y3 + 5 - y9=1/6
do you mean:
$\displaystyle 3y+5 -9y = \frac{1}{6}$
if so:
$\displaystyle 3y+5 -9y = 5 - 6y = \frac{1}{6} $
$\displaystyle 6y = 5-\frac{1}{6} = \frac{30}{6}-\frac{1}{6} = \frac{29}{6}$
$\displaystyle y = \frac{29}{36}$
OR do you mean:
$\displaystyle y^3+5-y^9 = \frac{1}{6}$
$\displaystyle \frac{y}{3}+5-\frac{y}{9} = \frac{1}{6}$
$\displaystyle 3y+5-y = \frac{9}{6}$ (multiply both sides by 9)
$\displaystyle 2y+5 = \frac{3}{2}$ (simplify)
$\displaystyle 2y = \frac{3}{2}-5$ (subtract 5 from both sides)
$\displaystyle 2y = \frac{-7}{2}$ (simplify)
$\displaystyle y = \frac{-7}{4} $(divide both sides by 2)