Question

How do you solve `\frac { x - 5} { x + 3} < 0`?

Answer 1

Only solution is `-3< x < 5` i.e. `x` lies between `-3` and `5`

Explanation:

`(x-5)/(x+3)<0` means `(x-5)/(x+3)` is negative

i.e. either `x-5<0` and `x+3>0` i.e. `x<5` and `x> -3`

i.e. `-3< x < 5`

or `x-5>0` and `x+3<0` i.e. `x>5` and `x< -3`, but this is not possible.

Hence only solution is `-3< x < 5` i.e. `x` lies between `-3` and `5`

Answer 2

Open interval (-3, 5)

Explanation:

Let me introduce the method of superimposing by using the double number line.
`f(x) = g(x)/(h(x)) = (x - 5)/(x + 3)` < 0
The first number line figures the variation of g(x) = x - 5
The second number line figures the variation of h(x) = x + 3
f(x) has the resulting sign of the product g(x).h(x)

-------------------------------- 0 ----------------------- 5 ++++++++++++++ h(x)
----------------- - 3 ++++++ 0+++++++++++++++5++++++++++++++ g(x)
+++++++++++ -3 ------------------------------------- 5 +++++++++++++ f(x)

By superimposing, we see that f(x) is negative (f(x) < 0) inside the open interval (-3, 5)