Question

EXERCISE 2A 1. Add. (1) 2 ab, -3ab, 5ab and 1/2ab​

Answer 1

Answer:

Given solution:

(ab+2b2+3b2−a2)−((12ab−10b2−18a2)+(9ab+12b2+14a2))

(ab+2b2+3b2−a2)−(21ab+2b2−4a2)

(ab+2b2+3b2−a2)−21ab−2b2+4a2

(ab+5b2−a2)−21ab−2b2+4a2

=3b2+3a2−20ab

Answer 2

Answer:

Given : 8(3a−2b)

2−10(3a−2b)

By talking (3a−2b) in the given question, we get,

(3a−2b)[8(3a−2b)−10]

By taking 2 as common in the second term,

(3−2b)2[4(3a−2b)−5]

So we get,

8(3a−2b)

2−10(3a−2b)=2(3a−2b)(12a−8b−5)

Step-by-step explanation

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