(8)/(x)+(9)/(2x+1) - add fractions

(8)/(x)+(9)/(2x+1) - step by step solution for the given fractions. Add fractions, full explanation.

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    Solution for the given fractions

    • 8/x + 9/(2*x+1) = ?
    • The common denominator of the two fractions is: x*(2*x+1)
    • 8/x = (8*(2*x+1))/(x*(2*x+1)) = (8*(2*x+1))/(x*(2*x+1))
    • 9/(2*x+1) = (9*x)/(x*(2*x+1)) = (9*x)/(x*(2*x+1))
    • Fractions adjusted to a common denominator
    • 8/x + 9/(2*x+1) = (8*(2*x+1))/(x*(2*x+1)) + (9*x)/(x*(2*x+1))
    • (8*(2*x+1))/(x*(2*x+1)) + (9*x)/(x*(2*x+1)) = (8*(2*x+1)+9*x)/(x*(2*x+1))
    • (8*(2*x+1)+9*x)/(x*(2*x+1)) = (8*(2*x+1)+9*x)/(x*(2*x+1))

    Solution for the given fractions

    $ \frac{8}{x }+\frac{ 9}{(2*x+1)} =? $

    The common denominator of the two fractions is: x*(2*x+1)

    $ \frac{8}{x }= \frac{(8*(2*x+1))}{(x*(2*x+1))} = \frac{(8*(2*x+1))}{(x*(2*x+1))} $

    $ \frac{9}{(2*x+1)} = \frac{(9*x)}{(x*(2*x+1))} = \frac{(9*x)}{(x*(2*x+1))} $

    Fractions adjusted to a common denominator

    $ \frac{8}{x }+\frac{ 9}{(2*x+1)} = \frac{(8*(2*x+1))}{(x*(2*x+1))} + \frac{(9*x)}{(x*(2*x+1))} $

    $ \frac{(8*(2*x+1))}{(x*(2*x+1))} + \frac{(9*x)}{(x*(2*x+1))} = \frac{(8*(2*x+1)+9*x)}{(x*(2*x+1))} $

    $ \frac{(8*(2*x+1)+9*x)}{(x*(2*x+1))} = \frac{(8*(2*x+1)+9*x)}{(x*(2*x+1))} $

    $ $

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