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anonymous

  • 5 years ago

how do you solve 6x-10 over 7 divided by 9x-15 over 21?

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  1. anonymous
    • 5 years ago
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    Well without it being equal to something you can't really 'solve' it...

  2. anonymous
    • 5 years ago
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    no there's a way to solve it because someone got the answer but i dont know what it is

  3. amistre64
    • 5 years ago
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    it can be reduced, simplified; but not really solved; its always gonna have an x in it unless that x factors out

  4. anonymous
    • 5 years ago
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    so what would the answer be and how would you get that answer?

  5. amistre64
    • 5 years ago
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    the other option is to plot the points on a graph that result from inputing values of x and the graph becomes the solution.... but without knowing tha tthis is a function with respect to x its rather redundant

  6. anonymous
    • 5 years ago
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    Is this the expression? \[(6x-10) \div {{9x - 15} \over 21}\]

  7. anonymous
    • 5 years ago
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    no.. 6x - 10 is over 7, but everything else is right

  8. anonymous
    • 5 years ago
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    \[{{6x-10} \over 7}\div {{9x - 15} \over 21}\]\[=2({{3x-5} \over 7}) \div 3({{3x-5} \over 7})\]\[={{2({{3x-5} \over 7})} \over {3({{3x-5} \over 7})}}\]\[ = {2 \over 3}\]

  9. anonymous
    • 5 years ago
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    Though you do need to stipulate that x cannot be equal to 5/3.

  10. anonymous
    • 5 years ago
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    one question.. how did you get both denominators 7?

  11. anonymous
    • 5 years ago
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    oh wait. I'm wrong.

  12. anonymous
    • 5 years ago
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    \[{{6x-10} \over 7}\div {{9x - 15} \over 21}\]\[=2({{3x-5} \over 7}) \div {3 \over 3}({{3x-5} \over 7})\]\[={{2({{3x-5} \over 7})} \over {1({{3x-5} \over 7})}} = 2,\ \forall x \ne {5 \over 3} \]

  13. anonymous
    • 5 years ago
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    Forgot I factored a 3 from the top and bottom of the fraction on the right.

  14. anonymous
    • 5 years ago
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    thx .can you help me with another problem?

  15. anonymous
    • 5 years ago
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    maybe

  16. anonymous
    • 5 years ago
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    Heres the problem...8-y over y squared + 4 minus y - 2 over y squared + 4

  17. anonymous
    • 5 years ago
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    It would be very helpful if you would write these up in the equation editor because there are a number of ways to interpret what you wrote there.

  18. anonymous
    • 5 years ago
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    oh i didn't know you could do that. because I'm kind of new to this website ....... 8-y/y squared + 4 minus y-2/y squared + 4? does that make a little more sense because I don't know how to do exponents so I just wrote it out

  19. anonymous
    • 5 years ago
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    \[{{8-y} \over y^2+ 4} - {{y-2} \over y^2+ 4} \] That?

  20. anonymous
    • 5 years ago
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    yah. how did u do that but anyway how do u solve it??

  21. anonymous
    • 5 years ago
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    Well they are over a common denominator already so just subtract the numerators.

  22. anonymous
    • 5 years ago
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    what answer would you get?

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