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Last complex fraction that I have to do! Please help :)
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|dw:1346758831513:dw|
and here starts the race.....
lol wow
\[\frac{ 1 }{ x-3 }+4x=\frac{ 1+4x(x-3) }{ x-3 }=\frac{ 1+4x^2-12x }{ x-3 }\]
hope you can do rest of the calculation :)
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yes i can! thank you! I just calculated it, and I got the answer: 5x-3/3(x-1)
shouldn't it be 3x??
check again!
|dw:1346759083401:dw|
\[\frac{ 4 }{ x }-\frac{ 1 }{ x-3 }= \frac{ 4(x-3)-x }{ x(x-3)}=\frac{ 4x-12-x }{ x^2-3x }=\frac{ 3x-12 }{ x^2-3x }\]
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does that help?? @kakrazz ??/ and @hartnn dude..where's the pied piper ??
|dw:1346759151619:dw|
lol....u are helping na....
oh i see.... :)
okay! thanks :)
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:) YW
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