Solve 1/1|1/6+1/7

$$\frac { 1 } { 1 } | \frac { 1 } { 6 } + \frac { 1 } { 7 }$$

Evaluate

$$\frac{13}{42}\approx 0.30952381$$

Solution Steps

Divide 1 by 1 to get 1.

$$1|\frac{1}{6}+\frac{1}{7}|$$

Least common multiple of 6 and 7 is 42. Convert \frac{1}{6} and \frac{1}{7} to fractions with denominator 42.

$$1|\frac{7}{42}+\frac{6}{42}|$$

Since \frac{7}{42} and \frac{6}{42} have the same denominator, add them by adding their numerators.

$$1|\frac{7+6}{42}|$$

Add 7 and 6 to get 13.

$$1|\frac{13}{42}|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of \frac{13}{42} is \frac{13}{42}.

$$1\times \frac{13}{42}$$

Multiply 1 and \frac{13}{42} to get \frac{13}{42}.

$$\frac{13}{42}$$

Factor

$$\frac{13}{2 \cdot 3 \cdot 7} = 0.30952380952380953$$