Solve For D 9920Bc

1. **State the problem:** Solve for $d$ in the equation $$\frac{1}{2} d + \frac{2}{3} = \frac{3}{4}$$. 2. **Isolate the term with $d$:** Subtract $\frac{2}{3}$ from both sides: $$\frac{1}{2} d + \frac{2}{3} - \frac{2}{3} = \frac{3}{4} - \frac{2}{3}$$ which simplifies to $$\frac{1}{2} d = \frac{3}{4} - \frac{2}{3}$$ 3. **Find common denominator and subtract fractions:** The common denominator of $4$ and $3$ is $12$. $$\frac{3}{4} = \frac{9}{12}, \quad \frac{2}{3} = \frac{8}{12}$$ So, $$\frac{1}{2} d = \frac{9}{12} - \frac{8}{12} = \frac{1}{12}$$ 4. **Solve for $d$ by dividing both sides by $\frac{1}{2}$:** $$d = \frac{\frac{1}{12}}{\frac{1}{2}} = \frac{1}{12} \times \frac{2}{1}$$ Show cancellation: $$d = \frac{\cancel{1}}{12} \times \frac{2}{\cancel{1}} = \frac{2}{12}$$ 5. **Simplify the fraction:** $$d = \frac{2}{12} = \frac{1}{6}$$ **Final answer:** $$d = \frac{1}{6}$$