One possible solution:
\(\Large\frac{x+2}{(x-7)(x+1)}\)
In order to write this expression, the denominator had to be zero if \(x\) were \( 7\) or \(-1\). Substituting 3 for \(x\) into the denominator simplifies to \(-16\). The numerator must solve for 5 when \(x = 3\).
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