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Thread: Find the solution

  1. #1
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    Find the solution

    Find the solution in R :

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  2. #2
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    Quote Originally Posted by dapore View Post
    Find the solution in R :

    I think you can see from the second equation that

    $\displaystyle y = 4 - x$.

    Substitute into the first equation:

    $\displaystyle x^3 + (4 - x)^3 = 28$

    $\displaystyle x^3 + 64 - 48x + 12x^2 - x^3 = 28$

    $\displaystyle 12x^2 - 48x + 36 = 0$

    $\displaystyle x^2 - 4x + 3 = 0$

    $\displaystyle (x - 1)(x - 3) = 0$

    $\displaystyle x = 1$ or $\displaystyle x = 3$.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    I think you can see from the second equation that

    $\displaystyle y = 4 - x$.

    Substitute into the first equation:

    $\displaystyle x^3 + (4 - x)^3 = 28$

    $\displaystyle x^3 + 64 - 48x + 12x^2 - x^3 = 28$

    $\displaystyle 12x^2 - 48x + 36 = 0$

    $\displaystyle x^2 - 4x + 3 = 0$

    $\displaystyle (x - 1)(x - 3) = 0$

    $\displaystyle x = 1$ or $\displaystyle x = 3$.
    x=1, y=4-1=3
    x=3, y=4-3=1
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  4. #4
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    Hello, dapore!

    Here is another approach.
    It's no shorter, but it's rather "cute".


    Find the solution in R: .$\displaystyle \begin{array}{cccc}x^3 + y^3 &=& 28 & [1] \\ x + y &=& 4 & [2] \end{array}$

    $\displaystyle \text{Cube [2]: }\;(x+y)^3 \:=\:4^3 \quad\Rightarrow\quad x^3 + 3x^2y + 3xy^2 + y^3 \:=\:64$

    $\displaystyle \text{and we have: }\;\underbrace{x^3 + y^3}_{\text{This is 28}} + 3xy\underbrace{(x+y)}_{\text{This is 4}} \:=\:64 \quad\Rightarrow\quad 28 + 3xy(4) \:=\:64$

    . . . . . . . . . $\displaystyle 12xy \:=\:36 \quad\Rightarrow\quad y \:=\:\tfrac{3}{x}\;\;[3]$


    Substitute into [2]: .$\displaystyle x + \tfrac{3}{x} \:=\:4 \quad\Rightarrow\quad x^2 - 4x + 3 \:=\:0$

    . . . . . . $\displaystyle (x-1)(x-3) \:=\:0 \quad\Rightarrow\quad x \:=\:1,\:3$


    Substitute into [3]: .$\displaystyle y \:=\:3,\:1$


    Solutions: .$\displaystyle (x,y) \:=\1,3),\3,1)$

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  5. #5
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    Thank you my friends
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  6. #6
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    $\displaystyle x^{3}+y^{3}=(x+y)\left( (x+y)^{2}-3xy \right)=4(16-3xy)=28,$ thus $\displaystyle xy=3.$ (1)

    Now put the second equation into (1) and get $\displaystyle 4x-x^2=3\implies x^2-4x+3=(x-1)(x-3)=0.$

    The solutions are $\displaystyle (x,y)=(1,3)$ and $\displaystyle (x,y)=(3,1).$
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