Solve (5/3-1/2):[2/7*(4+2/3)]=[(frac{21}{4}-3)*(2-frac{4}{5})-2]:x

$$( \frac { 5 } { 3 } - \frac { 1 } { 2 } ) : [ \frac { 2 } { 7 } \cdot ( 4 + \frac { 2 } { 3 } ) ] = [ ( \frac { 21 } { 4 } - 3 ) \cdot ( 2 - \frac { 4 } { 5 } ) - 2 ] : x$$

Solve for x

$$x=\frac{4}{5}=0.8$$

Steps for Solving Linear Equation

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$\frac{3}{4}x\left(\frac{5}{3}-\frac{1}{2}\right)=\left(\frac{21}{4}-3\right)\left(2-\frac{4}{5}\right)-2$$

Least common multiple of 3 and 2 is 6. Convert \frac{5}{3} and \frac{1}{2} to fractions with denominator 6.

$$\frac{3}{4}x\left(\frac{10}{6}-\frac{3}{6}\right)=\left(\frac{21}{4}-3\right)\left(2-\frac{4}{5}\right)-2$$

Since \frac{10}{6} and \frac{3}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3}{4}x\times \frac{10-3}{6}=\left(\frac{21}{4}-3\right)\left(2-\frac{4}{5}\right)-2$$

Subtract 3 from 10 to get 7.

$$\frac{3}{4}x\times \frac{7}{6}=\left(\frac{21}{4}-3\right)\left(2-\frac{4}{5}\right)-2$$

Multiply \frac{3}{4} times \frac{7}{6} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3\times 7}{4\times 6}x=\left(\frac{21}{4}-3\right)\left(2-\frac{4}{5}\right)-2$$

Do the multiplications in the fraction \frac{3\times 7}{4\times 6}.

$$\frac{21}{24}x=\left(\frac{21}{4}-3\right)\left(2-\frac{4}{5}\right)-2$$

Reduce the fraction \frac{21}{24} to lowest terms by extracting and canceling out 3.

$$\frac{7}{8}x=\left(\frac{21}{4}-3\right)\left(2-\frac{4}{5}\right)-2$$

Convert 3 to fraction \frac{12}{4}.

$$\frac{7}{8}x=\left(\frac{21}{4}-\frac{12}{4}\right)\left(2-\frac{4}{5}\right)-2$$

Since \frac{21}{4} and \frac{12}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7}{8}x=\frac{21-12}{4}\left(2-\frac{4}{5}\right)-2$$

Subtract 12 from 21 to get 9.

$$\frac{7}{8}x=\frac{9}{4}\left(2-\frac{4}{5}\right)-2$$

Convert 2 to fraction \frac{10}{5}.

$$\frac{7}{8}x=\frac{9}{4}\left(\frac{10}{5}-\frac{4}{5}\right)-2$$

Since \frac{10}{5} and \frac{4}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7}{8}x=\frac{9}{4}\times \frac{10-4}{5}-2$$

Subtract 4 from 10 to get 6.

$$\frac{7}{8}x=\frac{9}{4}\times \frac{6}{5}-2$$

Multiply \frac{9}{4} times \frac{6}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{7}{8}x=\frac{9\times 6}{4\times 5}-2$$

Do the multiplications in the fraction \frac{9\times 6}{4\times 5}.

$$\frac{7}{8}x=\frac{54}{20}-2$$

Reduce the fraction \frac{54}{20} to lowest terms by extracting and canceling out 2.

$$\frac{7}{8}x=\frac{27}{10}-2$$

Convert 2 to fraction \frac{20}{10}.

$$\frac{7}{8}x=\frac{27}{10}-\frac{20}{10}$$

Since \frac{27}{10} and \frac{20}{10} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7}{8}x=\frac{27-20}{10}$$

Subtract 20 from 27 to get 7.

$$\frac{7}{8}x=\frac{7}{10}$$

Multiply both sides by \frac{8}{7}, the reciprocal of \frac{7}{8}.

$$x=\frac{7}{10}\times \frac{8}{7}$$

Multiply \frac{7}{10} times \frac{8}{7} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{7\times 8}{10\times 7}$$

Cancel out 7 in both numerator and denominator.

$$x=\frac{8}{10}$$

Reduce the fraction \frac{8}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{5}$$