What is (1/(y^2)-9)/(1/y+3) ?

The solution to (1/(y^2)-9)/(1/y+3) is -(3y-1)/y

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simplify (1/(y^2)-9)/(1/y+3)

simplify

\frac{\frac{1}{y ^{2}} - 9}{\frac{1}{y} + 3}

- \frac{3 y - 1}{y}

Solution steps

\frac{\frac{1}{y ^{2}} - 9}{\frac{1}{y} + 3}

Join

\frac{1}{y ^{2}} - 9 : \frac{- 9 y ^{2} + 1}{y ^{2}}

= \frac{\frac{- 9 y ^{2} + 1}{y ^{2}}}{\frac{1}{y} + 3}

Join

\frac{1}{y} + 3 : \frac{3 y + 1}{y}

= \frac{\frac{- 9 y ^{2} + 1}{y ^{2}}}{\frac{3 y + 1}{y}}

Apply the fraction rule:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{( - 9 y ^{2} + 1 ) y}{y ^{2} ( 3 y + 1 )}

Cancel

\frac{( - 9 y ^{2} + 1 ) y}{y ^{2} ( 3 y + 1 )} : - \frac{3 y - 1}{y}

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

\frac{2}{2 x - 9} + \frac{1}{2 x - 9}

\frac{1}{x + 1} + \frac{1}{x}

+ 21 6^{- \frac{4}{3}}

\frac{x ^{2} - 9 x + 8}{x ^{2} - 12 x + 32}