Question
Solve the following inequation and represent the solution set on the number line :
$4 x-19<\frac{3 x}{5}-2 \leq \frac{-2}{5}+x, x \in R$
$4 x-19<\frac{3 x}{5}-2 \leq \frac{-2}{5}+x, x \in R$
✓
Answer
$4 x-19<\frac{3 x}{5}-2 \leq \frac{-2}{5}+x, x \in R $
$ \therefore 4 x-19<\frac{3 x}{5}-2$
$4 x-\frac{3 x}{5}<-2+19$
$ \frac{17 x}{5}<17$
$x < 5,$
and $\frac{3 x}{5}-2 \leq \frac{-2}{5}+x$
$\frac{3 x}{5}-x \leq \frac{-2}{5}+2$
$-2x \leq 8$
$x \geq - 4$
$\Rightarrow - 4 ≤ x ≤ 5$
$ \therefore 4 x-19<\frac{3 x}{5}-2$
$4 x-\frac{3 x}{5}<-2+19$
$ \frac{17 x}{5}<17$
$x < 5,$
and $\frac{3 x}{5}-2 \leq \frac{-2}{5}+x$
$\frac{3 x}{5}-x \leq \frac{-2}{5}+2$
$-2x \leq 8$
$x \geq - 4$
$\Rightarrow - 4 ≤ x ≤ 5$
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