Finding the values that $x$ can take $\left|\frac{-10}{x-3}\right|>\:5$

Question

$$\left|\frac{-10}{x-3}\right|>\:5$$

• Find the values that $x$ can take.

I know that

$$\left|\frac{-10}{x-3}\right|>\:5$$ and $$\left|\frac{-10}{x-3}\right|<\:-5$$

Answer 1

$$\left| \frac { -10 }{ x-3 } \right| >\: 5\\ \frac { 10 }{ \left| x-3 \right| } >5\\ \left| x-3 \right| <2\\ -2<x-3<2\\ 1<x<5\\ \left( 1;5 \right) -\left\{ 3 \right\} \\ \\ $$

Answer 2

$$\left|\frac{-10}{x-3}\right|>\:5$$ so $$10> 5|x-3| \Longrightarrow -2<x-3<2 \Longrightarrow 1<x<5; x\ne 3$$

Answer 3

simplifying and multiplying by $$|x-3|$$ for $$x\ne 3$$ we get $$2>|x-3|$$ this is equivalent to $$2>x-3$$ if $$x\geq 3$$ and $$2>-x+3$$ if $$x<3$$

Answer 4

First note that $x=3$ cannot be a part of the solution. That in mind, we have: $$|\frac{-10}{x-3}|>5 \implies \frac{10}{5}> |x-3| \implies -2< x -3 < 2 \implies x \in (1,5) - \{3\}$$