Question

If $xϵR$, find the solution set of each of the following inequations. Represent it on the number line.
$\frac{x-3}{x+4}<0$

Answer 1

By replacing the inequality sign with an equal sign, so that we can solve it like a normal equation.

$\frac{x-3}{x+4}=0$

Multiplying both sides by $x+4$ we get

$x-3=0$

Adding $3$ to both sides, we get $x=3$

Also note that $x$ is undefined at $-4$.

$x\ne -4$

From the values of $x$ above, we have these $3$ intervals to test.

$x<-4$

$-4<x<3$

$x>3$

Let us pick a test point for each interval.

For the interval $x<-4$

Lets' pick $x=-5$

Then $\frac{-5-3}{-5+4}<0$

By simplifying we get $8<0$ which is false

Lets' pick $-4<x<3$

Then $\frac{0-3}{0+4}<0$

By simplifying we get $-0.75<0$ which is true.

Lets' pick $-x>3$

Then $\frac{4-3}{4+4}<0$

By simplifying we get $0.125<0$ which is false

$\therefore -4<x<3$

Hence $x\in (-4,3)$