Explain (in words) how you would solve the following inequality using an interval table: 5/x+3 ≤ 2/x-1.

Answer 1

First of all, I will assume you meant:
5/(x+3) ≤ 2/(x-1) , ............ (that's sort of important)

here is how I would do this question:
critical values, x = -3 and x = 1, neither the LS or the RS exist for those.
I would consider
5/(x+3) = 2/(x-1)
5x-5 = 2x + 6
3x = 11
x = 11/3
so on a number line, (or your interval table), I would note
x = -3, x = 1 and x = 11/3, giving me 4 segments on the number line.
test for an arbitrary value from each segment in the inequality

for x < -3, say x = -8
is 5/-5 ≤ 2/-9 ?, YES, so x < -3 works
for -3 < x < 1 , say x = 0
is 5/3 ≤ -2 ?, NO
for 1 ≤ x 11/3 , let's pick x = 2
is 5/5 ≤ 2/(2 - 1) ?, YES , so 1 < x < 11/3

and finally,
for x > 11/3 , let's pick x = 7
is 5/10 ≤ 2/6 ? , NO

so I get x < -3 OR 1 < x < 11/3
now put that into interval notation, which I usually don't use (yes, I am old school)

You can also verify, by graphing
y < 5/(x+3) and y > 2/(x-1)
x = -3, x = 1 (those would be the asymptotes)

on Desmos to see , (google Desmos)