How do you solve #(1/4)+(1/x)=(1/3)#?

1 Answer
May 21, 2017

Answer:

See a solution process below:

Explanation:

First, we can multiply each side of the equation by #color(red)(12)color(blue)(x)# to eliminate the fractions while keeping the equation balanced. #color(red)(12)color(blue)(x)# is the Least Common Denominator for the three fractions:

#color(red)(12)color(blue)(x)((1/4) + (1/x)) = color(red)(12)color(blue)(x) xx (1/3)#

#(color(red)(12)color(blue)(x) xx (1/4)) + (color(red)(12)color(blue)(x) xx (1/x)) = cancel(color(red)(12))4color(blue)(x) xx (1/color(red)(cancel(color(black)(3))))#

#(cancel(color(red)(12))3color(blue)(x) xx (1/color(red)(cancel(color(black)(4))))) + (color(red)(12)cancel(color(blue)(x)) xx (1/color(blue)(cancel(color(black)(x))))) = 4x#

#3x + 12 = 4x#

Now, subtract #color(red)(3x)# from each side of the equation to solve for #x# while keeping the equation balanced:

#-color(red)(3x) + 3x + 12 = -color(red)(3x) + 4x#

#0 + 12 = (-color(red)(3) + 4)x#

#12 = 1x#

#12 = x#

#x = 12#