- To subtract fractions with like denominators, subtract the numerators and keep the common denominator: \(\frac{a}{d} - \frac{b}{d} = \frac{a-b}{d}\).
- For missing numerator problems (Section 7), the results are \(\frac{6}{25}\), \(\frac{6}{12}\), \(\frac{5}{15}\), \(\frac{4}{8}\), \(\frac{13}{25}\), and \(\frac{4}{10}\).
- For expressions in Section A, we simplify the resulting fractions to get: \(1/3\), \(9/40\), \(15/17\), \(101/10\), \(43/8\), and \(31/5\).
- Final answers: Section 7 boxes: 6, 6, 5, 4, 13, 4. Section A results: \(\frac{1}{3}\), \(\frac{9}{40}\), \(\frac{15}{17}\), \(\frac{101}{10}\), \(\frac{43}{8}\), and \(\frac{31}{5}\).
Explanation
- Analyzing Fraction Operations
To solve these problems, we use the fundamental rules for working with fractions that have a common denominator. For addition and subtraction, the rule is \(\frac{a}{d} \pm \frac{b}{d} = \frac{a \pm b}{d}\). If we are dealing with mixed numbers \(A \frac{a}{d} \pm B \frac{b}{d}\), we perform operations on the whole numbers and the fractions separately: \((A \pm B) + \frac{a \pm b}{d}\). If an operation results in a value that requires borrowing (such as subtracting a fraction from an integer), we express the integer as a fraction using the common denominator, for example, \(n = (n-1) + \frac{d}{d}\).
- Solving Section 7: Missing Numerators
In each equation, we determine the missing numerator by subtracting the numerators while keeping the denominator the same:
- \(\frac{14}{25} - \frac{8}{25} = \frac{14-8}{25} = \frac{6}{25}\) (Box = 6)
- \(\frac{7}{12} - \frac{1}{12} = \frac{7-1}{12} = \frac{6}{12} = \frac{1}{2}\) (Box = 6)
- \(\frac{12}{15} - \frac{7}{15} = \frac{12-7}{15} = \frac{5}{15} = \frac{1}{3}\) (Box = 5)
- \(\frac{7}{8} - \frac{3}{8} = \frac{7-3}{8} = \frac{4}{8} = \frac{1}{2}\) (Box = 4)
- \(\frac{17}{25} - \frac{4}{25} = \frac{17-4}{25} = \frac{13}{25}\) (Box = 13)
- \(\frac{7}{10} - \frac{3}{10} = \frac{7-3}{10} = \frac{4}{10} = \frac{2}{5}\) (Box = 4)
- Solving Section A: Operations with Fractions and Mixed Numbers
Now, we perform the arithmetic operations for each expression:
- \(\frac{5}{9} - \frac{2}{9} = \frac{3}{9} = \frac{1}{3}\)
- \(\frac{5}{40} + \frac{6}{40} - \frac{2}{40} = \frac{5+6-2}{40} = \frac{9}{40}\)
- \(\frac{19}{17} + \frac{2}{17} - \frac{6}{17} = \frac{19+2-6}{17} = \frac{15}{17}\)
- \(11 - \frac{9}{10} = 10 \frac{10}{10} - \frac{9}{10} = 10 \frac{1}{10} = \frac{101}{10}\)
- \(9 \frac{4}{8} - 4 \frac{1}{8} = (9-4) + \frac{4-1}{8} = 5 \frac{3}{8} = \frac{43}{8}\)
- \(15 \frac{7}{5} - 10 \frac{1}{5} = (15-10) + \frac{7-1}{5} = 5 + \frac{6}{5} = 5 + 1 \frac{1}{5} = 6 \frac{1}{5} = \frac{31}{5}\)
Examples
Understanding how to subtract and add fractions is essential in everyday tasks, such as cooking. If a recipe calls for \(\frac{3}{4}\) cup of flour, and you have already added \(\frac{1}{4}\) cup, you need to calculate how much more is required: \(\frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\) cup. Similarly, when working on construction projects or DIY repairs, knowing how to subtract fractions allows you to measure and cut materials accurately, ensuring everything fits perfectly as planned. Mastering these simple operations helps you navigate daily measurements with confidence and precision.