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Math Help - Solving a system involving products

  1. #1
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    Solving a system involving products

    The following is from a list of problems that, as far as I know, all have solutions. Unfortunately, I don't have a list of solutions.

    Solve the following system, where x > y:

    x + y + xy = 19

    xy(x + y) = 84

    The answer is x=4, y=3. I don't know if there is another solution, and I got it simply by inspection - is there an algebraic method to get the answer (or a reason why it isn't algebraically solvable)?

    Thanks!
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  2. #2
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    Hello, Mathnasium!

    I found an approach.
    Don't know if it's any better than Brute Force algebra.


    Solve the following system, where x > y:

    \begin{array}{ccc}x + y + xy &=& 19 \\ xy(x + y) &=& 84\end{array}
    Let a\:=\:x+y,\;\;b \:=\:xy

    Then we have: . \begin{array}{cccc}a + b &=& 19 &{\color{blue}[1]} \\ ab &=& 84 &{\color{blue}[2]} \end{array}

    From [1], we have: .  b \:=\:19-a

    Substitute into [2]: . a(19-a)\:=\:84\quad\Rightarrow\quad a^2-19a+84\:=\:0

    Hence: . (a-7)(a-12)\:=\:0\quad\Rightarrow\quad a \:=\:7,\,12 \quad\Rightarrow\quad b \:=\:12,\,7


    If a=7,\:b = 12, we have: . \begin{array}{ccc}x+y &=&7\\ xy &=&12 \end{array}

    . . which has the solution: . \boxed{x = 4,\:y = 3}


    If a= 12,\:b=7, we have: . \begin{array}{ccc}x+y &=&12 \\ xy&=& 7\end{array}

    . .which has the solution: . \boxed{x \:=\:6+\sqrt{29},\;\;y \:=\:6-\sqrt{29}}

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