Почему никто не помогает??????????????????

$1)\\ |x+3|-3x \leq 14-|2-x|\\ \left \{ {{x+3 \geq 0} \atop {2-x \geq 0}} \right. \\ \left \{ {{x \geq -3} \atop {x \leq 2}} \right. \\\\ x+3-3x \leq 14-2+x\\ 3-2x \leq 12+x\\ -3x \leq 9\\ x \geq -3$

2}} \right. \\\\ -(x-3)-3x \leq 14+(2-x)\\ x \neq |-\frac{13}{3};+oo)" alt=" \left \{ {{x+3<0} \atop {2-x<0}} \right. \\ \left \{ {{x<-3} \atop {x>2}} \right. \\\\ -(x-3)-3x \leq 14+(2-x)\\ x \neq |-\frac{13}{3};+oo)" align="absmiddle" class="latex-formula">
Решение первой системы будет
$|-3;+oo)$

0}} \right. \\ \left \{ {{-4x^2-5x+21 \leq 0} \atop {x<0}} \right. " alt="x^2-3x+1-\frac{x^3+x^2+3x-21}{x} \geq 3\\ \frac{-4x^2-2x+21}{x} \geq 3\\ \frac{-4x^2-2x+21-3x}{x} \geq 0\\ x \neq 0\\ \left \{ {{-4x^2-5x+21 \geq 0} \atop {x>0}} \right. \\ \left \{ {{-4x^2-5x+21 \leq 0} \atop {x<0}} \right. " align="absmiddle" class="latex-formula">

решение первой системы

\\ -3 \ \ \ \frac{7}{4}\\\\ x >0\\ (-oo;-3] \ U \ (0;\frac{7}{4}] \\\\ " alt="-4x^2-5x+21 \geq 0\\ D=25+4*4*21=19^2\\ x=-3\\ x=\frac{7}{4}\\ ---------------->\\ -3 \ \ \ \frac{7}{4}\\\\ x >0\\ (-oo;-3] \ U \ (0;\frac{7}{4}] \\\\ " align="absmiddle" class="latex-formula">

Объединяя решения
$-3$ $(0;\frac{7}{4}]$