Question

What is the sum of 9 over the sum of y and 4 + 6 over the difference of y and 3?

Answer 1

\[9/y+4+6/y+3\]

Answer 2

\[\frac{\frac{9}{y+4}+6}{y-3}\]

Answer 3

the 9/y+4 is separate from the 6/y+3

Answer 4

\[\frac{9}{y+4}+\frac{6}{y-3}\]

Answer 5

Yes, that's correct

Answer 6

so wat shud we do||????

Answer 7

step1; take common denominator

Answer 8

stay change flip?

Answer 9

\[\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}\]

Answer 10

Oh okay so 9(y+3)+6(y+4)/(y+4)(y+3)

Answer 11

\[{9\over {y+4}}+{6\over{y-3}}={{9(y-3)+6(y+4)}\over{(y+4)(y-3)}}\]

Answer 12

Okay, then we simplify, correct?

Answer 13

\[={3(5y-1)}\over{(y+4)(y-6)}\]

Answer 14

Then you distribute the 3

Answer 15

and get 15y-3

Answer 16

haha.. I actually un-distributed it...

Answer 17

whoops... haha