Решите пжлстЗ) и К)!!!!!!!!!!!!!!!!!!!!!!!!!

З) $\frac{7x-3}{6}-1+ \frac{2x+4}{3}=0$
$\frac{7x-3}{6}-1+ \frac{2x+4}{3}=0$
$\frac{7x-3-6+4x+8}{6}=0$
$\frac{11x-1}{6}=0$ |·6
$11x-1=0$
   $11x=1$
$x= \frac{1}{11}$

ж) $27y(5y-3)+2y=9y(15y-7)+6,4$
   $135y^{2}-81y+2y=135 y^{2}-63y+6,4$
   $135 y^{2}-81y+2y-135 y^{2}+63y-6,4=0$
   $-16y-6,4=0$
   $-16y=6,4$
   $y= -\frac{6,4}{16}$
   $y=6 \frac{4}{10} : (-\frac{16}{1})$
   $y=\frac{64}{10}$ · $( -\frac{1}{16} )$
   $y=- \frac{4}{10}$
   $y=- \frac{2}{5}$

Я не уверена в правильности решения.