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

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

Evaluate

$$\frac{3}{2}=1.5$$

Solution Steps

Convert 1 to fraction \frac{2}{2}.

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

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

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

Add 2 and 1 to get 3.

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

Convert 3 to fraction \frac{9}{3}.

$$\frac{\frac{3}{2}}{\frac{9}{3}-\frac{1}{3}}\left(4-\frac{4}{3}\right)$$

Since \frac{9}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{3}{2}}{\frac{9-1}{3}}\left(4-\frac{4}{3}\right)$$

Subtract 1 from 9 to get 8.

$$\frac{\frac{3}{2}}{\frac{8}{3}}\left(4-\frac{4}{3}\right)$$

Divide \frac{3}{2} by \frac{8}{3} by multiplying \frac{3}{2} by the reciprocal of \frac{8}{3}.

$$\frac{3}{2}\times \frac{3}{8}\left(4-\frac{4}{3}\right)$$

Multiply \frac{3}{2} times \frac{3}{8} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3\times 3}{2\times 8}\left(4-\frac{4}{3}\right)$$

Do the multiplications in the fraction \frac{3\times 3}{2\times 8}.

$$\frac{9}{16}\left(4-\frac{4}{3}\right)$$

Convert 4 to fraction \frac{12}{3}.

$$\frac{9}{16}\left(\frac{12}{3}-\frac{4}{3}\right)$$

Since \frac{12}{3} and \frac{4}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9}{16}\times \frac{12-4}{3}$$

Subtract 4 from 12 to get 8.

$$\frac{9}{16}\times \frac{8}{3}$$

Multiply \frac{9}{16} times \frac{8}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{9\times 8}{16\times 3}$$

Do the multiplications in the fraction \frac{9\times 8}{16\times 3}.

$$\frac{72}{48}$$

Reduce the fraction \frac{72}{48} to lowest terms by extracting and canceling out 24.

$$\frac{3}{2}$$

Factor

$$\frac{3}{2} = 1\frac{1}{2} = 1.5$$