5. \( \frac{2 y-3}{3}=\frac{y-7}{2} \)

To solve the equation \( \frac{2 y-3}{3}=\frac{y-7}{2} \), start by cross-multiplying to eliminate the fractions. This gives you: \[ 2(2y - 3) = 3(y - 7) \] Expanding both sides leads to: \[ 4y - 6 = 3y - 21 \] Now, isolate \( y \) by subtracting \( 3y \) from both sides and adding 6: \[ 4y - 3y = -21 + 6 \] This simplifies to: \[ y = -15 \] So, the solution to the equation is \( y = -15 \). Now, let's verify the solution by substituting \( y = -15 \) back into the original equation. First, substitute \( y = -15 \) into \( \frac{2(-15)-3}{3} \) which yields \( \frac{-30 - 3}{3} = \frac{-33}{3} = -11 \). Then substitute into \( \frac{-15 - 7}{2} \) which results in \( \frac{-22}{2} = -11 \). As both sides equal \(-11\), our solution \( y = -15 \) is correct!