b) \( \frac{7}{7 x+\frac{1}{2}}=\frac{1}{3} x+1 \)

To solve the equation \( \frac{7}{7 x+\frac{1}{2}}=\frac{1}{3} x+1 \), we start by eliminating the fraction on the left side. Multiply both sides by \( 7x + \frac{1}{2} \) to obtain: \[ 7 = \left( \frac{1}{3} x + 1 \right) \left( 7x + \frac{1}{2} \right) \] Expanding the right side gives us: \[ 7 = \frac{7}{3} x^2 + \frac{1}{2} x + 7x + \frac{1}{2} \] Rearranging terms and combining like terms leads to: \[ 0 = \frac{7}{3} x^2 + \left(\frac{1}{2} + 7\right)x + \frac{1}{2} - 7 \] Simplifying further: \[ 0 = \frac{7}{3} x^2 + \frac{15}{2}x - \frac{13}{2} \] Now, multiply through by \( 6 \) (the least common multiple of the denominators) to eliminate the fractions: \[ 0 = 14x^2 + 45x - 39 \] Next, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): With \( a = 14, b = 45, c = -39 \): First, calculate the discriminant: \[ b^2 - 4ac = 45^2 - 4(14)(-39) = 2025 + 2184 = 2209 \] Now, compute the roots: \[ x = \frac{-45 \pm \sqrt{2209}}{2 \cdot 14} \] Since \( \sqrt{2209} = 47 \): \[ x = \frac{-45 \pm 47}{28} \] This results in two possible solutions: 1. \( x = \frac{2}{28} = \frac{1}{14} \) 2. \( x = \frac{-92}{28} = -\frac{23}{7} \) Thus, the solutions to the original equation are \( x = \frac{1}{14} \) and \( x = -\frac{23}{7} \).