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

Question

23164 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-02T13:55:59

# color(blue)(p = (fq) / (q - f ) #

#1/p + 1/q = 1/f#

Isolating the term containing #color(blue)(p)#
#1/color(blue)(p) + 1/q = 1/f#

#1/color(blue)(p) = 1/f - 1/q#

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

#1/p = (1 * color(green)(q))/(f * color(green)(q)) - (1 * color(green)(f))/(q * color(green)(f))#

#1/p = q/(fq) - f/(fq)#

#1/p = (q - f ) /(fq) #

# color(blue)(p = (fq) / (q - f ) #