We are given three math problems involving fractions. We need to evaluate them and express the results as common fractions. (a) $\frac{5}{8} - \frac{3}{5}$ (b) $\frac{6}{7} \div 2\frac{1}{7}$ (c) $\frac{3}{4} + \frac{1}{4} \times \frac{2}{3}$

1. Problem Description

We are given three math problems involving fractions. We need to evaluate them and express the results as common fractions.

(a) $\frac{5}{8} - \frac{3}{5}$

(b) $\frac{6}{7} \div 2\frac{1}{7}$

(c) $\frac{3}{4} + \frac{1}{4} \times \frac{2}{3}$

2. Solution Steps

(a) $\frac{5}{8} - \frac{3}{5}$

To subtract these fractions, we need to find a common denominator. The least common multiple of 8 and 5 is

4

0. $\frac{5}{8} = \frac{5 \times 5}{8 \times 5} = \frac{25}{40}$

$\frac{3}{5} = \frac{3 \times 8}{5 \times 8} = \frac{24}{40}$

$\frac{5}{8} - \frac{3}{5} = \frac{25}{40} - \frac{24}{40} = \frac{25-24}{40} = \frac{1}{40}$

(b) $\frac{6}{7} \div 2\frac{1}{7}$

First, convert the mixed number to an improper fraction.

$2\frac{1}{7} = \frac{2 \times 7 + 1}{7} = \frac{14 + 1}{7} = \frac{15}{7}$

Now, divide the fractions:

$\frac{6}{7} \div \frac{15}{7} = \frac{6}{7} \times \frac{7}{15} = \frac{6 \times 7}{7 \times 15} = \frac{42}{105}$

We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is

2

1. $\frac{42}{105} = \frac{42 \div 21}{105 \div 21} = \frac{2}{5}$

(c) $\frac{3}{4} + \frac{1}{4} \times \frac{2}{3}$

According to the order of operations, we must perform multiplication before addition.

$\frac{1}{4} \times \frac{2}{3} = \frac{1 \times 2}{4 \times 3} = \frac{2}{12} = \frac{1}{6}$

Now, we add the fractions:

$\frac{3}{4} + \frac{1}{6}$. The least common multiple of 4 and 6 is

1

2. $\frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}$

$\frac{1}{6} = \frac{1 \times 2}{6 \times 2} = \frac{2}{12}$

$\frac{3}{4} + \frac{1}{6} = \frac{9}{12} + \frac{2}{12} = \frac{9+2}{12} = \frac{11}{12}$

3. Final Answer

(a) $\frac{1}{40}$

(b) $\frac{2}{5}$

(c) $\frac{11}{12}$