Solve 3/4*[7/3-(1/2+2/1*frac{1}{4})]

$$\frac { 3 } { 4 } \cdot [ \frac { 7 } { 3 } - ( \frac { 1 } { 2 } + \frac { 2 } { 1 } \cdot \frac { 1 } { 4 } ) ]$$

Evaluate

$$1$$

Solution Steps

Anything divided by one gives itself.

$$\frac{3}{4}\left(\frac{7}{3}-\left(\frac{1}{2}+2\times \frac{1}{4}\right)\right)$$

Multiply 2 and \frac{1}{4} to get \frac{2}{4}.

$$\frac{3}{4}\left(\frac{7}{3}-\left(\frac{1}{2}+\frac{2}{4}\right)\right)$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$\frac{3}{4}\left(\frac{7}{3}-\left(\frac{1}{2}+\frac{1}{2}\right)\right)$$

Since \frac{1}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

$$\frac{3}{4}\left(\frac{7}{3}-\frac{1+1}{2}\right)$$

Add 1 and 1 to get 2.

$$\frac{3}{4}\left(\frac{7}{3}-\frac{2}{2}\right)$$

Divide 2 by 2 to get 1.

$$\frac{3}{4}\left(\frac{7}{3}-1\right)$$

Convert 1 to fraction \frac{3}{3}.

$$\frac{3}{4}\left(\frac{7}{3}-\frac{3}{3}\right)$$

Since \frac{7}{3} and \frac{3}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3}{4}\times \frac{7-3}{3}$$

Subtract 3 from 7 to get 4.

$$\frac{3}{4}\times \frac{4}{3}$$

Cancel out \frac{3}{4} and its reciprocal \frac{4}{3}.

$$1$$

Factor

$$1$$