How do you solve for p in $\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$ $1/p+1/q=1/f$ ?

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How do you solve for p in $\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$ $1/p+1/q=1/f$ ?

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Answer 1 · 2016-06-02T13:55:59

$p = \frac{f q}{q - f}$ $color(blue)(p = (fq) / (q - f )$

Explanation:

$\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$ $1/p + 1/q = 1/f$

Isolating the term containing $p$ $color(blue)(p)$
$\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$ $1/color(blue)(p) + 1/q = 1/f$

$\frac{1}{p} = \frac{1}{f} - \frac{1}{q}$ $1/color(blue)(p) = 1/f - 1/q$

The L.C.M of the denominators of the terms on the L.H.S is $f q$ $color(green)(fq$

$\frac{1}{p} = \frac{1 \cdot q}{f \cdot q} - \frac{1 \cdot f}{q \cdot f}$ $1/p = (1 * color(green)(q))/(f * color(green)(q)) - (1 * color(green)(f))/(q * color(green)(f))$

$\frac{1}{p} = \frac{q}{f q} - \frac{f}{f q}$ $1/p = q/(fq) - f/(fq)$

$\frac{1}{p} = \frac{q - f}{f q}$ $1/p = (q - f ) /(fq)$

$p = \frac{f q}{q - f}$ $color(blue)(p = (fq) / (q - f )$

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How do you solve for p in $\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$ $1/p+1/q=1/f$ ?

1 Answer

Explanation:

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How do you solve for p in 1/p+1/q=1/f1p+1q=1f 1/p+1/q=1/f?

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How do you solve for p in $\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$ $1/p+1/q=1/f$ ?