Solve 153/7div123/49

$$15 \frac { 3 } { 7 } \div 1 \frac { 23 } { 49 }$$

Evaluate

$$\frac{21}{2}=10.5$$

Solution Steps

Divide \frac{15\times 7+3}{7} by \frac{1\times 49+23}{49} by multiplying \frac{15\times 7+3}{7} by the reciprocal of \frac{1\times 49+23}{49}.

$$\frac{\left(15\times 7+3\right)\times 49}{7\left(1\times 49+23\right)}$$

Cancel out 7 in both numerator and denominator.

$$\frac{7\left(3+7\times 15\right)}{23+49}$$

Multiply 7 and 15 to get 105.

$$\frac{7\left(3+105\right)}{23+49}$$

Add 3 and 105 to get 108.

$$\frac{7\times 108}{23+49}$$

Multiply 7 and 108 to get 756.

$$\frac{756}{23+49}$$

Add 23 and 49 to get 72.

$$\frac{756}{72}$$

Reduce the fraction \frac{756}{72} to lowest terms by extracting and canceling out 36.

$$\frac{21}{2}$$

Factor

$$\frac{3 \cdot 7}{2} = 10\frac{1}{2} = 10.5$$