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Solve for \(x\) in \(\frac{1}{x-4}+\frac{2}{x^{2}-16}=\frac{3}{x+4}\) [solved]

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Solve for \(x\) in \(\frac{1}{x-4}+\frac{2}{x^{2}-16}=\frac{3}{x+4}\)
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in Mathematics by ♦ Platinum (122,212 points)
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\(\frac{1}{x-4}+\frac{2}{x^{2}-16}=\frac{3}{x+4}\)
\(\frac{1}{x-4}+\frac{2}{(x-4)(x+4)}=\frac{3}{x+4}\)
\(\frac{1}{x-4} \cdot\left(\frac{x+4}{x+4}\right)+\frac{2}{(x-4)(x+4)}=\frac{3}{x+4} \cdot\left(\frac{x-4}{x-4}\right)\)
\(\frac{1(x+4)}{(x-4)(x+4)}+\frac{2}{(x-4)(x+4)}=\frac{3(x-4)}{(x+4)(x-4)}\)
\(1(x+4)+2=3(x-4)\)
\(x+6=3 x-12\)
\(x+18=3 x\)
\(18=2 x\)
\(9=x\)
by ♦ Platinum (122,212 points)
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