Answer
$\frac {7(y+1)}{7y+1}$
Work Step by Step
$\frac{\frac{7y^2+6y-1}{y+3}}{\frac{49y^2-1}{7y+21}}$
Side note:
$(1/2)/(3/5) = .5/.6 = 5/6$
$(1/2)/(3/5) = 1*5/2*3 = 5/6$
So, for a fraction $(a/b)/(c/d)$, where $a, b, c, d$ are non-zero,
$(a/b)/(c/d)$ = $a*d/b*c$ (end side note)
$\frac{\frac{7y^2+6y-1}{y+3}}{\frac{49y^2-1}{7y+21}}$
$\frac {7y^2+6y-1}{y+3}* \frac {7y+21}{49y^2-1}$
$\frac {(7y-1)(y+1)}{y+3}*\frac {7(y+3)}{(7y+1)(7y-1)}$
$\frac {(7y-1)(y+1)(7)(y+3)}{(y+3)(7y+1)(7y-1)}$
$\frac {(y+1)(7)(y+3)}{(y+3)(7y+1)}$
$\frac {(y+1)(7)}{(7y+1)}$
$\frac {7(y+1)}{7y+1}$
