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? asked in Science & MathematicsMathematics · 5 years ago

4/(n^2+4n)=3/(n^2+5n+4)?

solve for x

4 Answers

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  • Ash
    Lv 7
    5 years ago

    4/(n² + 4n) = 3 / (n² + 5n 4)

    4/n(n+4) = 3/(n+4)(n+1)

    (n+4) cancels from both denominator, provided (n+4)≠0

    4/n = 3/(n+1)

    4(n+1) = 3n

    4n + 4 = 3n

    n = -4

    However, we just mentioned earlier that n + 4 ≠ 0, that is n ≠ -4

    So there is no solution

  • Como
    Lv 7
    5 years ago

    Not an x in sight !!!!!

    4 / [ n (n + 4) ] = 3 / [ (n + 4)(n + 1) ]

    4 [ n + 1 ] = 3 n

    4n + 4 = 3n

    n = - 4

  • Philip
    Lv 6
    5 years ago

    4/(n^2 + 4n) = 3/(n^2 + 5n + 4), ie., 4(n+1)(n+4) = 3n(n+4). For (n+4) non-zero, 4n+4 = 3n, ie.,

    (n+4) = 0. CONTRADICTION !!! Therefore NO SOLUTION EXISTS.

  • DWRead
    Lv 7
    5 years ago

    4/(n²+4n) = 3/(n²+5n+4)

    4/(n(n+4)) = 3/((n+1)(n+4))

    4/n = 3/(n+1)

    4(n+1) = 3n

    4n + 4 = 3n

    n = -4

    However, (-4)²+4(-4) = 0, and division by zero is undefined. No solution.

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