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anonymous

  • one year ago

----♥Algebra 2 Help♥------ ♥How do I add these rational expressions O.o♥ (っ◔◡◔)っ (っ◔◡◔)っ (っ◔◡◔)っ (っ◔◡◔)っ

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  1. anonymous
    • one year ago
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  2. kobeni-chan
    • one year ago
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    I believe that you'll be using cross multiplication and the distributive property :)

  3. anonymous
    • one year ago
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    So 2x times 2x? and x+4 times x+1?

  4. kobeni-chan
    • one year ago
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    Yep

  5. anonymous
    • one year ago
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    I'm gonna solve that and show you c:

  6. kobeni-chan
    • one year ago
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    Okies ^_^

  7. kobeni-chan
    • one year ago
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    Oh wait...

  8. anonymous
    • one year ago
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    I looked up the answer on Wolfram and it looks different O.o

  9. anonymous
    • one year ago
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  10. kobeni-chan
    • one year ago
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    I forgot that you use cross multiplication to see if fractions are equal... >_< Since the numerator on the first fraction and the denominator on the second fraction are both 2x, I think you need to cancel them out.

  11. anonymous
    • one year ago
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    thats the answer, but i need to explain in sentcnces how i got there and ok c:

  12. anonymous
    • one year ago
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    the image above your reply is the answer

  13. kobeni-chan
    • one year ago
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    Ah ok. When I cross multiplied, I got 4x^2 for the first num. and second denom., and x^2 + 5x + 4 when multiplying the other values. It looks like in the answer, they added the two to get the answer's numerator.

  14. anonymous
    • one year ago
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    How would they get whats on the numerator?

  15. kobeni-chan
    • one year ago
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    I'm looking at this right now: http://www.purplemath.com/modules/rtnladd.htm About halfway down the page, under 'The process works similarly for rationals.', I think its talking about this kind of thing.

  16. anonymous
    • one year ago
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    ok i opened up the page, just waiting for it to load c:

  17. anonymous
    • one year ago
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    This is confusing .-. :x

  18. kobeni-chan
    • one year ago
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    Ok :) What I figured out so far is that in the answer, the denominator is 2x(x+4), which is the common denominator for 2x and x+4. You multiply x+4 (denom. of 1st fraction) by 2x, and do the same to its numerator. Then, you'll multiply 2x (denom. of 2nd fraction) by (x+4) (without distributing) and do that to the numerator too. Does that makes sense maybe?

  19. anonymous
    • one year ago
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    the denomator looks easy but i dont get how the numerator came out like that

  20. kobeni-chan
    • one year ago
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    Once the denominators are the same, you can just add the numerators like a regular adding problem. Like, 3/5 + 1/5 = 4/5. The numerator's just complicated-looking because there was multiplication in the numerators when making a common denominator.

  21. anonymous
    • one year ago
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    the message befor this one you said to add the denomator with the numerator?

  22. anonymous
    • one year ago
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    2x(x+4)

  23. kobeni-chan
    • one year ago
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    Well you're not adding the denominator to the numerator, if that's what you meant. Here: with the first fraction, you have 2x/(x+4). To make that denominator into 2x(x+4), you need to multiply the denominator by 2x. This will look like 2x(x+4). Since you multiplied a value in the denominator, you need to do the same to the numerator. So what will the first fraction be?

  24. anonymous
    • one year ago
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    2x(x+1) or would it be (x+1)2x

  25. kobeni-chan
    • one year ago
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    Well we're just looking at the first fraction for now. You need to multiply 2x (numerator of 1st fraction) by 2x, since we multiplied that fraction's denominator by it.

  26. anonymous
    • one year ago
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    this is making my mind hurt lol Ill get help on here tommorow c: or monday

  27. kobeni-chan
    • one year ago
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    Oh I'm sorry :( ok I hope you feel better and have better luck with someone else! lol

  28. anonymous
    • one year ago
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    Its ok algebra 2 is mind tripping sometimes lol

  29. kobeni-chan
    • one year ago
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    Lol I know that feel. I'm cramming to get work done by Wednesday ;-;

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