Solve (18.75+52/9)+(17/9-83/4)

$$(18.75+ 5 \frac { 2 } { 9 } )+( 1 \frac { 7 } { 9 } - 8 \frac { 3 } { 4 } )$$

Evaluate

$$17$$

Solution Steps

Multiply 5 and 9 to get 45.

$$18.75+\frac{45+2}{9}+\frac{1\times 9+7}{9}-\frac{8\times 4+3}{4}$$

Add 45 and 2 to get 47.

$$18.75+\frac{47}{9}+\frac{1\times 9+7}{9}-\frac{8\times 4+3}{4}$$

Convert decimal number 18.75 to fraction \frac{1875}{100}. Reduce the fraction \frac{1875}{100} to lowest terms by extracting and canceling out 25.

$$\frac{75}{4}+\frac{47}{9}+\frac{1\times 9+7}{9}-\frac{8\times 4+3}{4}$$

Least common multiple of 4 and 9 is 36. Convert \frac{75}{4} and \frac{47}{9} to fractions with denominator 36.

$$\frac{675}{36}+\frac{188}{36}+\frac{1\times 9+7}{9}-\frac{8\times 4+3}{4}$$

Since \frac{675}{36} and \frac{188}{36} have the same denominator, add them by adding their numerators.

$$\frac{675+188}{36}+\frac{1\times 9+7}{9}-\frac{8\times 4+3}{4}$$

Add 675 and 188 to get 863.

$$\frac{863}{36}+\frac{1\times 9+7}{9}-\frac{8\times 4+3}{4}$$

Multiply 1 and 9 to get 9.

$$\frac{863}{36}+\frac{9+7}{9}-\frac{8\times 4+3}{4}$$

Add 9 and 7 to get 16.

$$\frac{863}{36}+\frac{16}{9}-\frac{8\times 4+3}{4}$$

Least common multiple of 36 and 9 is 36. Convert \frac{863}{36} and \frac{16}{9} to fractions with denominator 36.

$$\frac{863}{36}+\frac{64}{36}-\frac{8\times 4+3}{4}$$

Since \frac{863}{36} and \frac{64}{36} have the same denominator, add them by adding their numerators.

$$\frac{863+64}{36}-\frac{8\times 4+3}{4}$$

Add 863 and 64 to get 927.

$$\frac{927}{36}-\frac{8\times 4+3}{4}$$

Reduce the fraction \frac{927}{36} to lowest terms by extracting and canceling out 9.

$$\frac{103}{4}-\frac{8\times 4+3}{4}$$

Multiply 8 and 4 to get 32.

$$\frac{103}{4}-\frac{32+3}{4}$$

Add 32 and 3 to get 35.

$$\frac{103}{4}-\frac{35}{4}$$

Since \frac{103}{4} and \frac{35}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{103-35}{4}$$

Subtract 35 from 103 to get 68.

$$\frac{68}{4}$$

Divide 68 by 4 to get 17.

$$17$$

Factor

$$17$$