Solve 1/3+5/6-1/9quad8/15

$$\frac { 1 } { 3 } + \frac { 5 } { 6 } - \frac { 1 } { 9 } \quad \frac { 8 } { 15 }$$

Evaluate

$$\frac{299}{270}\approx 1.107407407$$

Solution Steps

Least common multiple of 3 and 6 is 6. Convert \frac{1}{3} and \frac{5}{6} to fractions with denominator 6.

$$\frac{2}{6}+\frac{5}{6}-\frac{1}{9}\times \frac{8}{15}$$

Since \frac{2}{6} and \frac{5}{6} have the same denominator, add them by adding their numerators.

$$\frac{2+5}{6}-\frac{1}{9}\times \frac{8}{15}$$

Add 2 and 5 to get 7.

$$\frac{7}{6}-\frac{1}{9}\times \frac{8}{15}$$

Multiply \frac{1}{9} times \frac{8}{15} by multiplying numerator times numerator and denominator times denominator.

$$\frac{7}{6}-\frac{1\times 8}{9\times 15}$$

Do the multiplications in the fraction \frac{1\times 8}{9\times 15}.

$$\frac{7}{6}-\frac{8}{135}$$

Least common multiple of 6 and 135 is 270. Convert \frac{7}{6} and \frac{8}{135} to fractions with denominator 270.

$$\frac{315}{270}-\frac{16}{270}$$

Since \frac{315}{270} and \frac{16}{270} have the same denominator, subtract them by subtracting their numerators.

$$\frac{315-16}{270}$$

Subtract 16 from 315 to get 299.

$$\frac{299}{270}$$

Factor

$$\frac{13 \cdot 23}{2 \cdot 3 ^ {3} \cdot 5} = 1\frac{29}{270} = 1.1074074074074074$$