8) \( \frac{3}{f+1}-\frac{1}{2}=\frac{1}{3 f+3} \)
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To solve the equation \( \frac{3}{f+1}-\frac{1}{2}=\frac{1}{3f+3} \), first, observe that \( 3f + 3 = 3(f + 1) \). This allows us to reframe the equation: \[ \frac{3}{f+1} - \frac{1}{2} = \frac{1}{3(f + 1)} \] Now, let's find a common denominator for the left-hand side, which will be \( 2(f + 1) \): \[ \frac{3 \cdot 2}{2(f + 1)} - \frac{(f + 1)}{2(f + 1)} = \frac{6 - (f + 1)}{2(f + 1)} = \frac{5 - f}{2(f + 1)} \] So, the equation now looks like: \[ \frac{5 - f}{2(f + 1)} = \frac{1}{3(f + 1)} \] Next, eliminate the denominators by multiplying each side by \( 6(f + 1) \): \[ 6(5 - f) = 2 \] This simplifies to: \[ 30 - 6f = 2 \] Rearranging gives: \[ 28 = 6f \quad \Rightarrow \quad f = \frac{28}{6} = \frac{14}{3} \] Thus, the solution is: \[ f = \frac{14}{3} \]