Solve 16\div42-38+2\times12\div24\div6\times2+4

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Question

$$16 \div 42-38+2 \times 12 \div 24 \div 6 \times 2+4$$

Evaluate

$-\frac{233}{7}\approx -33.285714286$

Solution Steps

Reduce the fraction $\frac{16}{42}$ to lowest terms by extracting and canceling out $2$.

$$\frac{8}{21}-38+\frac{\frac{2\times 12}{24}}{6}\times 2+4$$

Convert $38$ to fraction $\frac{798}{21}$.

$$\frac{8}{21}-\frac{798}{21}+\frac{\frac{2\times 12}{24}}{6}\times 2+4$$

Since $\frac{8}{21}$ and $\frac{798}{21}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{8-798}{21}+\frac{\frac{2\times 12}{24}}{6}\times 2+4$$

Subtract $798$ from $8$ to get $-790$.

$$-\frac{790}{21}+\frac{\frac{2\times 12}{24}}{6}\times 2+4$$

Express $\frac{\frac{2\times 12}{24}}{6}$ as a single fraction.

$$-\frac{790}{21}+\frac{2\times 12}{24\times 6}\times 2+4$$

Cancel out $2\times 12$ in both numerator and denominator.

$$-\frac{790}{21}+\frac{1}{6}\times 2+4$$

Multiply $\frac{1}{6}$ and $2$ to get $\frac{2}{6}$.

$$-\frac{790}{21}+\frac{2}{6}+4$$

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

$$-\frac{790}{21}+\frac{1}{3}+4$$

Least common multiple of $21$ and $3$ is $21$. Convert $-\frac{790}{21}$ and $\frac{1}{3}$ to fractions with denominator $21$.

$$-\frac{790}{21}+\frac{7}{21}+4$$

Since $-\frac{790}{21}$ and $\frac{7}{21}$ have the same denominator, add them by adding their numerators.

$$\frac{-790+7}{21}+4$$

Add $-790$ and $7$ to get $-783$.

$$\frac{-783}{21}+4$$

Reduce the fraction $\frac{-783}{21}$ to lowest terms by extracting and canceling out $3$.

$$-\frac{261}{7}+4$$

Convert $4$ to fraction $\frac{28}{7}$.

$$-\frac{261}{7}+\frac{28}{7}$$

Since $-\frac{261}{7}$ and $\frac{28}{7}$ have the same denominator, add them by adding their numerators.

$$\frac{-261+28}{7}$$

Add $-261$ and $28$ to get $-233$.

$$-\frac{233}{7}$$

Show Solution

Factor

$-\frac{233}{7} = -33\frac{2}{7} = -33.285714285714285$