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Решите задачу:

$\left \{ {{(x+4)(x-2)<x^{2}-3x+7 } \atop {\frac{3x+3}{2}-2\geq3x}} \right.\\\\\left \{ {{x^{2}-2x+4x-8<x^{2}-3x+7} \atop {\frac{3x+3}{2}*2-2*2\geq3x*2}} \right.\\\\\left \{ {{x^{2}+2x-x^{2}+3x<7+8} \atop {3x+3-4\geq6x }} \right.\\\\\left \{ {{5x<15} \atop {3x-6x\geq 4-3}} \right.\\\\\left \{ {{x<3} \atop {-3x\geq1 }} \right. \\\\\left \{ {{x<3} \atop {x\leq-\frac{1}{3}}} \right.\\\\Otvet:\boxed{x\in(-\infty;-\frac{1}{3}]}$

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