(11) \( \left(\frac{1}{2} a+\frac{b}{3}\right)\left(4 a-\frac{3}{2} b\right) \)

To expand the expression \( \left(\frac{1}{2} a+\frac{b}{3}\right)\left(4 a-\frac{3}{2} b\right) \), we will use the distributive property (also known as the FOIL method for binomials). 1. **First terms**: \( \frac{1}{2} a \cdot 4 a = 2 a^2 \) 2. **Outer terms**: \( \frac{1}{2} a \cdot \left(-\frac{3}{2} b\right) = -\frac{3}{4} ab \) 3. **Inner terms**: \( \frac{b}{3} \cdot 4 a = \frac{4}{3} ab \) 4. **Last terms**: \( \frac{b}{3} \cdot \left(-\frac{3}{2} b\right) = -\frac{3}{6} b^2 = -\frac{1}{2} b^2 \) Now we combine these results: \[ 2a^2 + \left(-\frac{3}{4} ab + \frac{4}{3} ab\right) - \frac{1}{2} b^2 \] To combine the \( ab \) terms, we need a common denominator, which is 12: \(-\frac{3}{4} ab = -\frac{9}{12} ab\) \(\frac{4}{3} ab = \frac{16}{12} ab\) Now the \( ab \) terms combine as follows: \[ -\frac{9}{12} ab + \frac{16}{12} ab = \frac{7}{12} ab \] Putting it all together, we get: \[ 2a^2 + \frac{7}{12} ab - \frac{1}{2} b^2 \] Thus, the final expanded form of the expression is: \[ \boxed{2a^2 + \frac{7}{12} ab - \frac{1}{2} b^2} \]