Solve 3 6 1/4 + \{4 1/8 - (3/4 + 1/8)\}

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Question

$$6 \frac { 1 } { 4 } + \{ 4 \frac { 1 } { 8 } - ( \frac { 3 } { 4 } + \frac { 1 } { 8 } ) \}$$

Evaluate

$\frac{19}{2}=9.5$

Solution Steps

Multiply $6$ and $4$ to get $24$.

$$\frac{24+1}{4}+\frac{4\times 8+1}{8}-\left(\frac{3}{4}+\frac{1}{8}\right)$$

Add $24$ and $1$ to get $25$.

$$\frac{25}{4}+\frac{4\times 8+1}{8}-\left(\frac{3}{4}+\frac{1}{8}\right)$$

Multiply $4$ and $8$ to get $32$.

$$\frac{25}{4}+\frac{32+1}{8}-\left(\frac{3}{4}+\frac{1}{8}\right)$$

Add $32$ and $1$ to get $33$.

$$\frac{25}{4}+\frac{33}{8}-\left(\frac{3}{4}+\frac{1}{8}\right)$$

Least common multiple of $4$ and $8$ is $8$. Convert $\frac{3}{4}$ and $\frac{1}{8}$ to fractions with denominator $8$.

$$\frac{25}{4}+\frac{33}{8}-\left(\frac{6}{8}+\frac{1}{8}\right)$$

Since $\frac{6}{8}$ and $\frac{1}{8}$ have the same denominator, add them by adding their numerators.

$$\frac{25}{4}+\frac{33}{8}-\frac{6+1}{8}$$

Add $6$ and $1$ to get $7$.

$$\frac{25}{4}+\frac{33}{8}-\frac{7}{8}$$

Since $\frac{33}{8}$ and $\frac{7}{8}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{25}{4}+\frac{33-7}{8}$$

Subtract $7$ from $33$ to get $26$.

$$\frac{25}{4}+\frac{26}{8}$$

Reduce the fraction $\frac{26}{8}$ to lowest terms by extracting and canceling out $2$.

$$\frac{25}{4}+\frac{13}{4}$$

Since $\frac{25}{4}$ and $\frac{13}{4}$ have the same denominator, add them by adding their numerators.

$$\frac{25+13}{4}$$

Add $25$ and $13$ to get $38$.

$$\frac{38}{4}$$

Reduce the fraction $\frac{38}{4}$ to lowest terms by extracting and canceling out $2$.

$$\frac{19}{2}$$

Show Solution

Factor

$\frac{19}{2} = 9\frac{1}{2} = 9.5$