If 1/(a+b+7)=1/a+1/b+1/7 and a + b ≠ 0 and a + b ≠ 7, then the possibilities are:

If 1/(a+b+7)=1/a+1/b+1/7 and a + b ≠ 0 and a + b ≠ 7, then…

If $\frac{1}{a+b+7}=\frac{1}{a}+\frac{1}{b}+\frac{1}{7}$ and a + b $\neq$ 0 and a + b $\neq$ 7, then the possibilities are:

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Correct answer: D

Explanation

The correct answer is Option 4 i.e. a = -7 or b = -7.

1/(a + b + 7) = 1/a + 1/b + 1/7

⇒ 1/(a + b + 7) = (7b + 7a + ab)/(7ab)

⇒ 7ab = (a + b + 7)(7a + 7b + ab)

⇒ 7ab = 7a² + 7ab + a²b + 7ab + 7b² + ab² + 49a + 49b + 7ab

⇒ 7a² + 7b² + a²b + ab² + 14ab + 49a + 49b = 0

⇒ 7(a + b)² + (ab + 49)(a + b) = 0

⇒ (a + b)[7(a + b) + ab + 49] = 0

⇒ 7(a + b) + ab + 49 = 0

⇒ (a + 7)(b + 7) = 0

⇒ a = -7 or b = -7