Solve [(9/5+11/5)*5]+[(124/2-44/2):2]+frac{1998}{1998}=

$$[ ( \frac { 9 } { 5 } + \frac { 11 } { 5 } ) \times 5 ] + [ ( \frac { 124 } { 2 } - \frac { 44 } { 2 } ) : 2 ] + \frac { 1998 } { 1998 } =$$

Evaluate

$$41$$

Solution Steps

Divide 1998 by 1998 to get 1.

$$\left(\frac{9}{5}+\frac{11}{5}\right)\times 5+\frac{\frac{124}{2}-\frac{44}{2}}{2}+1$$

Since \frac{9}{5} and \frac{11}{5} have the same denominator, add them by adding their numerators.

$$\frac{9+11}{5}\times 5+\frac{\frac{124}{2}-\frac{44}{2}}{2}+1$$

Add 9 and 11 to get 20.

$$\frac{20}{5}\times 5+\frac{\frac{124}{2}-\frac{44}{2}}{2}+1$$

Divide 20 by 5 to get 4.

$$4\times 5+\frac{\frac{124}{2}-\frac{44}{2}}{2}+1$$

Multiply 4 and 5 to get 20.

$$20+\frac{\frac{124}{2}-\frac{44}{2}}{2}+1$$

Divide 124 by 2 to get 62.

$$20+\frac{62-\frac{44}{2}}{2}+1$$

Divide 44 by 2 to get 22.

$$20+\frac{62-22}{2}+1$$

Subtract 22 from 62 to get 40.

$$20+\frac{40}{2}+1$$

Divide 40 by 2 to get 20.

$$20+20+1$$

Add 20 and 20 to get 40.

$$40+1$$

Add 40 and 1 to get 41.

$$41$$

Factor

$$41$$