$$\frac{(t)[17(2-y)-5(y+12)]}{1-7y}=8$$
$t=\frac{4\left(7y-1\right)}{11y+13}$
$y\neq -\frac{13}{11}\text{ and }y\neq \frac{1}{7}$
$y=\frac{13t+4}{28-11t}$
$t\neq 0\text{ and }t\neq \frac{28}{11}$
