Subtract:

$(i) 5 a b + 4 a c - 6 a + 15 c - 15$ from $10 a b + 22 b c - 13 a - 14 b - 16 c - 24$

$(i i) 5 a^{2} b + 10 b^{2} c^{2} - 10 b^{2} c - 21 a b + 32$ from $50 - 20 b^{2} c^{2} - 4 b c - a c - 5 a b$ . [2 MARKS]

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Solution

Application: 1 Mark each

(i) $(10 a b + 22 b c - 13 a - 14 b - 16 c - 24) - (5 a b - 6 a + 15 c - 15 + 4 a c)$

$10 a b + 22 b c - 13 a - 14 b - 16 c - 24 - 5 a b + 6 a - 15 c + 15 - 4 a c$

$= 5 a b + 22 b c - 7 a - 14 b - 31 c - 9 - 4 a c$

(ii) $(50 - 20 b^{2} c^{2} - 4 b c - a c - 5 a b) - (32 + 10 b^{2} c^{2} - 21 a b + 5 a^{2} b - 10 b^{2} c)$

$50 - 20 b^{2} c^{2} - 4 b c - a c - 5 a b - 32 - 10 b^{2} c^{2} + 21 a b - 5 a^{2} b + 10 b^{2} c$

$= 18 - 30 b^{2} c^{2} - 4 b c - a c + 16 a b - 5 a^{2} b + 10 b^{2} c$

Suggest Corrections

Similar questions

(i) 5 a b + 4 a c - 6 a + 15 c - 15

10 a b + 22 b c - 13 a - 14 b - 16 c - 24

(i i) 5 a^{2} b + 10 b^{2} c^{2} - 10 b^{2} c - 21 a b + 32

50 - 20 b^{2} c^{2} - 4 b c - a c - 5 a b

When $5 a^{2} b + 10 b^{2} c^{2} - 21 a b + 32$ is subtracted from $50 - 20 b^{2} c^{2} - 4 b c - a c - 5 a b$ , we get

6 a^{2} - 7 b + 5 a b

2 a b

12 a^{3} b - 14 a b^{2} + 10 a b

12 a^{3} b - 14 a b^{2} + 10 a^{2} b^{2}

6 a^{2} - 7 b + 7 a b

12 a^{2} b - 7 a b^{2} + 10 a b

x^{2} + a x - (6 a^{2} - 5 a b + b^{2})

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