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JacksonJRB

  • one year ago

Solve for x in terms of y. Please help!

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  1. JacksonJRB
    • one year ago
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    \[y=\sqrt{\frac{ 4x-1 }{ x }}\]

  2. JacksonJRB
    • one year ago
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    Every new problem I look at seems to make less and less sense.

  3. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle y^2=\frac{4x-1}{x} }\) \(\large\color{black}{ \displaystyle y^2=\frac{4x}{x}-\frac{1}{x} }\) \(\large\color{black}{ \displaystyle y^2=4-\frac{1}{x} }\) can you take it from there?

  4. SolomonZelman
    • one year ago
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    all they are asking you to do, is to isolate the x (without performing incorrect operations).

  5. JacksonJRB
    • one year ago
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    \[x=\frac{ -1 }{ y^2-4 }\]

  6. SolomonZelman
    • one year ago
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    yes, and if you want you can simplify that just a bit, to get: \(\large\color{black}{ \displaystyle x=\frac{1}{4-y^2} }\)

  7. JacksonJRB
    • one year ago
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    Ah, thanks again

  8. SolomonZelman
    • one year ago
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    You are always welcome

  9. JacksonJRB
    • one year ago
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    There is another one that I will try on my own for now but I'll ask if I need help again

  10. SolomonZelman
    • one year ago
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    Ok:)

  11. JacksonJRB
    • one year ago
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    \[y=\frac{ 1 }{ x^2+2x }\] \[x^2+2x=\frac{ 1 }{ y }\]

  12. SolomonZelman
    • one year ago
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    the second equation is the next step that you took?

  13. JacksonJRB
    • one year ago
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    Yes

  14. SolomonZelman
    • one year ago
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    Ok, you can subtract -1/y from both sides, and do the quadratic formula. a=1 b=2 c=1/y

  15. SolomonZelman
    • one year ago
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    i mean c=-1/y

  16. JacksonJRB
    • one year ago
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    Ah, ok

  17. JacksonJRB
    • one year ago
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    ill try that

  18. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle x^2+2x\color{red}{-\frac{1}{y}}=\frac{1}{y}\color{red}{-\frac{1}{y}} }\) \(\large\color{black}{ \displaystyle x^2+2x-\frac{1}{y}=0 }\)

  19. SolomonZelman
    • one year ago
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    go on... :)

  20. JacksonJRB
    • one year ago
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    \[x=\frac{ -2+/-\sqrt{4+\frac{ 4 }{ y }} }{ }\]

  21. JacksonJRB
    • one year ago
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    over 2

  22. SolomonZelman
    • one year ago
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    you can do a \(\pm\) sign in latex by \pm (just a code advise) and yes, so far correct

  23. JacksonJRB
    • one year ago
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    \[x=-2\pm \sqrt{4y+4}\]

  24. JacksonJRB
    • one year ago
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    over 2 again

  25. SolomonZelman
    • one year ago
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    i don't think I followed you this time. How come your y came from denominator to numerator?

  26. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle x=\frac{-2\pm\sqrt{4-4(1)(-\frac{1}{y})}}{2} }\) \(\large\color{black}{ \displaystyle x=\frac{-2\pm\sqrt{4+4(\frac{1}{y})}}{2} }\) it should be like this

  27. JacksonJRB
    • one year ago
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    ah

  28. SolomonZelman
    • one year ago
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    it can be simplified ....

  29. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle x=\frac{-2\pm\sqrt{4+4(\frac{1}{y})}}{2} }\) \(\large\color{black}{ \displaystyle x=\frac{-2\pm2\sqrt{1+(\frac{1}{y})}}{2} }\) \(\large\color{black}{ \displaystyle x=-1\pm\sqrt{1+\frac{1}{y}} }\)

  30. SolomonZelman
    • one year ago
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    this is it, technically, although, if you feel like or need to you can rationalize the denominator

  31. JacksonJRB
    • one year ago
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    Don't think do

  32. JacksonJRB
    • one year ago
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    Thank you so much

  33. JacksonJRB
    • one year ago
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    *think so

  34. SolomonZelman
    • one year ago
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    yw

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