Solve 325/9-105/3+3=

$$3 \frac{ 25 }{ 9 } -10 \frac{ 5 }{ 3 } +3=$$

Evaluate

$$-\frac{26}{9}\approx -2.888888889$$

Solution Steps

Multiply 3 and 9 to get 27.

$$\frac{27+25}{9}-\frac{10\times 3+5}{3}+3$$

Add 27 and 25 to get 52.

$$\frac{52}{9}-\frac{10\times 3+5}{3}+3$$

Multiply 10 and 3 to get 30.

$$\frac{52}{9}-\frac{30+5}{3}+3$$

Add 30 and 5 to get 35.

$$\frac{52}{9}-\frac{35}{3}+3$$

Least common multiple of 9 and 3 is 9. Convert \frac{52}{9} and \frac{35}{3} to fractions with denominator 9.

$$\frac{52}{9}-\frac{105}{9}+3$$

Since \frac{52}{9} and \frac{105}{9} have the same denominator, subtract them by subtracting their numerators.

$$\frac{52-105}{9}+3$$

Subtract 105 from 52 to get -53.

$$-\frac{53}{9}+3$$

Convert 3 to fraction \frac{27}{9}.

$$-\frac{53}{9}+\frac{27}{9}$$

Since -\frac{53}{9} and \frac{27}{9} have the same denominator, add them by adding their numerators.

$$\frac{-53+27}{9}$$

Add -53 and 27 to get -26.

$$-\frac{26}{9}$$

Factor

$$-\frac{26}{9} = -2\frac{8}{9} = -2.888888888888889$$