Question

$\frac{1-\frac{9}{y^2}}{1-\frac{3}{y}}-\frac{3}{y}$, where $(y\ne 0)=$

• A. $\frac{y-3}{y}$
• B. $\frac{y+3}{y}$
• C. $3$
• D. $1$
• E. $3y-1$

Answer 1

Answer: (D)

$1$

The value of $\frac{1-\frac{9}{y^2}}{1-\frac{3}{y}}-\frac{3}{y}$

$=\frac{y^2-9}{y(y-3)}-\frac{3}{y}$

$=\frac{(y-3)(y+3)}{y(y-3)}-\frac{3}{y}$

$=\frac{y+3}{y}-\frac{3}{y}$

$=1+\frac{3}{y}-\frac{3}{y}$

$=1$