Question

Solve $\frac{2}{|x-3|}>5,xϵR$

Answer 1

Clearly, $x-3\ne 0$ and therefore, $x\ne 3$.

We have, $\frac{2}{|x-3|}>5$

Since $|x-3|$ is positive, we may multiply both sides of (i) by |x-3|
This gives

$2>5|x-3|$

$\iff \frac{2}{5}>|x-3|$

$\iff |x-3|<\frac{2}{5}$

$\iff \frac{-2}{5}<X-3<\frac{2}{5}[\because |x|<a\iff -a<x<a]$

$\iff \frac{-2}{5}<x-3andx-3<\frac{2}{5}$

$\iff \frac{-2}{5}+3<xandx<\frac{2}{5}+3$

$\iff \frac{13}{5}<xandx<\frac{17}{5}$

$\iff \frac{13}{5}<x<\frac{17}{5}$

Also, as shown above, $x\ne 3$.

$\therefore$ solution set = $\{xϵR:\frac{13}{5}<x<\frac{17}{5}\}-\left[3\right]$

$=(2.6,3.4)-\left\{3\right\}=(2.6,3)\cup (3,3.4)$