Question

Subtract the sum of 12ab - 10b2 - 18a2 and 9ab + 12b2 + 14a2 from the sum of ab + 2b2 and
3b2-a2
(a) 3a2 + 3b2 - 20ab
(b) -3b2 + 6a2 - 20ab
(C) -5a2 +7b2 + 22ab
(d) 6b2 - 3a2 + 20ab

Answer 1

Answer:

-2ab +3a² +27b²

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Step-by-step explanation:

first, we form the formula.

[(1ab + 2b²) + (3b² - a²)] - [(12ab - 10b² - 18a²) + (9ab + 12b² + 14a²)]

now we simplify this by opening brackets one by one.

= [1ab + 2b² + 3b² - 1a²] - [12ab - 10b² - 18a² + 9ab + 12b² + 14a²]

= 1ab + 2b² + 3b² - 1a² - 12ab + 10b² + 18a² + 9ab + 12b² - 14a²

now, we simplify it by putting it in order: grouping like terms together.

= (1ab - 12ab + 9ab) + (2b² + 3b² + 10b² + 12b²) + (-1a² + 18a² - 14a²)

= (-2ab) + (27b²) + (3a²)

= -2ab +3a² +27b²

Answer 2

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