Solve (575-3/7*153/4+22/35+144)

$$( 575 - \frac { 3 } { 7 } \times 15 \frac { 3 } { 4 } + 2 \frac { 2 } { 35 } + 144 )$$

Evaluate

$$\frac{100003}{140}\approx 714.307142857$$

Solution Steps

Multiply 15 and 4 to get 60.

$$575-\frac{3}{7}\times \frac{60+3}{4}+\frac{2\times 35+2}{35}+144$$

Add 60 and 3 to get 63.

$$575-\frac{3}{7}\times \frac{63}{4}+\frac{2\times 35+2}{35}+144$$

Multiply \frac{3}{7} times \frac{63}{4} by multiplying numerator times numerator and denominator times denominator.

$$575-\frac{3\times 63}{7\times 4}+\frac{2\times 35+2}{35}+144$$

Do the multiplications in the fraction \frac{3\times 63}{7\times 4}.

$$575-\frac{189}{28}+\frac{2\times 35+2}{35}+144$$

Reduce the fraction \frac{189}{28} to lowest terms by extracting and canceling out 7.

$$575-\frac{27}{4}+\frac{2\times 35+2}{35}+144$$

Convert 575 to fraction \frac{2300}{4}.

$$\frac{2300}{4}-\frac{27}{4}+\frac{2\times 35+2}{35}+144$$

Since \frac{2300}{4} and \frac{27}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2300-27}{4}+\frac{2\times 35+2}{35}+144$$

Subtract 27 from 2300 to get 2273.

$$\frac{2273}{4}+\frac{2\times 35+2}{35}+144$$

Multiply 2 and 35 to get 70.

$$\frac{2273}{4}+\frac{70+2}{35}+144$$

Add 70 and 2 to get 72.

$$\frac{2273}{4}+\frac{72}{35}+144$$

Least common multiple of 4 and 35 is 140. Convert \frac{2273}{4} and \frac{72}{35} to fractions with denominator 140.

$$\frac{79555}{140}+\frac{288}{140}+144$$

Since \frac{79555}{140} and \frac{288}{140} have the same denominator, add them by adding their numerators.

$$\frac{79555+288}{140}+144$$

Add 79555 and 288 to get 79843.

$$\frac{79843}{140}+144$$

Convert 144 to fraction \frac{20160}{140}.

$$\frac{79843}{140}+\frac{20160}{140}$$

Since \frac{79843}{140} and \frac{20160}{140} have the same denominator, add them by adding their numerators.

$$\frac{79843+20160}{140}$$

Add 79843 and 20160 to get 100003.

$$\frac{100003}{140}$$

Factor

$$\frac{100003}{2 ^ {2} \cdot 5 \cdot 7} = 714\frac{43}{140} = 714.3071428571428$$