Answer
$y=4$
Work Step by Step
$ y(y+8)=16(y-1)\qquad$...apply the distributive rule: $m(n\pm c)=mn\pm mc$
$ y^{2}+8y=16y-16\qquad$...add $(-16y+16)$ to both sides
$ y^{2}-8y+16=0\qquad$...recognize a perfect square trinomial.
$(A\pm B)^{2}=A^{2}\pm 2AB+B^{2}$
$(y-4)^{2}=0$
$(y-4)(y-4)=0\qquad$...apply the principle of zero products.
$y=4$
Type the equation into a graphing utility and see
that the graph intercepts the x-axis at $4$.
