Solve (2+1/4):(2-1/4)+1d

$$( 2 + \frac { 1 } { 4 } ) : ( 2 - \frac { 1 } { 4 } ) + 1 d$$

Evaluate

$$d = \frac{9}{7} = 1\frac{2}{7} = 1.2857142857142858$$

Solution Steps

Convert 2 to fraction \frac{8}{4}.

$$\frac{\frac{8}{4}+\frac{1}{4}}{2-\frac{1}{4}}+1d$$

Since \frac{8}{4} and \frac{1}{4} have the same denominator, add them by adding their numerators.

$$\frac{\frac{8+1}{4}}{2-\frac{1}{4}}+1d$$

Add 8 and 1 to get 9.

$$\frac{\frac{9}{4}}{2-\frac{1}{4}}+1d$$

Convert 2 to fraction \frac{8}{4}.

$$\frac{\frac{9}{4}}{\frac{8}{4}-\frac{1}{4}}+1d$$

Since \frac{8}{4} and \frac{1}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{9}{4}}{\frac{8-1}{4}}+1d$$

Subtract 1 from 8 to get 7.

$$\frac{\frac{9}{4}}{\frac{7}{4}}+1d$$

Divide \frac{9}{4} by \frac{7}{4} by multiplying \frac{9}{4} by the reciprocal of \frac{7}{4}.

$$\frac{9}{4}\times \frac{4}{7}+1d$$

Multiply \frac{9}{4} times \frac{4}{7} by multiplying numerator times numerator and denominator times denominator.

$$\frac{9\times 4}{4\times 7}+1d$$

Cancel out 4 in both numerator and denominator.

$$\frac{9}{7}+1d$$

For any term t, t\times 1=t and 1t=t.

$$\frac{9}{7}+d$$

Expand

$$d = \frac{9}{7} = 1\frac{2}{7} = 1.2857142857142858$$

Solution Steps

Convert 2 to fraction \frac{8}{4}.

$$\frac{\frac{8}{4}+\frac{1}{4}}{2-\frac{1}{4}}+1d$$

Since \frac{8}{4} and \frac{1}{4} have the same denominator, add them by adding their numerators.

$$\frac{\frac{8+1}{4}}{2-\frac{1}{4}}+1d$$

Add 8 and 1 to get 9.

$$\frac{\frac{9}{4}}{2-\frac{1}{4}}+1d$$

Convert 2 to fraction \frac{8}{4}.

$$\frac{\frac{9}{4}}{\frac{8}{4}-\frac{1}{4}}+1d$$

Since \frac{8}{4} and \frac{1}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{9}{4}}{\frac{8-1}{4}}+1d$$

Subtract 1 from 8 to get 7.

$$\frac{\frac{9}{4}}{\frac{7}{4}}+1d$$

Divide \frac{9}{4} by \frac{7}{4} by multiplying \frac{9}{4} by the reciprocal of \frac{7}{4}.

$$\frac{9}{4}\times \frac{4}{7}+1d$$

Multiply \frac{9}{4} times \frac{4}{7} by multiplying numerator times numerator and denominator times denominator.

$$\frac{9\times 4}{4\times 7}+1d$$

Cancel out 4 in both numerator and denominator.

$$\frac{9}{7}+1d$$

For any term t, t\times 1=t and 1t=t.

$$\frac{9}{7}+d$$