Solve 2 3 12 /[1 4 5 *\ 1 1 3 +[2 1 2 +1 1 3 -2 1 6 )\ of 1 2 3 ]

Question

$$2 \frac { 3 } { 12 } \times [ 1 \frac { 1 } { 5 } \times ( 1 \frac { 1 } { 3 } + [ 2 \frac { 1 } { 2 } + 1 \frac { 1 } { 3 } - 2 \frac { 1 } { 6 } ) ] 1 \frac { 2 } { 3 } ]$$

Evaluate

$\frac{27}{2}=13.5$

Solution Steps

Multiply $2$ and $12$ to get $24$.

$$\frac{24+3}{12}\times \frac{1\times 5+1}{5}\left(\frac{1\times 3+1}{3}+\frac{2\times 2+1}{2}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Add $24$ and $3$ to get $27$.

$$\frac{27}{12}\times \frac{1\times 5+1}{5}\left(\frac{1\times 3+1}{3}+\frac{2\times 2+1}{2}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Reduce the fraction $\frac{27}{12}$ to lowest terms by extracting and canceling out $3$.

$$\frac{9}{4}\times \frac{1\times 5+1}{5}\left(\frac{1\times 3+1}{3}+\frac{2\times 2+1}{2}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Multiply $1$ and $5$ to get $5$.

$$\frac{9}{4}\times \frac{5+1}{5}\left(\frac{1\times 3+1}{3}+\frac{2\times 2+1}{2}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Add $5$ and $1$ to get $6$.

$$\frac{9}{4}\times \frac{6}{5}\left(\frac{1\times 3+1}{3}+\frac{2\times 2+1}{2}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Multiply $\frac{9}{4}$ times $\frac{6}{5}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{9\times 6}{4\times 5}\left(\frac{1\times 3+1}{3}+\frac{2\times 2+1}{2}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Do the multiplications in the fraction $\frac{9\times 6}{4\times 5}$.

$$\frac{54}{20}\left(\frac{1\times 3+1}{3}+\frac{2\times 2+1}{2}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Reduce the fraction $\frac{54}{20}$ to lowest terms by extracting and canceling out $2$.

$$\frac{27}{10}\left(\frac{1\times 3+1}{3}+\frac{2\times 2+1}{2}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Multiply $1$ and $3$ to get $3$.

$$\frac{27}{10}\left(\frac{3+1}{3}+\frac{2\times 2+1}{2}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Add $3$ and $1$ to get $4$.

$$\frac{27}{10}\left(\frac{4}{3}+\frac{2\times 2+1}{2}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Multiply $2$ and $2$ to get $4$.

$$\frac{27}{10}\left(\frac{4}{3}+\frac{4+1}{2}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Add $4$ and $1$ to get $5$.

$$\frac{27}{10}\left(\frac{4}{3}+\frac{5}{2}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Least common multiple of $3$ and $2$ is $6$. Convert $\frac{4}{3}$ and $\frac{5}{2}$ to fractions with denominator $6$.

$$\frac{27}{10}\left(\frac{8}{6}+\frac{15}{6}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Since $\frac{8}{6}$ and $\frac{15}{6}$ have the same denominator, add them by adding their numerators.

$$\frac{27}{10}\left(\frac{8+15}{6}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Add $8$ and $15$ to get $23$.

$$\frac{27}{10}\left(\frac{23}{6}+\frac{1\times 3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Multiply $1$ and $3$ to get $3$.

$$\frac{27}{10}\left(\frac{23}{6}+\frac{3+1}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Add $3$ and $1$ to get $4$.

$$\frac{27}{10}\left(\frac{23}{6}+\frac{4}{3}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Least common multiple of $6$ and $3$ is $6$. Convert $\frac{23}{6}$ and $\frac{4}{3}$ to fractions with denominator $6$.

$$\frac{27}{10}\left(\frac{23}{6}+\frac{8}{6}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Since $\frac{23}{6}$ and $\frac{8}{6}$ have the same denominator, add them by adding their numerators.

$$\frac{27}{10}\left(\frac{23+8}{6}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Add $23$ and $8$ to get $31$.

$$\frac{27}{10}\left(\frac{31}{6}-\frac{2\times 6+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Multiply $2$ and $6$ to get $12$.

$$\frac{27}{10}\left(\frac{31}{6}-\frac{12+1}{6}\right)\times \frac{1\times 3+2}{3}$$

Add $12$ and $1$ to get $13$.

$$\frac{27}{10}\left(\frac{31}{6}-\frac{13}{6}\right)\times \frac{1\times 3+2}{3}$$

Since $\frac{31}{6}$ and $\frac{13}{6}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{27}{10}\times \frac{31-13}{6}\times \frac{1\times 3+2}{3}$$

Subtract $13$ from $31$ to get $18$.

$$\frac{27}{10}\times \frac{18}{6}\times \frac{1\times 3+2}{3}$$

Divide $18$ by $6$ to get $3$.

$$\frac{27}{10}\times 3\times \frac{1\times 3+2}{3}$$

Express $\frac{27}{10}\times 3$ as a single fraction.

$$\frac{27\times 3}{10}\times \frac{1\times 3+2}{3}$$

Multiply $27$ and $3$ to get $81$.

$$\frac{81}{10}\times \frac{1\times 3+2}{3}$$

Multiply $1$ and $3$ to get $3$.

$$\frac{81}{10}\times \frac{3+2}{3}$$

Add $3$ and $2$ to get $5$.

$$\frac{81}{10}\times \frac{5}{3}$$

Multiply $\frac{81}{10}$ times $\frac{5}{3}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{81\times 5}{10\times 3}$$

Do the multiplications in the fraction $\frac{81\times 5}{10\times 3}$.

$$\frac{405}{30}$$

Reduce the fraction $\frac{405}{30}$ to lowest terms by extracting and canceling out $15$.

$$\frac{27}{2}$$

Show Solution

Factor

$\frac{3 ^ {3}}{2} = 13\frac{1}{2} = 13.5$

Basic Math

Pre-Algebra

Algebra

Trigonometry

Calculus

Precalculus

Statistics

Finite Math

Linear Algebra

Chemistry

Example

Question

Evaluate

Solution Steps

Factor