anonymous
  • anonymous
What is the sum of 9 over the sum of y and 4 + 6 over the difference of y and 3?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[9/y+4+6/y+3\]
anonymous
  • anonymous
\[\frac{\frac{9}{y+4}+6}{y-3}\]
anonymous
  • anonymous
the 9/y+4 is separate from the 6/y+3

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anonymous
  • anonymous
\[\frac{9}{y+4}+\frac{6}{y-3}\]
anonymous
  • anonymous
Yes, that's correct
anonymous
  • anonymous
so wat shud we do||????
anonymous
  • anonymous
step1; take common denominator
anonymous
  • anonymous
stay change flip?
anonymous
  • anonymous
\[\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}\]
anonymous
  • anonymous
Oh okay so 9(y+3)+6(y+4)/(y+4)(y+3)
anonymous
  • anonymous
\[{9\over {y+4}}+{6\over{y-3}}={{9(y-3)+6(y+4)}\over{(y+4)(y-3)}}\]
anonymous
  • anonymous
Okay, then we simplify, correct?
anonymous
  • anonymous
\[={3(5y-1)}\over{(y+4)(y-6)}\]
anonymous
  • anonymous
Then you distribute the 3
anonymous
  • anonymous
and get 15y-3
anonymous
  • anonymous
haha.. I actually un-distributed it...
anonymous
  • anonymous
whoops... haha

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