4/x+5>1/2x+3

Answer 1

assuming the usual carelessness with parentheses,
4/(x+5) > 1/(2x+3)
you can work out four separate solutions, choosing which expressions are positive or negative, or take a shortcut via
16/(x+5)^2 > 1/(2x+3)^2
16(2x+3)^2 > (x+5)^2
64x^2 + 192x + 144 > x^2 + 10x + 25
63x^2 + 182x + 119 > 0
9x^2 + 26x + 13 > 0
(9x+17)(x+1) > 0
x is in (-∞,-17/9) U (-1.∞)
But this does not take into account the fact that x ≠ -5 , -3/2
So the real solution is
(-∞,-5) U (-5,-17/9) U (-1.∞)

Answer 2

I usually do these the following way:

consider the actual equation 4/(x+5) = 1/(2x+3)
easy to solve and x = -1
also x ≠ -5, -3/2
So we have the following critical parts of the domain for x
x < -5
-5 < x < -3/2
-3/2 < x < -1 , and
x > -1

pick a value in each section
for x < -5, let x = -6 .... is 4/-1 > 1/-9 ?? , NO
-5 < x < -3/2, let x = -3 ... is 4/2 > 1/-3 ?? , YES
-3/2 < x < -1, let x = -1.1 ... is 4/3.9 > 1/.8 ??, NO
finally, x > -1, let x = 10, ..... is 4/15 > 1/23 ?? YES

so we have:
-5 < x < -3/2 OR x > -1

graphing y = 4/(x+5) and y= 1/(2x+3) using DESMOS shows this is correct.
By squaring both sides, sometimes solutions are introduced which do
not satisfy the original relation. That is why all solutions must be verified after squaring.

The value of x = -17/9 does not show up when graphing

Answer 3

well said; always check your answers against the original equations.
Unfettered cleverness will get you in trouble.

Answer 4

Can't beat being unfettered no matter what !