How do you solve for do in #(hi)/(ho)=-(di)/(do) #?

1 Answer
Sep 3, 2016

Answer:

#(do) = -(ho xx di)/(hi)#

Explanation:

If an equation has one fraction on each side, you can get rid of the denominators by cross-multiplying.

#(hi)/(ho) = -(di)/(color(red)(do))#

#hixxcolor(red)(do) = -ho xx di " "larr div "(hi)#

#color(red)(do) = -(ho xx di)/(hi)#

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If an equation has one fraction on each side, you can invert the whole equation. This will put #do# in the numerator.

#(ho)/(hi) = - (color(red)(do))/(di)" "larr xx -di#

#-(ho xx di)/(hi) =color(red)(do)#

#color(red)(do) = -(ho xx di)/(hi)#