How do you solve #(1/3)-(x/5)=(1/2)-(x/10)#?

1 Answer
May 18, 2017

Answer:

#x=-5/3#

Explanation:

Whenever you have fractions in an equation, you can get rid of them in the very beginning.

Do this by multiplying each fraction by the LCM of the denominators.

This means, find a number which ALL the denominators can divide into. In this case it is #color(magenta)(30)#

#(color(magenta)(30)xx1)/3-(color(magenta)(30xx) x)/5 = (color(magenta)(30xx)1)/2-(color(magenta)(30xx)x)/10#

Now cancel all the denominators with the #color(magenta)(30# then simplify.

#(color(magenta)(cancel30^10)xx1)/cancel3-(color(magenta)(cancel30^6xx) x)/cancel5 = (color(magenta)(cancel30^15xx)1)/cancel2-(color(magenta)(cancel30^3xx)x)/cancel10#

#color(white)(.....)10color(white)(..........)-6xcolor(white)(.....)=color(white)(.....)15color(white)(...........)-3x#

Now solve the equation as usual- there are no fractions.

#10-6x = 15-3x" "# add #6x# and subtract #15# on both sides

#10-15 = -3x+6x#

#-5 = 3x#

#x=-5/3#