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FeoNeo33

  • 2 years ago

Rational Expressions and Functions: http://i47.tinypic.com/sbua04.png

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  1. nbouscal
    • 2 years ago
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    I would start by multiplying both sides of the equation by 3x+9 and see what happens :)

  2. FeoNeo33
    • 2 years ago
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    Thanks, I'll see if that gets me the answer.

  3. ParthKohli
    • 2 years ago
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    I have a suggestion here. Shall I spit it out?

  4. FeoNeo33
    • 2 years ago
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    Nah, don't work, it's a fill-in answer. @ParthKohli Sure.

  5. ParthKohli
    • 2 years ago
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    Multiply both the numerator and denominator of \(\Large {2 \over x + 3}\).That'll get you two like terms :)

  6. ParthKohli
    • 2 years ago
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    Multiply both numerator and denominator by 3*

  7. FeoNeo33
    • 2 years ago
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    okay, bottom is 9, top is 6. Is there another step?

  8. FeoNeo33
    • 2 years ago
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    I do see the nine and nine, like terms.

  9. ParthKohli
    • 2 years ago
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    \(\color{Black}{\Rightarrow \Large{1 \over 3x + 9}- {3(2) \over 3(x + 3)} }\)

  10. FeoNeo33
    • 2 years ago
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    I see, but still confusing.

  11. ParthKohli
    • 2 years ago
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    I don't find anything confusing here. \(\color{Black}{\Rightarrow \Large {x \over y} = {ax \over ay}}\) This is what I am doing here.

  12. FeoNeo33
    • 2 years ago
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    I know, but this is one of the first units, I'm really, for some reason, having trouble with. Okay, you say, 'Multiply both numerator and denominator by 3*', Okay then what do I do?

  13. ParthKohli
    • 2 years ago
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    You'll get two like terms :)

  14. ParthKohli
    • 2 years ago
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    First simplify the thing I gave you

  15. FeoNeo33
    • 2 years ago
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    \[1/3x + 9 - 6 / x9\] ^^Don't know how to really format it, but the 6 and 9 is there.

  16. ParthKohli
    • 2 years ago
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    No no

  17. FeoNeo33
    • 2 years ago
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    Or am I missing something

  18. ParthKohli
    • 2 years ago
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    \(\color{Black}{\Rightarrow 3(x + 3) = 3x + 9 }\)

  19. FeoNeo33
    • 2 years ago
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    okay look, we both are going in circles, but probably my fault, but I see the like terms, but what do I do now? :L lol...

  20. ParthKohli
    • 2 years ago
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    just operate on the numerators.

  21. FeoNeo33
    • 2 years ago
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    Nope, couldn't get anything. Okay, I don't want to waste anymore of your precious time, so I'ma ask for a straight up answer? This pellet is just hard for me, point blank.

  22. FeoNeo33
    • 2 years ago
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    or tell me next steps, I don't mind learning it.

  23. ParthKohli
    • 2 years ago
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    Wolfram if you want the straight answer.

  24. FeoNeo33
    • 2 years ago
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    lol i just want to learn the steps, but I don't want to waste ur time tutoring me.

  25. ParthKohli
    • 2 years ago
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    \(\color{Black}{\Rightarrow \Large {1 \over 3x+ 9} -{6 \over 3x + 9} = 2 }\) \(\color{Black}{\Rightarrow \Large {-5 \over 3x + 9} = 2 }\) Multiply both sides by 3x + 9 \(\color{Black}{\Rightarrow-5 = 2(3x+ 9) }\) \(\color{Black}{\Rightarrow -5 = 6x + 18 }\) \(\color{Black}{\Rightarrow -23 = 6x }\) \(\color{Black}{\Rightarrow x = {-23 \over 6} }\)

  26. FeoNeo33
    • 2 years ago
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    k i was good up till the part with the Multiply both sides by 3x + 9... but Ima let you go, I promise, Im not here just for the answer. Ima learn this stuff, I don't want to be a leecher my whole life. Thanks, bye.

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