\( \left. \begin{array} { l } { \frac { 3 } { 4 } - \frac { 2 } { 5 } = \frac { ( 5 ) ( 3 ) } { ( 5 ) ( 4 ) } - \frac { ( 4 ) ( 2 ) } { ( 4 ) ( 5 ) } = \frac { 15 } { 20 } - \frac { 8 } { 20 } = \frac { } { 20 } } \\ { \frac { 1 } { 4 } + \frac { 1 } { 2 } = \frac { 1 } { 4 } + \frac { ( L ) ( 1 ) } { ( ) ( 2 ) } = \frac { 1 } { 4 } + \frac { 2 } { 4 } = - } \end{array} \right. \)

Let's solve the two equations step by step. 1. **First Equation:** \[ \frac{3}{4} - \frac{2}{5} \] To subtract these fractions, we need a common denominator. The least common multiple of 4 and 5 is 20. Convert \(\frac{3}{4}\) and \(\frac{2}{5}\) to have a denominator of 20: \[ \frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20} \] \[ \frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20} \] Now we can subtract: \[ \frac{15}{20} - \frac{8}{20} = \frac{15 - 8}{20} = \frac{7}{20} \] 2. **Second Equation:** \[ \frac{1}{4} + \frac{1}{2} \] Again, we need a common denominator. The least common multiple of 4 and 2 is 4. Convert \(\frac{1}{2}\) to have a denominator of 4: \[ \frac{1}{2} = \frac{1 \times 2}{2 \times 2} = \frac{2}{4} \] Now we can add: \[ \frac{1}{4} + \frac{2}{4} = \frac{1 + 2}{4} = \frac{3}{4} \] Putting it all together, we have: - The result of the first equation is \(\frac{7}{20}\). - The result of the second equation is \(\frac{3}{4}\). Thus, the completed equations are: \[ \frac{3}{4} - \frac{2}{5} = \frac{7}{20} \] \[ \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \]