How do you solve: [ (x-2) / (x+3) ] < [ (x+1) / (x) ]?

1 Answer
Oct 20, 2015

Answer:

#x in (-1/2, + oo) "\" {0}#

Explanation:

The inequality given to you looks like this

#(x-2)/(x+3) < (x+1)/x#

Right from the start, you know that any solution interval must not contain the values of #x# that will make the two denominators equal to zero.

More specifically, you need to have

#x+3 != 0 implies x != -3" "# and #" "x != 0#

With that in mind, use the common denominator of the two fractions, which is equal to #(x+3) * x#, to get rid of the denominators.

More specifically, multiply the first fraction by #1 = x/x# and the second fraction by #1 = (x+3)/(x+3)#.

This will get you

#(x-2)/(x+3) * x/x < (x+1)/x * (x+3)/(x+3)#

#(x(x-2))/(x(x+3)) < ((x+1)(x+3))/(x(x+3))#

This is equivalent to

#x(x-2) < (x+1)(x+3)#

Expand the parantheses to get

#color(red)(cancel(color(black)(x^2))) - 2x < color(red)(cancel(color(black)(x^2))) + x + 3x + 3#

Rearrange the inequality to isolate #x# on one side

#-6x < 3 implies x > 3/((-6)) <=> x > -1/2#

This means that any value of #x# that is greater than #-1/2#, except #x = 0#, will be a solution to the original inequality.

Therefore, the solution interval will be #x in (-1/2, + oo) "\" {0}#