$$\left|\frac{-10}{x-3}\right|>\:5$$
- Find the values that $x$ can take.
I know that
$$\left|\frac{-10}{x-3}\right|>\:5$$ and $$\left|\frac{-10}{x-3}\right|<\:-5$$
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2$\left \lvert \frac{-10}{x-3} \right \rvert$ is not less then $-5$... – Staki42 Dec 18 '17 at 16:13
4 Answers
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$$\left| \frac { -10 }{ x-3 } \right| >\: 5\\ \frac { 10 }{ \left| x-3 \right| } >5\\ \left| x-3 \right| <2\\ -2<x-3<2\\ 1<x<5\\ \left( 1;5 \right) -\left\{ 3 \right\} \\ \\ $$
haqnatural
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$$\left|\frac{-10}{x-3}\right|>\:5$$ so $$10> 5|x-3| \Longrightarrow -2<x-3<2 \Longrightarrow 1<x<5; x\ne 3$$
nonuser
- 90,026
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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 $$x<3$$
Dr. Sonnhard Graubner
- 95,283
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First note that $x=3$ cannot be a part of the solution. That in mind, we have: $$|\frac{-10}{x-3}|>5 \implies \frac{10}{5}> |x-3| \implies -2< x -3 < 2 \implies x \in (1,5) - \{3\}$$