Помогите решить систему, пожалуйста

$|y-3|-3|y|-2(y+1)^2 \geq 1\\ y \geq 3\\ y \geq 0\\\\ 1)y\in(0;3)\\ 3-y-3y-2(y+1)^2 \geq 1\\ 3-4y-2(y+1)^2 \geq 1\\ -4y-2(y+1)^2 \geq -2\\ 2y+(y+1)^2 \leq 1\\ y^2+4y \leq 0\\ y(y+4) \leq 0\\ y\in[-4;0]\\\\ 2)\\ y\in[3;\infty)\\ y-3-3y-2(y+1)^2 \geq 1\\ -2y-3-2(y^2+2y+1) \geq 1\\ -2y-2y^2-4y-2 \geq 4\\ -2y^2-6y-6 \geq 0\\ y \neq D\\\\ y<3\\ y<0\\\\ 3)\\ y\in( -\infty;0) \\ 3-y+3y-2(y+1)^2 \geq 1\\ 3+2y-2(y^2+2y+1) \geq 1\\ y-(y^2+2y+1) \geq -1\\ -y^2-y \geq 0\\ y\in[0;-1]$
Объединяя получим $y\in[-1;0]$ .

$4^{|x^2-8x+12|-log_{4}7}=7^{2y-1}\\ 4^{|x^2-8x+12|}=49^{y} \\\\ x^2-8x+12 \geq 0\\ x\in(-\infty;-2 ] \cup [ -6 ; \infty) \\ y=(x^2-8x+12)*log_{7}2\\\\ x^2-8x+12<0\\ x\in(2;6)\\ y=-(x^2-8x+12)*log_{7}2\\\\ \left \{ {{(x^2-8x+12)*log_{7}2 \geq -1} \atop {(x^2-8x+12)*log_{7}2 \leq 0}} \right.\\\\ \left \{ {{-(x^2-8x+12)*log_{7}2 \geq -1} \atop {-(x^2-8x+12)*log_{7}2 \leq 0}} \right.\\\\$
решая получаем
решения $x=2;y=0\\ x=6;y=0$