Solve ence . ( ii) 1 7 12 -[ 7 2 -\ 8-(5 1 3 - overline 3-2 1 2 )\ ] 6

Question

$$1 \frac { 7 } { 12 } - [ \frac { 7 } { 2 } - \{ 8 - ( 5 \frac { 1 } { 3 } - 3 - 2 \frac { 1 } { 2 } ) \} ]$$

Evaluate

$\frac{25}{4}=6.25$

Solution Steps

Multiply $1$ and $12$ to get $12$.

$$\frac{12+7}{12}-\left(\frac{7}{2}-\left(8-\left(\frac{5\times 3+1}{3}-3-\frac{2\times 2+1}{2}\right)\right)\right)$$

Add $12$ and $7$ to get $19$.

$$\frac{19}{12}-\left(\frac{7}{2}-\left(8-\left(\frac{5\times 3+1}{3}-3-\frac{2\times 2+1}{2}\right)\right)\right)$$

Multiply $5$ and $3$ to get $15$.

$$\frac{19}{12}-\left(\frac{7}{2}-\left(8-\left(\frac{15+1}{3}-3-\frac{2\times 2+1}{2}\right)\right)\right)$$

Add $15$ and $1$ to get $16$.

$$\frac{19}{12}-\left(\frac{7}{2}-\left(8-\left(\frac{16}{3}-3-\frac{2\times 2+1}{2}\right)\right)\right)$$

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

$$\frac{19}{12}-\left(\frac{7}{2}-\left(8-\left(\frac{16}{3}-\frac{9}{3}-\frac{2\times 2+1}{2}\right)\right)\right)$$

Since $\frac{16}{3}$ and $\frac{9}{3}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{19}{12}-\left(\frac{7}{2}-\left(8-\left(\frac{16-9}{3}-\frac{2\times 2+1}{2}\right)\right)\right)$$

Subtract $9$ from $16$ to get $7$.

$$\frac{19}{12}-\left(\frac{7}{2}-\left(8-\left(\frac{7}{3}-\frac{2\times 2+1}{2}\right)\right)\right)$$

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

$$\frac{19}{12}-\left(\frac{7}{2}-\left(8-\left(\frac{7}{3}-\frac{4+1}{2}\right)\right)\right)$$

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

$$\frac{19}{12}-\left(\frac{7}{2}-\left(8-\left(\frac{7}{3}-\frac{5}{2}\right)\right)\right)$$

Least common multiple of $3$ and $2$ is $6$. Convert $\frac{7}{3}$ and $\frac{5}{2}$ to fractions with denominator $6$.

$$\frac{19}{12}-\left(\frac{7}{2}-\left(8-\left(\frac{14}{6}-\frac{15}{6}\right)\right)\right)$$

Since $\frac{14}{6}$ and $\frac{15}{6}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{19}{12}-\left(\frac{7}{2}-\left(8-\frac{14-15}{6}\right)\right)$$

Subtract $15$ from $14$ to get $-1$.

$$\frac{19}{12}-\left(\frac{7}{2}-\left(8-\left(-\frac{1}{6}\right)\right)\right)$$

The opposite of $-\frac{1}{6}$ is $\frac{1}{6}$.

$$\frac{19}{12}-\left(\frac{7}{2}-\left(8+\frac{1}{6}\right)\right)$$

Convert $8$ to fraction $\frac{48}{6}$.

$$\frac{19}{12}-\left(\frac{7}{2}-\left(\frac{48}{6}+\frac{1}{6}\right)\right)$$

Since $\frac{48}{6}$ and $\frac{1}{6}$ have the same denominator, add them by adding their numerators.

$$\frac{19}{12}-\left(\frac{7}{2}-\frac{48+1}{6}\right)$$

Add $48$ and $1$ to get $49$.

$$\frac{19}{12}-\left(\frac{7}{2}-\frac{49}{6}\right)$$

Least common multiple of $2$ and $6$ is $6$. Convert $\frac{7}{2}$ and $\frac{49}{6}$ to fractions with denominator $6$.

$$\frac{19}{12}-\left(\frac{21}{6}-\frac{49}{6}\right)$$

Since $\frac{21}{6}$ and $\frac{49}{6}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{19}{12}-\frac{21-49}{6}$$

Subtract $49$ from $21$ to get $-28$.

$$\frac{19}{12}-\frac{-28}{6}$$

Reduce the fraction $\frac{-28}{6}$ to lowest terms by extracting and canceling out $2$.

$$\frac{19}{12}-\left(-\frac{14}{3}\right)$$

The opposite of $-\frac{14}{3}$ is $\frac{14}{3}$.

$$\frac{19}{12}+\frac{14}{3}$$

Least common multiple of $12$ and $3$ is $12$. Convert $\frac{19}{12}$ and $\frac{14}{3}$ to fractions with denominator $12$.

$$\frac{19}{12}+\frac{56}{12}$$

Since $\frac{19}{12}$ and $\frac{56}{12}$ have the same denominator, add them by adding their numerators.

$$\frac{19+56}{12}$$

Add $19$ and $56$ to get $75$.

$$\frac{75}{12}$$

Reduce the fraction $\frac{75}{12}$ to lowest terms by extracting and canceling out $3$.

$$\frac{25}{4}$$

Show Solution

Factor

$\frac{5 ^ {2}}{2 ^ {2}} = 6\frac{1}{4} = 6.25$

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