Solve 92/5div11/3=

$$9 \frac { 2 } { 5 } \div 1 \frac { 1 } { 3 } =$$

Evaluate

$$\frac{141}{20}=7.05$$

Solution Steps

Divide \frac{9\times 5+2}{5} by \frac{1\times 3+1}{3} by multiplying \frac{9\times 5+2}{5} by the reciprocal of \frac{1\times 3+1}{3}.

$$\frac{\left(9\times 5+2\right)\times 3}{5\left(1\times 3+1\right)}$$

Multiply 9 and 5 to get 45.

$$\frac{\left(45+2\right)\times 3}{5\left(1\times 3+1\right)}$$

Add 45 and 2 to get 47.

$$\frac{47\times 3}{5\left(1\times 3+1\right)}$$

Multiply 47 and 3 to get 141.

$$\frac{141}{5\left(1\times 3+1\right)}$$

Multiply 1 and 3 to get 3.

$$\frac{141}{5\left(3+1\right)}$$

Add 3 and 1 to get 4.

$$\frac{141}{5\times 4}$$

Multiply 5 and 4 to get 20.

$$\frac{141}{20}$$

Factor

$$\frac{3 \cdot 47}{2 ^ {2} \cdot 5} = 7\frac{1}{20} = 7.05$$