1/7(7/8y+7)-3/4(2/9y+1 7/9)=1/2

Решите задачу:

$\frac{1}{7}( \frac{7}{8}y+7)- \frac{3}{4}( \frac{2}{9}y+1 \frac{7}{9})= \frac{1}{2}$
$\frac{1}{8}y+1 - \frac{3}{4}( \frac{2}{9}y+\frac{16}{9})= \frac{1}{2}$
$\frac{1}{8}y+1 - \frac{1}{6}y-\frac{4}{3}= \frac{1}{2}$
$\frac{1}{8}y - \frac{1}{6}y= \frac{1}{2}-1+\frac{4}{3}$
$- \frac{1}{4}y = \frac{5}{6}$
$y = \frac{5}{6}:(- \frac{1}{24})$
$y = \frac{5}{6}*(- \frac{24}{1})$
$x=-20$