Solve 1/3div1/2-16div1/4+2

$$\frac{ 1 }{ 3 } \div \frac{ 1 }{ 2 } -16 \div \frac{ 1 }{ 4 } +2$$

Evaluate

$$-\frac{184}{3}\approx -61.333333333$$

Solution Steps

Divide \frac{1}{3} by \frac{1}{2} by multiplying \frac{1}{3} by the reciprocal of \frac{1}{2}.

$$\frac{1}{3}\times 2-\frac{16}{\frac{1}{4}}+2$$

Multiply \frac{1}{3} and 2 to get \frac{2}{3}.

$$\frac{2}{3}-\frac{16}{\frac{1}{4}}+2$$

Divide 16 by \frac{1}{4} by multiplying 16 by the reciprocal of \frac{1}{4}.

$$\frac{2}{3}-16\times 4+2$$

Multiply 16 and 4 to get 64.

$$\frac{2}{3}-64+2$$

Convert 64 to fraction \frac{192}{3}.

$$\frac{2}{3}-\frac{192}{3}+2$$

Since \frac{2}{3} and \frac{192}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2-192}{3}+2$$

Subtract 192 from 2 to get -190.

$$-\frac{190}{3}+2$$

Convert 2 to fraction \frac{6}{3}.

$$-\frac{190}{3}+\frac{6}{3}$$

Since -\frac{190}{3} and \frac{6}{3} have the same denominator, add them by adding their numerators.

$$\frac{-190+6}{3}$$

Add -190 and 6 to get -184.

$$-\frac{184}{3}$$

Factor

$$-\frac{184}{3} = -61\frac{1}{3} = -61.333333333333336$$