Question

Solve the inequation:
$\frac{5x+1}{7}-4\left(x+\frac{2}{5}\right)\le 1\frac{3}{5}+\frac{3x-1}{7},x\in R$

Answer 1

Solution

$\frac{5x+1}{7}-4(x+\frac{2}{5})\le 1\frac{3}{5}+\frac{3x-1}{7},xϵR$$

$\frac{5x+1}{7}-\left(\frac{9x+8}{5}\right)\le \frac{8}{5}+\frac{3x-1}{7}$

$\frac{5x+1}{7}-\frac{(3x-1)}{7}\le \frac{8}{5}+(4x+\frac{8}{5})$

$\frac{2x+2}{7}\le \frac{16}{5}+4x$

$O\le \frac{4x+16}{5}-\frac{(2x+2)}{7}$

$O\le 140x+112-10x-10$

$O\le 130x+102$

$x\ge \frac{-102}{130}$

$x\ge \frac{-51}{65}$