Question

rewrite the expression completely.
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\\(\frac{\frac{y^2 - 9}{y - 3}}{\frac{y^2 - 9}{y + 3}}\\)
\\(\frac{y^2 - 9}{y - 3} cdot \frac{y + 3}{y^2 - 9}\\)

Explanation:

Step1: Factor difference of squares

$y^2 - 9 = (y-3)(y+3)$

Step2: Substitute factored forms

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

Step3: Cancel common factors

Cancel $(y-3)$ in first fraction, cancel $(y+3)$ across fractions:
$\frac{\cancel{(y-3)}(y+3)}{\cancel{y-3}} \cdot \frac{y+3}{(y-3)\cancel{(y+3)}} = \frac{y+3}{y-3}$
(Note: $y
eq 3, -3$ to avoid division by zero)

Answer:

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