How do you simplify #(4x+2)/(5x) + (10x-1)/(20x) + (2x-3)/(4x)#?

1 Answer
Oct 18, 2015

#(9x-2)/(5x)#

Explanation:

Your goal here is to find the common denominator for all those fractions and multiply each one so that you can get them to have the same denominator.

This will in turn allow you to add the resulting numerators and simplify the expression.

So, your three denominators are

#5 * x" "#, #" "20 * x" "#, and #" "4 * x#

Notice that you can rewrite #20x# as

#20x = 5 * 4 * x#

This means that you can use this as the common denominator for the three fractions, and multiply the first one by #1 = 4/4# and the second one by #1 = 5/5# to get them to have the denominator #20x#.

The expression can thus be written as

#(4x+2)/(5x) * 4/4 + (10x-1)/(20x) + (2x-3)/(4x) * 5/5#

#((4x + 2) * 4)/(20x) + (10x-1)/(20x) + ((2x-3) * 5)/(20x)#

Now that the fractions have the same denominator, add the numerators like you normally would

#(4x + 2) * 4 + 10x - 1 + (2x-3) * 5#

#16x + 8 + 10x - 1 + 10x - 15#

#36x - 8 = 4 * (9x-2)#

The expression can be simplified to

#(color(red)(cancel(color(black)(4))) * (9x-2))/(color(red)(cancel(color(black)(4))) * 5x) = color(green)( (9x-2)/(5x))#