Solve {c}{(1/3+|-1/4|)*6/7-|-2|}{(1,1)}

$$\left. \begin{array} { c } { ( \frac { 1 } { 3 } + | - \frac { 1 } { 4 } | ) \cdot \frac { 6 } { 7 } - | - 2 | } \\ { ( 1,1 ) } \end{array} \right.$$

Sort

$$-1,5,\ 1,1$$

Solution Steps

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{1}{4} is \frac{1}{4}.

$$sort(\left(\frac{1}{3}+\frac{1}{4}\right)\times \frac{6}{7}-|-2|;1,1)$$

Least common multiple of 3 and 4 is 12. Convert \frac{1}{3} and \frac{1}{4} to fractions with denominator 12.

$$sort(\left(\frac{4}{12}+\frac{3}{12}\right)\times \frac{6}{7}-|-2|;1,1)$$

Since \frac{4}{12} and \frac{3}{12} have the same denominator, add them by adding their numerators.

$$sort(\frac{4+3}{12}\times \frac{6}{7}-|-2|;1,1)$$

Add 4 and 3 to get 7.

$$sort(\frac{7}{12}\times \frac{6}{7}-|-2|;1,1)$$

Multiply \frac{7}{12} times \frac{6}{7} by multiplying numerator times numerator and denominator times denominator.

$$sort(\frac{7\times 6}{12\times 7}-|-2|;1,1)$$

Cancel out 7 in both numerator and denominator.

$$sort(\frac{6}{12}-|-2|;1,1)$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

$$sort(\frac{1}{2}-|-2|;1,1)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -2 is 2.

$$sort(\frac{1}{2}-2;1,1)$$

Convert 2 to fraction \frac{4}{2}.

$$sort(\frac{1}{2}-\frac{4}{2};1,1)$$

Since \frac{1}{2} and \frac{4}{2} have the same denominator, subtract them by subtracting their numerators.

$$sort(\frac{1-4}{2};1,1)$$

Subtract 4 from 1 to get -3.

$$sort(-\frac{3}{2};1,1)$$

Convert decimal numbers in the list -\frac{3}{2};1,1 to fractions.

$$-\frac{3}{2};\frac{11}{10}$$

Least common denominator of the numbers in the list -\frac{3}{2};\frac{11}{10} is 10. Convert numbers in the list to fractions with denominator 10.

$$-\frac{15}{10};\frac{11}{10}$$

Evaluate

$$-1,5,\ 1,1$$