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Help me please! I need help on my algebra II. How would I do this question? [(2)/(x^2-9)]-[(3x)/(x^2-5x+6)]
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\[\frac{ 2 }{ x ^{2}-9 }-\frac{ 3x }{ x ^{2}-5x+6 }=\frac{ 2 }{ x ^{2}-3^{2} }-\frac{ 3x }{ x ^{2}-3x-2x+6 }\]
\[=\frac{ 2 }{ (x+3)(x-3) }-\frac{ 3x }{ x(x-3)-2(x-3) }=\frac{ 2 }{ (x+3)(x-3) }-\frac{ 3x }{ (x-2)(x-3) }\]
\[=\frac{ 1 }{ (x-3) } [\frac{ 2 }{ x+3 }-\frac{ 3x }{ x-2 } ]\]
\[=\frac{ 1 }{ (x-3) }[\frac{ 2(x-2)-3x(x+3) }{ (x+3)(x-2) }]=\frac{ -3x ^{2}-7x-4 }{ (x-3)(x+3)(x-2) }\]
@neer2890 Thank you! How do I give you a medal?
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\[=\frac{ -3x ^{2}-3x-4x-4 }{ (x-2)(x-3)(x+3) }=\frac{ -3x(x+1)-4(x+1) }{ (x-2)(x-3)(x+3) }\]
\[=\frac{ (-3x-4)(x+1) }{ (x-2)(x-3)(x+3) }\] you're welcome...:)
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