What is (\frac{16)/(x-2)}{4/(x+1)+6/x} ?

The solution to (\frac{16)/(x-2)}{4/(x+1)+6/x} is (8x(x+1))/((x-2)(5x+3))

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simplify (\frac{16)/(x-2)}{4/(x+1)+6/x}

simplify

\frac{\frac{16}{x - 2}}{\frac{4}{x + 1} + \frac{6}{x}}

\frac{8 x ( x + 1 )}{( x - 2 ) ( 5 x + 3 )}

Solution steps

\frac{\frac{16}{x - 2}}{\frac{4}{x + 1} + \frac{6}{x}}

Apply the fraction rule:

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

= \frac{16}{( x - 2 ) ( \frac{4}{x + 1} + \frac{6}{x} )}

Simplify

(x - 2) (\frac{4}{x + 1} + \frac{6}{x}) : \frac{( x - 2 ) ( 10 x + 6 )}{x ( x + 1 )}

= \frac{16}{\frac{( x - 2 ) ( 10 x + 6 )}{x ( x + 1 )}}

Apply the fraction rule:

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

= \frac{16 x ( x + 1 )}{( x - 2 ) ( 10 x + 6 )}

Cancel

\frac{16 x ( x + 1 )}{( x - 2 ) ( 10 x + 6 )} : \frac{8 x ( x + 1 )}{( x - 2 ) ( 5 x + 3 )}

= \frac{8 x ( x + 1 )}{( x - 2 ) ( 5 x + 3 )}

\frac{15 x ^{2} + 30 x}{5 x ^{2}}

\frac{1}{0 + 1}

4^{l o g_{2} (n) + 1}

\frac{3}{2 3}

What is (\frac{16)/(x-2)}{4/(x+1)+6/x} ?
The solution to (\frac{16)/(x-2)}{4/(x+1)+6/x} is (8x(x+1))/((x-2)(5x+3))