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anonymous
 5 years ago
how do you solve 6x10 over 7 divided by 9x15 over 21?
anonymous
 5 years ago
how do you solve 6x10 over 7 divided by 9x15 over 21?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well without it being equal to something you can't really 'solve' it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no there's a way to solve it because someone got the answer but i dont know what it is

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it can be reduced, simplified; but not really solved; its always gonna have an x in it unless that x factors out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what would the answer be and how would you get that answer?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this the expression? \[(6x10) \div {{9x  15} \over 21}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no.. 6x  10 is over 7, but everything else is right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{{6x10} \over 7}\div {{9x  15} \over 21}\]\[=2({{3x5} \over 7}) \div 3({{3x5} \over 7})\]\[={{2({{3x5} \over 7})} \over {3({{3x5} \over 7})}}\]\[ = {2 \over 3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Though you do need to stipulate that x cannot be equal to 5/3.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one question.. how did you get both denominators 7?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{{6x10} \over 7}\div {{9x  15} \over 21}\]\[=2({{3x5} \over 7}) \div {3 \over 3}({{3x5} \over 7})\]\[={{2({{3x5} \over 7})} \over {1({{3x5} \over 7})}} = 2,\ \forall x \ne {5 \over 3} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Forgot I factored a 3 from the top and bottom of the fraction on the right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thx .can you help me with another problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Heres the problem...8y over y squared + 4 minus y  2 over y squared + 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i didn't know you could do that. because I'm kind of new to this website ....... 8y/y squared + 4 minus y2/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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{{8y} \over y^2+ 4}  {{y2} \over y^2+ 4} \] That?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yah. how did u do that but anyway how do u solve it??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well they are over a common denominator already so just subtract the numerators.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what answer would you get?
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