How do you solve #(4x)/15-5/6=2/3# by clearing the fractions?

Question

874 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-07T14:58:21

#x = 5 5/8 = 5.625#

When you have an equation with fractions, you can get rid of the denominators by multiplying each fraction by the LCM to the denominators can cancel.

#(4x)/15 -5/6 = 2/3" "larr LCM = 30#

#(color(blue)(30xx)4x)/15 -(color(blue)(30xx)5)/6 = (color(blue)(30xx)2)/3#

#(color(blue)(cancel30^2xx)4x)/cancel15 -(color(blue)(cancel30^5xx)5)/cancel6 = (color(blue)(cancel30^10xx)2)/cancel3#

#8x-25 = 20#

#8x = 20+25#

#8x = 45#

#x = 5 5/8 = 5.625#