Solve each equation. Remember to check for extraneous solutions. \( \frac{1}{2 x}+\frac{7}{2 x^{2}+10 x}=\frac{2}{x^{2}+5 x} \)

To solve the equation \( \frac{1}{2x} + \frac{7}{2x^{2} + 10x} = \frac{2}{x^{2} + 5x} \), we first notice that \( 2x^{2} + 10x \) can be factored as \( 2x(x + 5) \), and \( x^{2} + 5x \) can be factored as \( x(x + 5) \). Rewriting the equation gives us: \[ \frac{1}{2x} + \frac{7}{2x(x + 5)} = \frac{2}{x(x + 5)}. \] Now, we find a common denominator, which is \( 2x(x + 5) \): \[ \frac{x + 5}{2x(x + 5)} + \frac{7}{2x(x + 5)} = \frac{4}{2x(x + 5)}. \] Combining the fractions on the left, we have: \[ \frac{x + 5 + 7}{2x(x + 5)} = \frac{4}{2x(x + 5)}. \] This simplifies to: \[ \frac{x + 12}{2x(x + 5)} = \frac{4}{2x(x + 5)}. \] Multiplying both sides by \( 2x(x + 5) \) (noting \( x \neq 0 \) and \( x \neq -5 \)): \[ x + 12 = 4. \] Solving for \( x \): \[ x = 4 - 12 = -8. \] Now let's check for extraneous solutions. Substituting \( x = -8 \) back into the original equation: \[ \frac{1}{2(-8)} + \frac{7}{2(-8)^2 + 10(-8)} = \frac{2}{(-8)^2 + 5(-8)}. \] Calculating each term: - LHS: \( \frac{1}{-16} + \frac{7}{128 - 80} = \frac{-1}{16} + \frac{7}{48} \). To add these, we find a common denominator (48): \[ \frac{-3}{48} + \frac{7}{48} = \frac{4}{48} = \frac{1}{12}. \] - RHS: \( \frac{2}{64 - 40} = \frac{2}{24} = \frac{1}{12}. \) Since LHS = RHS, our solution is confirmed: \[ \boxed{-8}. \]