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adding fractions with x's problem below
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\[\frac{ x+1 }{ (x+3)(x-3) } + \frac{ 4(x-3) }{ (x-3)(x+3)} + \frac{ (x-1)(x-3) }{ (3-x)(x+3) }\]
Since we have the same denominator, add up all the numerators.
but the last one is different its (3-x) instead of (x-3) so i have a negative 3 and a positive 3
Ahh hm.
hi u got same denominator just add em up
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Think you can multiply by -1 but that's just a hunch.
lol dan -_- no....
so take -1 out from the denominator of the rightmost fraction...
oh i seee
Yah i know that how i did previous problems, but then wouldn't the negative one effect the (x+3) as well?
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no @bekkah323
souvik, show me this!!
|dw:1373224244857:dw|
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