Verified answer

[tex]\displaystyle 1)\frac{1}{3}x > 2\\ \\3*\frac{1}{3}x > 3*2\\ \\x > 6=x \in (6, + \infty) \\\\2)2-7x > 0\\-7x > -2\\\\x < \frac{2}{7}= x \in (- \infty,\frac{2}{7})\\ \\3)6(y-1,5)-3,4 > 4y-2,4\\ 6y-9-3,4 > 4y-2,4\\6y-12,4 > 4y-2,4\\6y-4y > -2,4+12,4\\2y > 10\\y > 5 = y \in (5, + \infty)[/tex]