Solve 72/3+(6-24/7):22/5-4x=

$$7 \frac { 2 } { 3 } + ( 6 - 2 \frac { 4 } { 7 } ) : 2 \frac { 2 } { 5 } - 4 x =$$

Evaluate

$$\frac{191}{21}-4x$$

Solution Steps

Multiply 7 and 3 to get 21.

$$\frac{21+2}{3}+\frac{6-\frac{2\times 7+4}{7}}{\frac{2\times 5+2}{5}}-4x$$

Add 21 and 2 to get 23.

$$\frac{23}{3}+\frac{6-\frac{2\times 7+4}{7}}{\frac{2\times 5+2}{5}}-4x$$

Multiply 2 and 7 to get 14.

$$\frac{23}{3}+\frac{6-\frac{14+4}{7}}{\frac{2\times 5+2}{5}}-4x$$

Add 14 and 4 to get 18.

$$\frac{23}{3}+\frac{6-\frac{18}{7}}{\frac{2\times 5+2}{5}}-4x$$

Convert 6 to fraction \frac{42}{7}.

$$\frac{23}{3}+\frac{\frac{42}{7}-\frac{18}{7}}{\frac{2\times 5+2}{5}}-4x$$

Since \frac{42}{7} and \frac{18}{7} have the same denominator, subtract them by subtracting their numerators.

$$\frac{23}{3}+\frac{\frac{42-18}{7}}{\frac{2\times 5+2}{5}}-4x$$

Subtract 18 from 42 to get 24.

$$\frac{23}{3}+\frac{\frac{24}{7}}{\frac{2\times 5+2}{5}}-4x$$

Multiply 2 and 5 to get 10.

$$\frac{23}{3}+\frac{\frac{24}{7}}{\frac{10+2}{5}}-4x$$

Add 10 and 2 to get 12.

$$\frac{23}{3}+\frac{\frac{24}{7}}{\frac{12}{5}}-4x$$

Divide \frac{24}{7} by \frac{12}{5} by multiplying \frac{24}{7} by the reciprocal of \frac{12}{5}.

$$\frac{23}{3}+\frac{24}{7}\times \frac{5}{12}-4x$$

Multiply \frac{24}{7} times \frac{5}{12} by multiplying numerator times numerator and denominator times denominator.

$$\frac{23}{3}+\frac{24\times 5}{7\times 12}-4x$$

Do the multiplications in the fraction \frac{24\times 5}{7\times 12}.

$$\frac{23}{3}+\frac{120}{84}-4x$$

Reduce the fraction \frac{120}{84} to lowest terms by extracting and canceling out 12.

$$\frac{23}{3}+\frac{10}{7}-4x$$

Least common multiple of 3 and 7 is 21. Convert \frac{23}{3} and \frac{10}{7} to fractions with denominator 21.

$$\frac{161}{21}+\frac{30}{21}-4x$$

Since \frac{161}{21} and \frac{30}{21} have the same denominator, add them by adding their numerators.

$$\frac{161+30}{21}-4x$$

Add 161 and 30 to get 191.

$$\frac{191}{21}-4x$$

Expand

$$\frac{191}{21}-4x$$

Solution Steps

Multiply 7 and 3 to get 21.

$$\frac{21+2}{3}+\frac{6-\frac{2\times 7+4}{7}}{\frac{2\times 5+2}{5}}-4x$$

Add 21 and 2 to get 23.

$$\frac{23}{3}+\frac{6-\frac{2\times 7+4}{7}}{\frac{2\times 5+2}{5}}-4x$$

Multiply 2 and 7 to get 14.

$$\frac{23}{3}+\frac{6-\frac{14+4}{7}}{\frac{2\times 5+2}{5}}-4x$$

Add 14 and 4 to get 18.

$$\frac{23}{3}+\frac{6-\frac{18}{7}}{\frac{2\times 5+2}{5}}-4x$$

Convert 6 to fraction \frac{42}{7}.

$$\frac{23}{3}+\frac{\frac{42}{7}-\frac{18}{7}}{\frac{2\times 5+2}{5}}-4x$$

Since \frac{42}{7} and \frac{18}{7} have the same denominator, subtract them by subtracting their numerators.

$$\frac{23}{3}+\frac{\frac{42-18}{7}}{\frac{2\times 5+2}{5}}-4x$$

Subtract 18 from 42 to get 24.

$$\frac{23}{3}+\frac{\frac{24}{7}}{\frac{2\times 5+2}{5}}-4x$$

Multiply 2 and 5 to get 10.

$$\frac{23}{3}+\frac{\frac{24}{7}}{\frac{10+2}{5}}-4x$$

Add 10 and 2 to get 12.

$$\frac{23}{3}+\frac{\frac{24}{7}}{\frac{12}{5}}-4x$$

Divide \frac{24}{7} by \frac{12}{5} by multiplying \frac{24}{7} by the reciprocal of \frac{12}{5}.

$$\frac{23}{3}+\frac{24}{7}\times \frac{5}{12}-4x$$

Multiply \frac{24}{7} times \frac{5}{12} by multiplying numerator times numerator and denominator times denominator.

$$\frac{23}{3}+\frac{24\times 5}{7\times 12}-4x$$

Do the multiplications in the fraction \frac{24\times 5}{7\times 12}.

$$\frac{23}{3}+\frac{120}{84}-4x$$

Reduce the fraction \frac{120}{84} to lowest terms by extracting and canceling out 12.

$$\frac{23}{3}+\frac{10}{7}-4x$$

Least common multiple of 3 and 7 is 21. Convert \frac{23}{3} and \frac{10}{7} to fractions with denominator 21.

$$\frac{161}{21}+\frac{30}{21}-4x$$

Since \frac{161}{21} and \frac{30}{21} have the same denominator, add them by adding their numerators.

$$\frac{161+30}{21}-4x$$

Add 161 and 30 to get 191.

$$\frac{191}{21}-4x$$