Question

what must be subtracted from 3a^2-6ab-3b^2-1 to get 4a2-7ab+4b^2+1
(with steps ) and correct answer will be the brainliest

Answer 1

4a^2-7ab+4b^2+1 -

3a^2 -6ab +3b^2-1

a^2 -ab + b^2 +2

a^2 -ab + b^2 + 2

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Sabharish G

Answer 2

Answer:

-a² + ab - 7b² - 2

Step-by-step explanation:

Let the number be x.

A/Q,

(3a² - 6ab - 3b² - 1) - x = 4a² - 7ab + 4b² + 1

→ (3a² - 6ab - 3b² - 1) - (4a² - 7ab + 4b² + 1) = x

→ (3a² - 6ab - 3b² - 1) - 4a² + 7ab - 4b² - 1 = x

→ 3a² - 4a² - 6ab + 7ab - 3b² - 4b² - 1 - 1 = x

→ -a² + ab - 7b² - 2 = x

→ x = -a² + ab - 7b² - 2

Therefore the required number is -a² + ab - 7b² - 2.

Common Mistakes done by students:

1. Whenever it's given subtract a from b then do b - a. For example subtract 2 from 10 then we should do 10 - 2 = 8
2. Change of sign takes place while subtracting. For example (y + z) - (y - z) = y + z - y + z = 2z
3. Group the like terms together.

Thanks !

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