(2x)/(X+1)+(X-2)/(x+4) - addition of fractions

(2x)/(X+1)+(X-2)/(x+4) - step by step solution for the given fractions. Addition of fractions, full explanation.

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    Solution for the given fractions

    • (2*x)/(X+1) + (X-2)/(x+4) = ?
    • The common denominator of the two fractions is: (X+1)*(x+4)
    • (2*x)/(X+1) = (2*x*(x+4))/((X+1)*(x+4)) = (2*x*(x+4))/((X+1)*(x+4))
    • (X-2)/(x+4) = ((X-2)*(X+1))/((x+4)*(X+1)) = ((X-2)*(X+1))/((X+1)*(x+4))
    • Fractions adjusted to a common denominator
    • (2*x)/(X+1) + (X-2)/(x+4) = (2*x*(x+4))/((X+1)*(x+4)) + ((X-2)*(X+1))/((X+1)*(x+4))
    • (2*x*(x+4))/((X+1)*(x+4)) + ((X-2)*(X+1))/((X+1)*(x+4)) = (2*x*(x+4)+(X-2)*(X+1))/((X+1)*(x+4))
    • (2*x*(x+4)+(X-2)*(X+1))/((X+1)*(x+4)) = (2*x*(x+4)+(X-2)*(X+1))/((X+1)*(x+4))

    Solution for the given fractions

    $ \frac{(2*x)}{(X+1)} + \frac{(X-2)}{(x+4)} =? $

    The common denominator of the two fractions is: (X+1)*(x+4)

    $ \frac{(2*x)}{(X+1)} = \frac{(2*x*(x+4))}{((X+1)*(x+4))} = \frac{(2*x*(x+4))}{((X+1)*(x+4))} $

    $ \frac{(X-2)}{(x+4)} = \frac{((X-2)*(X+1))}{((x+4)*(X+1))} = \frac{((X-2)*(X+1))}{((X+1)*(x+4))} $

    Fractions adjusted to a common denominator

    $ \frac{(2*x)}{(X+1)} + \frac{(X-2)}{(x+4)} = \frac{(2*x*(x+4))}{((X+1)*(x+4))} + \frac{((X-2)*(X+1))}{((X+1)*(x+4))} $

    $ \frac{(2*x*(x+4))}{((X+1)*(x+4))} + \frac{((X-2)*(X+1))}{((X+1)*(x+4))} = \frac{(2*x*(x+4)+(X-2)*(X+1))}{((X+1)*(x+4))} $

    $ \frac{(2*x*(x+4)+(X-2)*(X+1))}{((X+1)*(x+4))} = \frac{(2*x*(x+4)+(X-2)*(X+1))}{((X+1)*(x+4))} $

    $ $

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