Find the difference and simplify: 2/(x^2+3x) - 3/(x^2-9)

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Find the difference and simplify: 2/(x^2+3x) - 3/(x^2-9)

Mathematics
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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 ?

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Yes
I think he went straight to the common denominator (:
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|
hmm no how did you get that ?
show ur work plz so i can find ot the mistakes :=)
I just subtracted 2-3 and got -1 and used the common denominator. I actually dont know what im doing but im trying
\[\frac{2}{x(x+3)}\times\frac{x-3}{x-3}-\frac{3}{(x-3)(x+3)}\times\frac{x}{x}\]
hm no you can't do that
Explain please?
just like i said multiply the numerator of first fraction with the `denominator ` of 2nd fraction so \[\huge\rm \frac{ 2 }{ x(x+3) } -\frac{ 3 }{ (x+3)(x-3) }\] x(x+3)(x-3) is a common denominator so multiply first fraction with x-3 and 2nd fraction with x
Would the numerator be x+6??
\[\huge\rm \frac{ 2(x-3)-3(x) }{ x (x+3)(x-3)}\] distribute the parentheses by 2 and then combine like terms
2(x-3) = ?
2x-6
yes right so \[\huge\rm \frac{\color{Red}{ 2x-6-3x} }{ x (x+3)(x-3)}\] now combine like terms
Would it he -x-6 / x(x+3)(x-3) ?!?!
yes right "=)
Omg thank you so much!
np :=)

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