Solve 1-(25/42-2/7)+3/5

$$1-( \frac{ 25 }{ 42 } - \frac{ 2 }{ 7 } )+ \frac{ 3 }{ 5 }$$

Evaluate

$$\frac{271}{210}\approx 1.29047619$$

Solution Steps

Least common multiple of 42 and 7 is 42. Convert \frac{25}{42} and \frac{2}{7} to fractions with denominator 42.

$$1-\left(\frac{25}{42}-\frac{12}{42}\right)+\frac{3}{5}$$

Since \frac{25}{42} and \frac{12}{42} have the same denominator, subtract them by subtracting their numerators.

$$1-\frac{25-12}{42}+\frac{3}{5}$$

Subtract 12 from 25 to get 13.

$$1-\frac{13}{42}+\frac{3}{5}$$

Convert 1 to fraction \frac{42}{42}.

$$\frac{42}{42}-\frac{13}{42}+\frac{3}{5}$$

Since \frac{42}{42} and \frac{13}{42} have the same denominator, subtract them by subtracting their numerators.

$$\frac{42-13}{42}+\frac{3}{5}$$

Subtract 13 from 42 to get 29.

$$\frac{29}{42}+\frac{3}{5}$$

Least common multiple of 42 and 5 is 210. Convert \frac{29}{42} and \frac{3}{5} to fractions with denominator 210.

$$\frac{145}{210}+\frac{126}{210}$$

Since \frac{145}{210} and \frac{126}{210} have the same denominator, add them by adding their numerators.

$$\frac{145+126}{210}$$

Add 145 and 126 to get 271.

$$\frac{271}{210}$$

Factor

$$\frac{271}{2 \cdot 3 \cdot 5 \cdot 7} = 1\frac{61}{210} = 1.2904761904761906$$