Question

The least of $x$ which satisfies the inequation $-3\le \frac{4x-3}{5}\le 3$ is

Answer 1

$-3\le \frac{4x-3}{5}\le 3$

$\frac{4x-3}{5}\ge -3$ and $\frac{4x-3}{5}\le 3$

$\frac{4x-3}{5}+3\ge 0$ $\frac{4x-3}{5}-3\le 0$

$\frac{4x-12}{5}\ge 0$ $\frac{4x-12}{5}\le 0$

$\frac{4(x-3)}{5}\ge 0$ $2x-9\le 0$

$x\in [3,\infty )$ $x\in (-\infty ,\frac{9}{2}]$

$x\in [3,\frac{9}{2}]$

Least value of $x=3$.