Page 98 - Start Up Mathematics_8 (Non CCE)

3
5
3
5
= 2x – 3x + 12x – 3x – 7x 5
3
3
5
5
5
= (2x – 3x – 7x ) + (–3x + 12x )
= (–8x ) + (9x ) (Regrouping like terms)
5
3
5
= –8x + 9x 3
(b) 4a(a – b) – 3b(a – c) + c(b – c)
= {(4a ¥ a) – (4a ¥ b)} – {(3b ¥ a) – (3b ¥ c)} + {(c ¥ b) – (c ¥ c)}
2
= 4a + (–4ab – 3ab) + (3bc + bc) – c 2
2
= 4a – 7ab + 4bc – c 2
2
2
2
3
Example 17: Add using the column method: x (3x – 2y), x(2x – 5xy), 4(x + x y)
2
2
2
2
3
2
2
Solution: x (3x – 2y) = (x ¥ 3x) – (x ¥ 2y) = 3 ¥ (x ¥ x) – 2x y = 3x – 2x y ...(1)
2
2
2
3
2
x(2x – 5xy) = (x ¥ 2x ) – (x ¥ 5xy) = 2 ¥ (x ¥ x ) – 5 ¥ (x ¥ x) ¥ y = 2x – 5x y ...(2)
2
3
3
2
2
3
and 4(x + x y) = 4 ¥ x + 4 ¥ x y = 4x + 4x y ...(3)
Adding (1), (2) and (3)
3
2
3x – 2x y
2
3
2x – 5x y
2
3
+ 4x + 4x y
3
2
9x – 3x y
3
2
2
4
4
Example 18: Subtract 2(a – 3a ) from 2a (a – a) – 3a(a + 2a).
4
3
2
2
4
Solution: 2a (a – a) – 3a(a + 2a) – 2(a – 3a )
2
4
4
2
3
2
= {(2a ¥ a ) – (2a ¥ a)} – {(3a ¥ a ) + (3a ¥ 2a)} – {(2 ¥ a ) – (2 ¥ 3a )}
2
4
= {2 ¥ (a 2 + 3 ) – 2 ¥ (a 2 + 1 )} – {3 ¥ a 1 + 4 + 6 ¥ (a 1 + 1 )} – {2a – 6a }
2
3
4
2
5
5
= (2a – 2a ) – (3a + 6a ) – (2a – 6a )
3
4
5
5
2
= 2a – 2a – 3a – 6a – 2a + 6a 2
2
5
5
3
2
4
= (2a – 3a ) – 2a – 2a + (–6a + 6a )
4
4
3
5
5
= –a – 2a – 2a + 0 = –a – 2a – 2a 3
EXERCISE 5.4
1. Find the following products:
2
3
3
3
2
2
2
(a) 2x(y + z) (b) a (a b + b a) (c) –4p ¥ (2pq + 3p ) (d) –10x (x + y )
4 Ê 1 3 ˆ 3 3 2 3 2
22
(e) ab ¥ Á ab - b (f) pq(4p – 9q) (g) 12p (6p – 5q ) (h) (8a b) ¥ (b + c )
˜
5 Ë 10 7 ¯ 7
2. Simplify the following:
2 ˆ
2
2
2
(a) 4x + 3x(4x – 3y) – 10xy (b) (abc) ¥ (4b + 9a ) – Ê Á Ë 3 ab ¥ ( ac+ )
˜
¯
2
1 Ê 2 1 2 ˆ 7 2 Ê 2 2 ˆ 2 2 2
2
(c) x Á x + y ˜ - x Á x + 4y ˜ + 12 (d) a (b – c) – b (c – a) – c (a – b)
2
10 Ë 5 4 ¯ 2 Ë 7 ¯
3 2 2 3 3 1 2 2 2 2 3 3
(e) y (y – 1) – y(y – 1) + y (y + y) (f) a (a + 1) – a (a + 1) – a(a – a)
2 4 4
3
2
2
2
2
2
2
2
2
(g) a (1 – 3b ) – 3b(b – 4a b) + a(ab – 2a) (h) q p(q – p ) + qp (4qp – 2q ) – q p(1 – 2p)
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