$\begin{array}{l}\frac{7}{4}-\frac{1}{2}=\frac{7-2}{5} \\ -\frac{5}{3}-\frac{4}{5} \frac{25-12}{15}=\frac{13}{15} \\ -\frac{3}{4}-\frac{1}{4}=\frac{3-1}{4}=\frac{2}{4}=1 \\ \frac{5}{4}-\frac{3}{8} \frac{18-3=15}{8} \\ \text { a, } \frac{2}{3}-\frac{1}{3} \frac{2-1=1}{3} \\ 6, \frac{1}{2}-\frac{1}{3} \\ \text { s } \frac{2}{3}-\frac{1}{2}\end{array}$
The task requires computing the result of two fraction subtraction expressions: \(\frac{1}{2} - \frac{1}{3}\) (label 'b.') and \(\frac{2}{3} - \frac{1}{2}\) (label 'c.'). To subtract fractions with different denominators, we find a common denominator, convert each fraction to an equivalent form with this common denominator, and then subtract the numerators while keeping the common denominator constant.
To solve \(\frac{1}{2} - \frac{1}{3}\), we first determine a common denominator for 2 and 3, which is 6. Converting the fractions:
\(\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}\)
\(\frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6}\)
Subtracting these:
\(\frac{3}{6} - \frac{2}{6} = \frac{3 - 2}{6} = \frac{1}{6}\)
Thus, the result for 'b.' is \(\frac{1}{6}\).
To solve \(\frac{2}{3} - \frac{1}{2}\), we find a common denominator for 3 and 2, which is 6. Converting the fractions:
\(\frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6}\)
\(\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}\)
Subtracting these:
\(\frac{4}{6} - \frac{3}{6} = \frac{4 - 3}{6} = \frac{1}{6}\)
Thus, the result for 'c.' is \(\frac{1}{6}\).
Fractions are used in everyday life for tasks like cooking and measurement. For example, if a recipe calls for \(\frac{2}{3}\) of a cup of flour and you already used \(\frac{1}{2}\) of a cup, you can calculate the remaining amount needed by performing the subtraction \(\frac{2}{3} - \frac{1}{2} = \frac{1}{6}\) of a cup. This simple application helps ensure recipes are followed accurately and resources are measured correctly.