Solve 45/9-(1/9+31/3)

$$4 \frac { 5 } { 9 } - ( \frac { 1 } { 9 } + 3 \frac { 1 } { 3 } )$$

Evaluate

$$\frac{10}{9}\approx 1.111111111$$

Solution Steps

Multiply 4 and 9 to get 36.

$$\frac{36+5}{9}-\left(\frac{1}{9}+\frac{3\times 3+1}{3}\right)$$

Add 36 and 5 to get 41.

$$\frac{41}{9}-\left(\frac{1}{9}+\frac{3\times 3+1}{3}\right)$$

Multiply 3 and 3 to get 9.

$$\frac{41}{9}-\left(\frac{1}{9}+\frac{9+1}{3}\right)$$

Add 9 and 1 to get 10.

$$\frac{41}{9}-\left(\frac{1}{9}+\frac{10}{3}\right)$$

Least common multiple of 9 and 3 is 9. Convert \frac{1}{9} and \frac{10}{3} to fractions with denominator 9.

$$\frac{41}{9}-\left(\frac{1}{9}+\frac{30}{9}\right)$$

Since \frac{1}{9} and \frac{30}{9} have the same denominator, add them by adding their numerators.

$$\frac{41}{9}-\frac{1+30}{9}$$

Add 1 and 30 to get 31.

$$\frac{41}{9}-\frac{31}{9}$$

Since \frac{41}{9} and \frac{31}{9} have the same denominator, subtract them by subtracting their numerators.

$$\frac{41-31}{9}$$

Subtract 31 from 41 to get 10.

$$\frac{10}{9}$$

Factor

$$\frac{2 \cdot 5}{3 ^ {2}} = 1\frac{1}{9} = 1.1111111111111112$$