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cuban_mind

  • 3 years ago

(x/x-3)^2-2(x/x-3)-15=0

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  1. ZeHanz
    • 3 years ago
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    If you write it in proper math notation:\[\left( \frac{ x }{ x-3 } \right)^2-2\left( \frac{ x }{ x-3 } \right)-15=0\]you see that it looks like a quadratic equation... So set p=x/(x-3) to get:\[p^2-2p-15=0\] Solve for p. Then remember that p=x/(x-3) and solve that for x!

  2. cuban_mind
    • 3 years ago
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    Yeah whenever I do that I get x=5 and x=-3 but I don't know how to go from there

  3. ZeHanz
    • 3 years ago
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    But you get p=5 and p=-3, so you still have to do solve\[\frac{ x }{ x-3 }=5\]and\[\frac{ x }{ x-3 }=-3\]I will get you started with these:\[\frac{ x }{ x-3 }=\frac{ 5 }{ 1 }\]You see two equal fractions. In general this can be dealt with the following way:\[\frac{ a }{ b }=\frac{ c }{ d } \Leftrightarrow ad=bc\]So do the multiplications and then solve for x.

  4. CR-Jennifer
    • 3 years ago
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    X=9/2 and x= 5/2

  5. ZeHanz
    • 3 years ago
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    Hold on, let me do it myself... ;)

  6. ZeHanz
    • 3 years ago
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    I've found different solutions, I'm afraid...\[x \cdot 1=5 \cdot (x-3) \Leftrightarrow x=5x-15 \Leftrightarrow 4x=15\]so x=15/4. Maybe you forgot to multiply 5 and -3?

  7. ZeHanz
    • 3 years ago
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    Second one: x*1=-3*(x-3), so x=-3x+9, 4x=9, x=9/4 So I've got 15/4 and 9/4

  8. cuban_mind
    • 3 years ago
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    ok let me see how you did it lol

  9. ZeHanz
    • 3 years ago
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    I just did...!

  10. ZeHanz
    • 3 years ago
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    Do you understand what went wrong?

  11. cuban_mind
    • 3 years ago
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    No, I don't see how you get 5/1

  12. cuban_mind
    • 3 years ago
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    oh wait! I get it nvm!

  13. ZeHanz
    • 3 years ago
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    Welcome!

  14. cuban_mind
    • 3 years ago
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    Thank you so much! I was so lost! lol

  15. ZeHanz
    • 3 years ago
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    Next time, you won't be!

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