Given $\frac{1}{f} = (n - 1) (\frac{a}{1} + \frac{1}{b}),$ make 'n' as the subject of the formula. Hence, find 'n' if $f = 1.2, a = 12$ and $b = 3$ ,

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Solution

$\frac{1}{f} = (n - 1) (\frac{a}{1} + \frac{1}{b})$
Or $n f a + n f b = a b + f a + f b$
Or $n f (a + b) = a b + f (a + b)$
Or $n = \frac{a b}{f (a + b)} + 1$
Given $f = 1.2, a = 12, b = 3$
So $n = \frac{12 \times 3}{1.2 (12 + 3)} + 1 = \frac{36}{18} + 1 = 3$

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