Solve 2/8<1/8+3/24<3/8

$$\frac { 2 } { 8 } < \frac { 1 } { 8 } + \frac { 3 } { 24 } < \frac { 3 } { 8 }$$

Verify

Solution Steps

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$\frac{1}{4}<\frac{1}{8}+\frac{3}{24}\text{ and }\frac{1}{8}+\frac{3}{24}<\frac{3}{8}$$

Reduce the fraction \frac{3}{24} to lowest terms by extracting and canceling out 3.

$$\frac{1}{4}<\frac{1}{8}+\frac{1}{8}\text{ and }\frac{1}{8}+\frac{3}{24}<\frac{3}{8}$$

Since \frac{1}{8} and \frac{1}{8} have the same denominator, add them by adding their numerators.

$$\frac{1}{4}<\frac{1+1}{8}\text{ and }\frac{1}{8}+\frac{3}{24}<\frac{3}{8}$$

Add 1 and 1 to get 2.

$$\frac{1}{4}<\frac{2}{8}\text{ and }\frac{1}{8}+\frac{3}{24}<\frac{3}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$\frac{1}{4}<\frac{1}{4}\text{ and }\frac{1}{8}+\frac{3}{24}<\frac{3}{8}$$

Compare \frac{1}{4} and \frac{1}{4}.

$$\text{false}\text{ and }\frac{1}{8}+\frac{3}{24}<\frac{3}{8}$$

Reduce the fraction \frac{3}{24} to lowest terms by extracting and canceling out 3.

$$\text{false}\text{ and }\frac{1}{8}+\frac{1}{8}<\frac{3}{8}$$

Since \frac{1}{8} and \frac{1}{8} have the same denominator, add them by adding their numerators.

$$\text{false}\text{ and }\frac{1+1}{8}<\frac{3}{8}$$

Add 1 and 1 to get 2.

$$\text{false}\text{ and }\frac{2}{8}<\frac{3}{8}$$

Compare \frac{2}{8} and \frac{3}{8}.

$$\text{false}\text{ and }\text{true}$$

The conjunction of \text{false} and \text{true} is \text{false}.

$$\text{false}$$