Question 390360
 (3x)/(x^2+2x-8)=(1/(x-2))+(x/(x+4))
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(3x)/[(x+4)(x-2)] = [(x+4+x(x-2)]/[(x-2)(x+4)]
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Multiply thru by (x-2)(x+4) to get:
3x = x+4 + x^2-2x
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3x = x^2-x+4
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x^2-4x+4 = 0
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Factor to get:
(x-2)^2 = 9
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Solution x = 2
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But x cannot be 2 because (x-2) is one of the denominators
of the original problem.
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Conclusion:
No solution
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Cheers,
Stan H.
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