Solve 31/4/2frac{1{4}}=191/2/11frac{1{4}}

$$\frac { 3 \frac { 1 } { 4 } } { 2 \frac { 1 } { 4 } } = \frac { 19 \frac { 1 } { 2 } } { 11 \frac { 1 } { 4 } }$$

Verify

Solution Steps

Divide \frac{3\times 4+1}{4} by \frac{2\times 4+1}{4} by multiplying \frac{3\times 4+1}{4} by the reciprocal of \frac{2\times 4+1}{4}.

$$\frac{\left(3\times 4+1\right)\times 4}{4\left(2\times 4+1\right)}=\frac{\frac{19\times 2+1}{2}}{\frac{11\times 4+1}{4}}$$

Cancel out 4 in both numerator and denominator.

$$\frac{1+3\times 4}{1+2\times 4}=\frac{\frac{19\times 2+1}{2}}{\frac{11\times 4+1}{4}}$$

Multiply 3 and 4 to get 12.

$$\frac{1+12}{1+2\times 4}=\frac{\frac{19\times 2+1}{2}}{\frac{11\times 4+1}{4}}$$

Add 1 and 12 to get 13.

$$\frac{13}{1+2\times 4}=\frac{\frac{19\times 2+1}{2}}{\frac{11\times 4+1}{4}}$$

Multiply 2 and 4 to get 8.

$$\frac{13}{1+8}=\frac{\frac{19\times 2+1}{2}}{\frac{11\times 4+1}{4}}$$

Add 1 and 8 to get 9.

$$\frac{13}{9}=\frac{\frac{19\times 2+1}{2}}{\frac{11\times 4+1}{4}}$$

Divide \frac{19\times 2+1}{2} by \frac{11\times 4+1}{4} by multiplying \frac{19\times 2+1}{2} by the reciprocal of \frac{11\times 4+1}{4}.

$$\frac{13}{9}=\frac{\left(19\times 2+1\right)\times 4}{2\left(11\times 4+1\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{13}{9}=\frac{2\left(1+2\times 19\right)}{1+4\times 11}$$

Multiply 2 and 19 to get 38.

$$\frac{13}{9}=\frac{2\left(1+38\right)}{1+4\times 11}$$

Add 1 and 38 to get 39.

$$\frac{13}{9}=\frac{2\times 39}{1+4\times 11}$$

Multiply 2 and 39 to get 78.

$$\frac{13}{9}=\frac{78}{1+4\times 11}$$

Multiply 4 and 11 to get 44.

$$\frac{13}{9}=\frac{78}{1+44}$$

Add 1 and 44 to get 45.

$$\frac{13}{9}=\frac{78}{45}$$

Reduce the fraction \frac{78}{45} to lowest terms by extracting and canceling out 3.

$$\frac{13}{9}=\frac{26}{15}$$

Least common multiple of 9 and 15 is 45. Convert \frac{13}{9} and \frac{26}{15} to fractions with denominator 45.

$$\frac{65}{45}=\frac{78}{45}$$

Compare \frac{65}{45} and \frac{78}{45}.

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