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Find the difference and simplify: 2/(x^2+3x) - 3/(x^2-9)
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\[\huge\rm \frac{ 2 }{ x^2+3x }-\frac{ 3 }{ x^2-9 }\] factor the denominator
X(x+3)(x-3) ??
for x^2-9 apply the difference of squares \[\huge\rm a^2-b^2=(a+b)(a-b)\] do you mean x(x+3)(x-3) is a common denominator ?
Yes
I think he went straight to the common denominator (:
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true >.<
Would the answer be -1/ x(x+3)(x-3) ??
alright then when we find common denominator we should multiply the numerator of 1st fraction with the denominator of the 2nd fraction and multiply the numerator of 2nd fraction with the denominator of first fraction (just like cross multiplication ) here is an example |dw:1442889917081:dw|
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