Solve the equation. \[ 2=\frac{7}{2 x-1}+\frac{-3}{(2 x-1)^{2}} \]

To solve the equation: \[ 2 = \frac{7}{2x - 1} + \frac{-3}{(2x - 1)^2} \] **Step 1: Combine the Fractions** First, combine the terms on the right side: \[ 2 = \frac{7(2x - 1) - 3}{(2x - 1)^2} \] Simplify the numerator: \[ 2 = \frac{14x - 7 - 3}{(2x - 1)^2} = \frac{14x - 10}{(2x - 1)^2} \] **Step 2: Clear the Denominator** Multiply both sides by \((2x - 1)^2\) to eliminate the denominator: \[ 2(2x - 1)^2 = 14x - 10 \] **Step 3: Expand and Simplify** Expand \((2x - 1)^2\): \[ (2x - 1)^2 = 4x^2 - 4x + 1 \] Substitute back into the equation: \[ 2(4x^2 - 4x + 1) = 14x - 10 \] Expand the left side: \[ 8x^2 - 8x + 2 = 14x - 10 \] Move all terms to one side: \[ 8x^2 - 22x + 12 = 0 \] **Step 4: Solve the Quadratic Equation** Divide the entire equation by 2 for simplicity: \[ 4x^2 - 11x + 6 = 0 \] Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -11\), and \(c = 6\): \[ x = \frac{11 \pm \sqrt{121 - 96}}{8} = \frac{11 \pm \sqrt{25}}{8} = \frac{11 \pm 5}{8} \] This gives two solutions: \[ x = \frac{16}{8} = 2 \quad \text{and} \quad x = \frac{6}{8} = \frac{3}{4} \] **Step 5: Verify the Solutions** Ensure that neither solution makes the denominator zero: - For \(x = 2\): \(2(2) - 1 = 3 \neq 0\) - For \(x = \frac{3}{4}\): \(2\left(\frac{3}{4}\right) - 1 = \frac{1}{2} \neq 0\) Both solutions are valid. **Final Answer:** Both real solutions are \(x = 2\) and \(x = \dfrac{3}{4}\).