How do you solve #g=(6y-p)/y# for #y#?

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Answer 1 · 2016-10-18T20:00:05

Apply the property #a/b = m/n -> a xx n = b xx m#.

#gy = 6y - p#

Collect all the terms containing #y# on one side of the equation.

#p = 6y - gy#

#p = y(6 - g)#

#p/(6- g) = y#

Hopefully this helps!