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anonymous

  • 5 years ago

Find the solution set to each equation. x+7/x+4 = x+1/x-2 Please help!!

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  1. anonymous
    • 5 years ago
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    There are no real solutions for x.

  2. anonymous
    • 5 years ago
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    Oh this equation is no solution? cause it'd be x+7=0/x+4=0 right

  3. anonymous
    • 5 years ago
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    Yes. well, which class are u and where are u studying?

  4. anonymous
    • 5 years ago
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    West Coast University. It's called Algebra for college students. I have this huge test at 7am :(

  5. anonymous
    • 5 years ago
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    All the very best.

  6. anonymous
    • 5 years ago
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    I have another question, have time? & thanks!

  7. anonymous
    • 5 years ago
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    Please ask. If I acn, I will help u.

  8. anonymous
    • 5 years ago
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    It's really easy but I keep messing up on one part, & just need someone to explain it. Reduce to the lowest terms. (2-x)(x+1)(x+3) I know for the (x+1)(x+3) it is x^2+3x+1x+3

  9. anonymous
    • 5 years ago
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    After that I just get stuck and mess up my #'s so do I distribute the (2-x) to the original or to x^2+3x+1x+3

  10. anonymous
    • 5 years ago
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    Answer= -x^3 - 2x^2 + 5x - 6 (After u get x^2+4x+3, multiply this by 2-x)

  11. anonymous
    • 5 years ago
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    \[(2-x)(x+1)(x+3)=(2-x)(x^2+4x+3)=2x^2+8x+6-x^3-4x^2-3x=-x^3-2x^2+5x+6\]

  12. anonymous
    • 5 years ago
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    oh okay so my question do I start off with 2 for example 2(x^2)+2(4x)+2(5x)-2(6) get what I'm saying, that part totally throws me off, I know it's easy but I'm stuck :/

  13. anonymous
    • 5 years ago
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    I got it!

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