Solve {l}{12*4+1/2+1/3}{+1.+4/98}{-(1/5+frac{7}{6}+frac{1}{10})}

$$\left. \begin{array} { l } { 12 \times 4 + \frac { 1 } { 2 } + \frac { 1 } { 3 } } \\ { + 1 . + \frac { 4 } { 98 } } \\ { - ( \frac { 1 } { 5 } + \frac { 7 } { 6 } + \frac { 1 } { 10 } ) } \end{array} \right.$$

Evaluate

$$\frac{299}{6}\approx 49.833333333$$

Solution Steps

Multiply 12 and 4 to get 48.

$$48+\frac{1}{2}+\frac{1}{3}+1$$

Convert 48 to fraction \frac{96}{2}.

$$\frac{96}{2}+\frac{1}{2}+\frac{1}{3}+1$$

Since \frac{96}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

$$\frac{96+1}{2}+\frac{1}{3}+1$$

Add 96 and 1 to get 97.

$$\frac{97}{2}+\frac{1}{3}+1$$

Least common multiple of 2 and 3 is 6. Convert \frac{97}{2} and \frac{1}{3} to fractions with denominator 6.

$$\frac{291}{6}+\frac{2}{6}+1$$

Since \frac{291}{6} and \frac{2}{6} have the same denominator, add them by adding their numerators.

$$\frac{291+2}{6}+1$$

Add 291 and 2 to get 293.

$$\frac{293}{6}+1$$

Convert 1 to fraction \frac{6}{6}.

$$\frac{293}{6}+\frac{6}{6}$$

Since \frac{293}{6} and \frac{6}{6} have the same denominator, add them by adding their numerators.

$$\frac{293+6}{6}$$

Add 293 and 6 to get 299.

$$\frac{299}{6}$$

Factor

$$\frac{13 \cdot 23}{2 \cdot 3} = 49\frac{5}{6} = 49.833333333333336$$