To solve the expression \(1 \frac{3}{5} \cdot 2 \frac{4}{7} + 2 \frac{4}{7} \cdot 1 \frac{1}{10} - 1 \frac{7}{10} \cdot 1 \frac{4}{7}\), we can start by converting each mixed number into an improper fraction.
Convert mixed numbers to improper fractions:
- \(1 \frac{3}{5} = \frac{1 \cdot 5 + 3}{5} = \frac{8}{5}\)
- \(2 \frac{4}{7} = \frac{2 \cdot 7 + 4}{7} = \frac{18}{7}\)
- \(1 \frac{1}{10} = \frac{1 \cdot 10 + 1}{10} = \frac{11}{10}\)
- \(1 \frac{7}{10} = \frac{1 \cdot 10 + 7}{10} = \frac{17}{10}\)
- \(1 \frac{4}{7} = \frac{1 \cdot 7 + 4}{7} = \frac{11}{7}\)
Substitute these values into the expression: \[ \frac{8}{5} \cdot \frac{18}{7} + \frac{18}{7} \cdot \frac{11}{10} - \frac{17}{10} \cdot \frac{11}{7} \]
Calculate each term:
First term: \[ \frac{8}{5} \cdot \frac{18}{7} = \frac{8 \cdot 18}{5 \cdot 7} = \frac{144}{35} \]
Second term: \[ \frac{18}{7} \cdot \frac{11}{10} = \frac{18 \cdot 11}{7 \cdot 10} = \frac{198}{70} = \frac{99}{35} \quad \text{(simplified)} \]
Third term: \[ \frac{17}{10} \cdot \frac{11}{7} = \frac{17 \cdot 11}{10 \cdot 7} = \frac{187}{70} = \frac{93.5}{35} \quad \text{(for easy subtraction later, convert since the denominators are not same)} \]
Now rewrite the subtraction term with a common denominator as well: \[ \frac{93.5}{35} = \frac{187}{70} \] We’ll have all terms over 35, thus: \[ \frac{187}{35} \]
Now combine like fractions: \[ \frac{144}{35} + \frac{99}{35} - \frac{93.5}{35} \Rightarrow \frac{243.5}{35} = \frac{491}{70} \]
Finally, simplify the result.
To convert back to a mixed number: \(\frac{491}{70} = 7 \frac{1}{70}\)
The final answer is \(7 \frac{1}{70}\).