Solve 19+238/20-(71)|3=

$$19 + \frac { 238 } { 20 } - ( 71 ) | 3 =$$

Evaluate

$$-\frac{1821}{10}=-182.1$$

Solution Steps

Reduce the fraction \frac{238}{20} to lowest terms by extracting and canceling out 2.

$$19+\frac{119}{10}-71|3|$$

Convert 19 to fraction \frac{190}{10}.

$$\frac{190}{10}+\frac{119}{10}-71|3|$$

Since \frac{190}{10} and \frac{119}{10} have the same denominator, add them by adding their numerators.

$$\frac{190+119}{10}-71|3|$$

Add 190 and 119 to get 309.

$$\frac{309}{10}-71|3|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 3 is 3.

$$\frac{309}{10}-71\times 3$$

Multiply 71 and 3 to get 213.

$$\frac{309}{10}-213$$

Convert 213 to fraction \frac{2130}{10}.

$$\frac{309}{10}-\frac{2130}{10}$$

Since \frac{309}{10} and \frac{2130}{10} have the same denominator, subtract them by subtracting their numerators.

$$\frac{309-2130}{10}$$

Subtract 2130 from 309 to get -1821.

$$-\frac{1821}{10}$$

Factor

$$-\frac{1821}{10} = -182\frac{1}{10} = -182.1$$