆ -100 = –10 – 10 = -90 -10 = -100
= 10 ¥ Á ˜ ¯ = 9 9 9 9
Ë 9
LHS = RHS
Hence verified.
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Example 14: Multiply x – 3xy by 2x y.
Solution: Horizontal method Column method
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(2x y) ¥ (x – 3xy ) x – 3xy (First multiply x by 2x y
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= (2x y) ¥ (x ) – (2x y)(3xy ) ¥ 2x y and then 3xy by 2x y)
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= 2 ¥ (x ¥ x ) ¥ y – (2 ¥ 3) ¥ (x ¥ x) ¥ (y ¥ y ) 2x y – 6x y
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4 3
= 2x y – 6x y
Example 15: Find the following product. Also find the value for a = 2 and b = 1.15.
0.01a ¥ (0.1a + 0.001b)
Solution: 0.01a ¥ (0.1a + 0.001b) = (0.01a) ¥ (0.1a) + (0.01a) ¥ (0.001b)
= (0.01 ¥ 0.1) ¥ (a ¥ a) + (0.01 ¥ 0.001) ¥ (a ¥ b)
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= 0.001a + 0.00001ab
Putting a = 2, b = 1.15 in final result, we get
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0.001a + 0.00001ab = (0.001) ¥ (2) + (0.00001) ¥ (2 ¥ 1.15)
= (0.001) ¥ 4 + (0.00001) ¥ 2.30
= 0.004 + 0.0000230
= 0.0040230
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Example 16: Simplify: (a) x (2x – 3) + 3x (4x – x ) – 7x 5 (b) 4a(a – b) – 3b(a – c) + c(b – c)
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Solution: (a) x (2x – 3) + 3x (4x – x ) – 7x = (x ¥ 2x ) – (x ¥ 3) + (3x ¥ 4x) – (3x ¥ x ) – 7x 5
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= 2 ¥ (x ¥ x ) – 3x + (3 ¥ 4) ¥ (x ¥ x) – 3 ¥ (x ¥ x ) – 7x 5
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