Proceed as for an equation (Attention: the inequality is reversed, if the two members are multiplied by a negative number !!) Or: Reduce to $ax + b $ in the first member, $ 0 $ in the 2nd member, then use the discussion of the sign of$ T = a x + b $ ------------------------------- If $ a \lt 0 $: $ T \lt 0 $ for $ x \gt \frac {-b} {a} $ $ T \gt 0 $ for $ x \lt \frac {-b} {a} $ ------------------------------- If $ a \gt 0 $ :: $ T \gt 0 $ for $ x \gt \frac {-b} {a} $ $ T \lt 0 $ for $ x \lt \frac {-b} {a} $ ------------------------------- If $ a = 0 $ :: $ T $ has the sign of $ b $ -------------------------------

Got it !

Solve:

$\LARGE 8x-6\gt 5+7x$	$\LARGE x\gt11$, $\LARGE S=]11,+\infty [$
$\LARGE 12-5x \gt x-60$	$\LARGE x\lt12$, $\LARGE S=]-\infty,12 [$
$\LARGE \frac{x}{2}+ 4 \gt \frac{2x}{3}- \frac{x}{8}$	$\LARGE x\lt96$, $\LARGE S=]-\infty,96 [$
$\LARGE 3-4(5-x) \leq 2x+5$	$\LARGE S=]-\infty,11]$
$\LARGE \frac{x-2}{3}- \frac{1-x}{3}\geq 0$	$\LARGE S = ]\frac{3}{2}.+\infty[ $
$\LARGE \frac{x}{3} -\frac{4-x}{4}\gt 5$	>$\LARGE S = ]\frac{72}{7}.+\infty[ $
$\LARGE 2(x+1)\lt 3+2x$	$\LARGE S = \mathbb R $
$\LARGE 3(\frac{x}{2}-1)\gt \frac{3}{2}x- \frac{7}{3}$	$\LARGE S = \emptyset$