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anonymous
 5 years ago
((x2)/(x3))((x+1)(x+3))+((x9(/x^29))
anonymous
 5 years ago
((x2)/(x3))((x+1)(x+3))+((x9(/x^29))

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think the best way to do this is to just multiply out the first two terms, and for the third term, factor the denominator to (x3)(x+3).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you multiply it? I dont understand this. it looks like subtraction to me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh sorry, I didn't see the / in the first term. Ok, for the second term, (x+1)(x+3), multiply this out Then for the third term, (x9)/(x^29), factor the denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0uh oh i typed this out wrong its supp((x2)/(x3))((x+1)/(x+3))+((x9)/(x^29))osed to be

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x2}{x3}  \frac{x+1}{x+3} + \frac{x9}{x^2 9}\] Recognize that \(x^2  9 = (x+3)(x3)\) and put them all under a common denominator of \(x^29\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Over a common denominator rather.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i need to multiply the first two fractions by (x+3) or x3?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Multiply the first by (x+3) on top and bottom, and the second by x3 on top and bottom.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, was the second term also a fraction? Remember when you learned how to add fractions with different denominators? Use the same logic. Multiply the first fraction by (x+3)/(x+3) and the second by (x3)/(x3) so that they all have a common denominator. Basically what polpak is saying but it is worth noting that this is just like what you do when you are adding normal fractions with different denominators.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh. I may have mistyped it. Was the middle term a fraction? If not it'd be \[\frac{x2}{x3} +(x+1)(x+3) +\frac{x9}{x^29}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But you still want to get them over a common denominator of \(x^29\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0polpak, she posted that they were all fractions a few minutes ago. I missed it too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you were correct the first time i mistyped it... thank you both for your help! its greatly appreciated

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i came up with 0(x+3)(x3) is this correct? i somehow dont think it is

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im sorry 0/(x3)(x+3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got something else. Did you forget that the second fraction is being subtracted? I almost forgot that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no i subtracted it but it all seemed to cancel out. what am i doing wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x2}{x3}  \frac{x+1}{x+3} + \frac{x9}{x^29} \] \[= \frac{(x2)(x+3)}{x^29}  \frac{(x+1)(x3)}{x^29} + \frac{x9}{x^29} \] \[= \frac{[x^2+x6]  [x^22x 3] + [x9]}{x^29} \] \[= \frac{x^2x^2+x + 2x + x 6+3 9}{x^29} \] \[= \frac{4x12}{x^29} \] \[= \frac{4(x3)}{x^29} \] \[= \frac{4(x3)}{(x+3)(x3)} \] \[= \frac{4}{x+3} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where'd you go wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And hopefully you can follow that mess.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i didnt change the signs in the second fraction AND i can follow it
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