Question

Find the value of $y\frac{(y-3)}{12y^2}\times \frac{72y^2}{y^2-3^2}$

• A. $\frac{y^2}{y+3}$
• B. $\frac{y(y-3)}{y+3}$
• C. $\frac{6y^2}{\left(y\right)y+3}$
• D. $\frac{6y}{y+3}$

Answer 1

The correct option is D $\frac{6y}{y+3}$

$y\frac{(y-3)}{12y^2}\times \frac{72y^2}{y^2-3^2}$
We know that $a^2-b^2=(a+b)(a-b)$
So, $\frac{(y-3)}{12y}\times \frac{72y^2}{(y+3)(y-3)}$
On simplifying we get,
$=$ $\frac{6y}{y+3}$