#1-1/(2x + 3) > (x+1)/(x-1)#
#1-1/(2x + 3) > (x-1 + 2)/(x-1)#
#1-1/(2x + 3) > (x-1)/(x-1) + 2/(x-1)#
#1-1/(2x + 3) > 1 + 2/(x-1)#
#0 > 2/(x-1)+1/(2x + 3)#
#2/(x-1)+1/(2x + 3) < 0#
Get an LCD.
#(2/(x-1))((2x+3)/(2x+3))+1/(2x + 3)((x-1)/(x-1)) < 0#
Multiply the numerators out.
#(4x+6)/((x-1)(2x+3))+(x-1)/((2x + 3)(x-1)) < 0#
Add the numerators to simplify.
#(5x+5)/((x-1)(2x+3)) < 0#
Factor the numerator completely.
#(5(x+1))/((x-1)(2x+3)) < 0#
Each factor in the numerator and denominator is linear and distinct (not raised to any power). The zeros of the numerator and denominator, in order from left to right, are:
#-3/2, -1, 1#.
When #x > 1#, the left side of the last inequality is positive.
When #-1 < x < 1#, the left side of the last inequality is negative.
When #-3/2 < x < -1#, the left side of the last inequality is positive.
When #x < -3/2#, the left side of the last inequality is negative.
We want to know when the left side is negative. This occurs on the compound interval #(-oo,-3/2)U(-1,1)#.
This is the solution to the original problem.
