How do you solve #2/5x=3# by clearing the fractions?

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Answer 1 · 2017-09-29T16:49:05

See a solution process below:

First, multiply each side of the equation by #color(red)(5)# to clear the fractions and keep the equation balanced:

#color(red)(5) xx 2/5x = color(red)(5) xx 3#

#cancel(color(red)(5)) xx 2/color(red)(cancel(color(black)(5)))x = 15#

#2x = 15#

Now, divide each side of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:

#(2x)/color(red)(2) = 15/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = 15/2#

#x = 15/2#

Or

#x = 7.5#