c) \( \frac{3}{4 y-1}=\frac{2}{4 y+1} \)

To solve the equation \( \frac{3}{4y-1} = \frac{2}{4y+1} \), we can cross-multiply to eliminate the fractions: \[ 3(4y + 1) = 2(4y - 1) \] Expanding both sides gives: \[ 12y + 3 = 8y - 2 \] Now, let's get all the terms involving \( y \) on one side and the constant terms on the other: \[ 12y - 8y = -2 - 3 \] This simplifies to: \[ 4y = -5 \] Dividing both sides by 4, we find: \[ y = -\frac{5}{4} \] To check if this solution works in the original equation, substitute \( y \) back into the fractions: 1. Calculate \( 4y - 1 \) and \( 4y + 1 \): - \( 4(-\frac{5}{4}) - 1 = -5 - 1 = -6 \) - \( 4(-\frac{5}{4}) + 1 = -5 + 1 = -4 \) 2. Substitute those into the original fractions: - Left side: \( \frac{3}{-6} = -\frac{1}{2} \) - Right side: \( \frac{2}{-4} = -\frac{1}{2} \) Both sides are equal, confirming that \( y = -\frac{5}{4} \) is indeed the correct solution! Feel free to ask if you have any further questions.