Solve (15-43/20)-(12-38/15)

$$(15-4 \frac{ 3 }{ 20 } )-(12-3 \frac{ 8 }{ 15 } )$$

Evaluate

$$\frac{143}{60}\approx 2.383333333$$

Solution Steps

Multiply 4 and 20 to get 80.

$$15-\frac{80+3}{20}-\left(12-\frac{3\times 15+8}{15}\right)$$

Add 80 and 3 to get 83.

$$15-\frac{83}{20}-\left(12-\frac{3\times 15+8}{15}\right)$$

Convert 15 to fraction \frac{300}{20}.

$$\frac{300}{20}-\frac{83}{20}-\left(12-\frac{3\times 15+8}{15}\right)$$

Since \frac{300}{20} and \frac{83}{20} have the same denominator, subtract them by subtracting their numerators.

$$\frac{300-83}{20}-\left(12-\frac{3\times 15+8}{15}\right)$$

Subtract 83 from 300 to get 217.

$$\frac{217}{20}-\left(12-\frac{3\times 15+8}{15}\right)$$

Multiply 3 and 15 to get 45.

$$\frac{217}{20}-\left(12-\frac{45+8}{15}\right)$$

Add 45 and 8 to get 53.

$$\frac{217}{20}-\left(12-\frac{53}{15}\right)$$

Convert 12 to fraction \frac{180}{15}.

$$\frac{217}{20}-\left(\frac{180}{15}-\frac{53}{15}\right)$$

Since \frac{180}{15} and \frac{53}{15} have the same denominator, subtract them by subtracting their numerators.

$$\frac{217}{20}-\frac{180-53}{15}$$

Subtract 53 from 180 to get 127.

$$\frac{217}{20}-\frac{127}{15}$$

Least common multiple of 20 and 15 is 60. Convert \frac{217}{20} and \frac{127}{15} to fractions with denominator 60.

$$\frac{651}{60}-\frac{508}{60}$$

Since \frac{651}{60} and \frac{508}{60} have the same denominator, subtract them by subtracting their numerators.

$$\frac{651-508}{60}$$

Subtract 508 from 651 to get 143.

$$\frac{143}{60}$$

Factor

$$\frac{11 \cdot 13}{2 ^ {2} \cdot 3 \cdot 5} = 2\frac{23}{60} = 2.3833333333333333$$