What is (\frac{x^2-1)/(9x)}{(x^2+7x-8)/(4x)} ?

The solution to (\frac{x^2-1)/(9x)}{(x^2+7x-8)/(4x)} is (4(x+1))/(9(x+8))

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simplify (\frac{x^2-1)/(9x)}{(x^2+7x-8)/(4x)}

simplify

\frac{\frac{x ^{2} - 1}{9 x}}{\frac{x ^{2} + 7 x - 8}{4 x}}

\frac{4 ( x + 1 )}{9 ( x + 8 )}

Solution steps

\frac{\frac{x ^{2} - 1}{9 x}}{\frac{x ^{2} + 7 x - 8}{4 x}}

Apply the fraction rule:

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

= \frac{( x ^{2} - 1 ) \cdot 4 x}{9 x ( x ^{2} + 7 x - 8 )}

Cancel the common factor:

x

= \frac{( x ^{2} - 1 ) \cdot 4}{9 ( x ^{2} + 7 x - 8 )}

Factor

x^{2} - 1 : (x + 1) (x - 1)

= \frac{( x + 1 ) ( x - 1 ) \cdot 4}{9 ( x ^{2} + 7 x - 8 )}

Factor

x^{2} + 7 x - 8 : (x - 1) (x + 8)

= \frac{( x + 1 ) ( x - 1 ) \cdot 4}{9 ( x - 1 ) ( x + 8 )}

Cancel the common factor:

x - 1

= \frac{( x + 1 ) \cdot 4}{9 ( x + 8 )}

= \frac{4 ( x + 1 )}{9 ( x + 8 )}

\frac{1}{2} x^{3}

\frac{7 - 17 i}{2 - 3 i}

\frac{1}{x - 1}

\frac{2}{3} \times 5

(\frac{16}{49})^{- \frac{1}{2}}

What is (\frac{x^2-1)/(9x)}{(x^2+7x-8)/(4x)} ?
The solution to (\frac{x^2-1)/(9x)}{(x^2+7x-8)/(4x)} is (4(x+1))/(9(x+8))