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



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

Jonask
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\[\frac{\frac{9}{y+4}+6}{y3}\]

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

Jonask
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\[\frac{9}{y+4}+\frac{6}{y3}\]

PedroSoccer
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Yes, that's correct

Jonask
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so wat shud we do????

Kanwar245
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step1; take common denominator

PedroSoccer
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stay change flip?

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

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

electrokid
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\[{9\over {y+4}}+{6\over{y3}}={{9(y3)+6(y+4)}\over{(y+4)(y3)}}\]

PedroSoccer
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Okay, then we simplify, correct?

electrokid
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\[={3(5y1)}\over{(y+4)(y6)}\]

PedroSoccer
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Then you distribute the 3

PedroSoccer
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and get 15y3

electrokid
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haha.. I actually undistributed it...

PedroSoccer
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whoops... haha