Solve 17/9div11/3

$$1 \frac { 7 } { 9 } \div 1 \frac { 1 } { 3 }$$

Evaluate

$$\frac{4}{3}\approx 1.333333333$$

Solution Steps

Divide \frac{1\times 9+7}{9} by \frac{1\times 3+1}{3} by multiplying \frac{1\times 9+7}{9} by the reciprocal of \frac{1\times 3+1}{3}.

$$\frac{\left(1\times 9+7\right)\times 3}{9\left(1\times 3+1\right)}$$

Cancel out 3 in both numerator and denominator.

$$\frac{7+9}{3\left(1+3\right)}$$

Add 7 and 9 to get 16.

$$\frac{16}{3\left(1+3\right)}$$

Add 1 and 3 to get 4.

$$\frac{16}{3\times 4}$$

Multiply 3 and 4 to get 12.

$$\frac{16}{12}$$

Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.

$$\frac{4}{3}$$

Factor

$$\frac{2 ^ {2}}{3} = 1\frac{1}{3} = 1.3333333333333333$$