What is (\frac{x^2-16)/(x-2)}{(x^2+3x-4)/(x-8)} ?

The solution to (\frac{x^2-16)/(x-2)}{(x^2+3x-4)/(x-8)} is ((x-4)(x-8))/((x-2)(x-1))

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simplify (\frac{x^2-16)/(x-2)}{(x^2+3x-4)/(x-8)}

simplify

\frac{\frac{x ^{2} - 16}{x - 2}}{\frac{x ^{2} + 3 x - 4}{x - 8}}

\frac{( x - 4 ) ( x - 8 )}{( x - 2 ) ( x - 1 )}

Solution steps

\frac{\frac{x ^{2} - 16}{x - 2}}{\frac{x ^{2} + 3 x - 4}{x - 8}}

Apply the fraction rule:

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

= \frac{( x ^{2} - 16 ) ( x - 8 )}{( x - 2 ) ( x ^{2} + 3 x - 4 )}

Cancel

\frac{( x ^{2} - 16 ) ( x - 8 )}{( x - 2 ) ( x ^{2} + 3 x - 4 )} : \frac{( x - 4 ) ( x - 8 )}{( x - 2 ) ( x - 1 )}

= \frac{( x - 4 ) ( x - 8 )}{( x - 2 ) ( x - 1 )}

\frac{n}{m ^{2}} + \frac{3}{mn} + \frac{2}{m}

- 6 \cdot (- 2)

\frac{\frac{5 x}{x + 3}}{\frac{x ^{2} - 8 x + 15}{x ^{2} - 9}}

\frac{a - 1}{a - 2} - \frac{a - 2}{a + 3} + \frac{1}{a - 1}

(3 x - 2 y)^{4}

What is (\frac{x^2-16)/(x-2)}{(x^2+3x-4)/(x-8)} ?
The solution to (\frac{x^2-16)/(x-2)}{(x^2+3x-4)/(x-8)} is ((x-4)(x-8))/((x-2)(x-1))