Solve 21/4+55/8*11/3-13/4div1frac{1}{2}

$$2 \frac { 1 } { 4 } + 5 \frac { 5 } { 8 } \times 1 \frac { 1 } { 3 } - 1 \frac { 3 } { 4 } \div 1 \frac { 1 } { 2 }$$

Evaluate

$$\frac{103}{12}\approx 8.583333333$$

Solution Steps

Multiply 2 and 4 to get 8.

$$\frac{8+1}{4}+\frac{5\times 8+5}{8}\times \frac{1\times 3+1}{3}-\frac{\frac{1\times 4+3}{4}}{\frac{1\times 2+1}{2}}$$

Add 8 and 1 to get 9.

$$\frac{9}{4}+\frac{5\times 8+5}{8}\times \frac{1\times 3+1}{3}-\frac{\frac{1\times 4+3}{4}}{\frac{1\times 2+1}{2}}$$

Multiply 5 and 8 to get 40.

$$\frac{9}{4}+\frac{40+5}{8}\times \frac{1\times 3+1}{3}-\frac{\frac{1\times 4+3}{4}}{\frac{1\times 2+1}{2}}$$

Add 40 and 5 to get 45.

$$\frac{9}{4}+\frac{45}{8}\times \frac{1\times 3+1}{3}-\frac{\frac{1\times 4+3}{4}}{\frac{1\times 2+1}{2}}$$

Multiply 1 and 3 to get 3.

$$\frac{9}{4}+\frac{45}{8}\times \frac{3+1}{3}-\frac{\frac{1\times 4+3}{4}}{\frac{1\times 2+1}{2}}$$

Add 3 and 1 to get 4.

$$\frac{9}{4}+\frac{45}{8}\times \frac{4}{3}-\frac{\frac{1\times 4+3}{4}}{\frac{1\times 2+1}{2}}$$

Multiply \frac{45}{8} times \frac{4}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{9}{4}+\frac{45\times 4}{8\times 3}-\frac{\frac{1\times 4+3}{4}}{\frac{1\times 2+1}{2}}$$

Do the multiplications in the fraction \frac{45\times 4}{8\times 3}.

$$\frac{9}{4}+\frac{180}{24}-\frac{\frac{1\times 4+3}{4}}{\frac{1\times 2+1}{2}}$$

Reduce the fraction \frac{180}{24} to lowest terms by extracting and canceling out 12.

$$\frac{9}{4}+\frac{15}{2}-\frac{\frac{1\times 4+3}{4}}{\frac{1\times 2+1}{2}}$$

Least common multiple of 4 and 2 is 4. Convert \frac{9}{4} and \frac{15}{2} to fractions with denominator 4.

$$\frac{9}{4}+\frac{30}{4}-\frac{\frac{1\times 4+3}{4}}{\frac{1\times 2+1}{2}}$$

Since \frac{9}{4} and \frac{30}{4} have the same denominator, add them by adding their numerators.

$$\frac{9+30}{4}-\frac{\frac{1\times 4+3}{4}}{\frac{1\times 2+1}{2}}$$

Add 9 and 30 to get 39.

$$\frac{39}{4}-\frac{\frac{1\times 4+3}{4}}{\frac{1\times 2+1}{2}}$$

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

$$\frac{39}{4}-\frac{\left(1\times 4+3\right)\times 2}{4\left(1\times 2+1\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{39}{4}-\frac{3+4}{2\left(1+2\right)}$$

Add 3 and 4 to get 7.

$$\frac{39}{4}-\frac{7}{2\left(1+2\right)}$$

Add 1 and 2 to get 3.

$$\frac{39}{4}-\frac{7}{2\times 3}$$

Multiply 2 and 3 to get 6.

$$\frac{39}{4}-\frac{7}{6}$$

Least common multiple of 4 and 6 is 12. Convert \frac{39}{4} and \frac{7}{6} to fractions with denominator 12.

$$\frac{117}{12}-\frac{14}{12}$$

Since \frac{117}{12} and \frac{14}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{117-14}{12}$$

Subtract 14 from 117 to get 103.

$$\frac{103}{12}$$

Factor

$$\frac{103}{2 ^ {2} \cdot 3} = 8\frac{7}{12} = 8.583333333333334$$