Odpowiedź :
[tex]zad.a\\\\-6x-4\leq -13\\\\-6x\leq -13+4\\\\-6x\leq -9~~\mid \div (-6)\\\\x\geq 1,5 ~~\Rightarrow ~~x\in <1,5 ; + \infty )\\\\zad.b\\\\3(x-1)\geq x+5\\\\3x-3\geq x+5\\\\3x-x\geq 5+3\\\\2x\geq 8~~\mid \div 2\\\\x\geq 4~~\Rightarrow ~~x\in <4 ; + \infty )\\\\zad.c\\\\\dfrac{x-3}{2} <\dfrac{x+2}{3} ~~\mid \cdot 6\\\\3(x-3)<2(x+2)\\\\3x-9<2x+4\\\\3x-2x<4+9\\\\x<13~~\Rightarrow ~~x\in (-\infty ;13)\\\\zad.4\\\\3(2-\frac{1}{6} x)\geq -0,5\\\\6-\frac{1}{2} x\geq -\frac{1}{2}[/tex]
[tex]-\frac{1}{2} x\geq -\frac{1}{2} -6\\\\-\frac{1}{2} x\geq -6\frac{1}{2} \\\\-\frac{1}{2} x\geq -\frac{13}{2} ~~\mid \div (-\frac{1}{2} )\\\\x\leq 13 ~~\Rightarrow ~~x\in (-\infty ;13)[/tex]