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kitsune0724

how do you solve 3/x+1 +2/x+3 =2? Please show me step by step. Thank you.

  • one year ago
  • one year ago

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  1. Calcmathlete
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    \[\text{Is this the question?}\frac3{x + 1} + \frac{2}{x + 3} = 2\]

    • one year ago
  2. kitsune0724
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    yes please.

    • one year ago
  3. Calcmathlete
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    \[\frac3{x + 1} + \frac{2}{x + 3} = 2\]\[\frac{3}{x + 1} \times \frac{x + 3}{x + 3} + \frac{2}{x + 3} \times \frac{x + 1}{x + 1} = \frac{2}{1} \times \frac{(x + 1)(x + 3)}{(x + 1)(x + 3)}\]\[\frac{3(x + 3) + 2(x + 1)}{x^{2} + 4x + 3} = \frac{ 2(x^{2} + 4x + 3)}{x^{2} + 4x + 3}\]You can just drop the denominator now. \[3(x + 3) + 2(x + 1) = 2(x^{2} + 4x + 3)\]Can you solve it now?

    • one year ago
  4. kitsune0724
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    let me see...

    • one year ago
  5. Calcmathlete
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    Did you get it yet?

    • one year ago
  6. kitsune0724
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    the next step please?

    • one year ago
  7. Calcmathlete
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    \[3x + 9 + 2x + 2 = 2x^{2} + 8x + 6\]\[2x^{2} + 3x - 5 = 0\]

    • one year ago
  8. Calcmathlete
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    Can you solve the quadratic?

    • one year ago
  9. kitsune0724
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    yes.

    • one year ago
  10. Calcmathlete
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    What did you get?

    • one year ago
  11. kitsune0724
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    i'm still doing it.

    • one year ago
  12. kitsune0724
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    \[3\pm i \sqrt{31} \over 4\] is that right?

    • one year ago
  13. Calcmathlete
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    No. \[3x + 9 + 2x + 2 = 2x^{2} + 8x + 6\]\[2x^{2} + 3x - 5 = 0\]\[2x^{2} -2x + 5x - 5 = 0\]\[2x(x - 1) + 5(x - 1) = 0\]\[(2x + 5)(x - 1) = 0\]\[x = -\frac{5}{2}, 1\]

    • one year ago
  14. Calcmathlete
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    Do you see how I got that?

    • one year ago
  15. kitsune0724
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    yes.

    • one year ago
  16. kitsune0724
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    thank you for your help.

    • one year ago
  17. Calcmathlete
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    np :)

    • one year ago
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