How do you solve #5/(x+2)>5/x + 2/3x#?

1 Answer
Jun 1, 2017

To work with fractions, our first goal needs to be to find a common denominator.

Let's combine the components on the right of the sign. To do that, we need them to have the same denominator.

#5/x+(2x)/3#

Let's multiply #5/x# by #3/3# and #(2x)/3# by #x/x#:

#3/3 xx 5/x + (2x)/3 xx x/x#

combine

#(3 xx 5)/(3 xx x) + (2 xx x xx x)/(3 xx x)#

simplify

#(15)/(3x) + (2x^2)/(3x)#

combine

#(15+ 2x^2)/(3x)#

Now let's put that back into the expression:L

#5/(x+2) > (15+ 2x^2)/(3x)#

Now, we need to get the same denominator again. Luckily, we already know how to do that!

#(3x)/(3x) xx 5/(x+2) > (15+ 2x^2)/(3x) xx (x+2)/(x+2)#

#(3 xx x xx 5)/(3 xx x xx (x+2)) > ((15+2x^2) xx (x+2))/(3 xx x xx (x+2))#

simplify

#(15x)/(3x^2+6x) > (15x+30+2x^3+4x^2)/(3x^2+6x)#

multiply both sides by #(3x^2+6x)#

#cancel((3x^2+6x)) xx (15x)/cancel((3x^2+6x)) > (15x+30+2x^3+4x^2)/cancel((3x^2+6x)) xx cancel((3x^2+6x))#

#15x > 2x^3+4x^2+15x+30#

Subtract #15x# on both sides

#0 > 2x^3+4x^2+30#

factor out a #2#

#0 > 2(x^3+2x^2+15)#

divide by #2# on both sides

#0 > x^3+2x^2+15#