Solve (5/7-1/49):x=[3/2-(6/15+frac{8}{15})]:(frac{3}{4}+frac{7}{8}-frac{2}{5})

$$( \frac { 5 } { 7 } - \frac { 1 } { 49 } ) : x = [ \frac { 3 } { 2 } - ( \frac { 6 } { 15 } + \frac { 8 } { 15 } ) ] : ( \frac { 3 } { 4 } + \frac { 7 } { 8 } - \frac { 2 } { 5 } )$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

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.

$$\frac{5}{7}-\frac{1}{49}=\frac{40}{49}x\left(\frac{3}{2}-\left(\frac{6}{15}+\frac{8}{15}\right)\right)$$

Least common multiple of 7 and 49 is 49. Convert \frac{5}{7} and \frac{1}{49} to fractions with denominator 49.

$$\frac{35}{49}-\frac{1}{49}=\frac{40}{49}x\left(\frac{3}{2}-\left(\frac{6}{15}+\frac{8}{15}\right)\right)$$

Since \frac{35}{49} and \frac{1}{49} have the same denominator, subtract them by subtracting their numerators.

$$\frac{35-1}{49}=\frac{40}{49}x\left(\frac{3}{2}-\left(\frac{6}{15}+\frac{8}{15}\right)\right)$$

Subtract 1 from 35 to get 34.

$$\frac{34}{49}=\frac{40}{49}x\left(\frac{3}{2}-\left(\frac{6}{15}+\frac{8}{15}\right)\right)$$

Reduce the fraction \frac{6}{15} to lowest terms by extracting and canceling out 3.

$$\frac{34}{49}=\frac{40}{49}x\left(\frac{3}{2}-\left(\frac{2}{5}+\frac{8}{15}\right)\right)$$

Least common multiple of 5 and 15 is 15. Convert \frac{2}{5} and \frac{8}{15} to fractions with denominator 15.

$$\frac{34}{49}=\frac{40}{49}x\left(\frac{3}{2}-\left(\frac{6}{15}+\frac{8}{15}\right)\right)$$

Since \frac{6}{15} and \frac{8}{15} have the same denominator, add them by adding their numerators.

$$\frac{34}{49}=\frac{40}{49}x\left(\frac{3}{2}-\frac{6+8}{15}\right)$$

Add 6 and 8 to get 14.

$$\frac{34}{49}=\frac{40}{49}x\left(\frac{3}{2}-\frac{14}{15}\right)$$

Least common multiple of 2 and 15 is 30. Convert \frac{3}{2} and \frac{14}{15} to fractions with denominator 30.

$$\frac{34}{49}=\frac{40}{49}x\left(\frac{45}{30}-\frac{28}{30}\right)$$

Since \frac{45}{30} and \frac{28}{30} have the same denominator, subtract them by subtracting their numerators.

$$\frac{34}{49}=\frac{40}{49}x\times \frac{45-28}{30}$$

Subtract 28 from 45 to get 17.

$$\frac{34}{49}=\frac{40}{49}x\times \frac{17}{30}$$

Multiply \frac{40}{49} times \frac{17}{30} by multiplying numerator times numerator and denominator times denominator.

$$\frac{34}{49}=\frac{40\times 17}{49\times 30}x$$

Do the multiplications in the fraction \frac{40\times 17}{49\times 30}.

$$\frac{34}{49}=\frac{680}{1470}x$$

Reduce the fraction \frac{680}{1470} to lowest terms by extracting and canceling out 10.

$$\frac{34}{49}=\frac{68}{147}x$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{68}{147}x=\frac{34}{49}$$

Multiply both sides by \frac{147}{68}, the reciprocal of \frac{68}{147}.

$$x=\frac{34}{49}\times \frac{147}{68}$$

Multiply \frac{34}{49} times \frac{147}{68} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{34\times 147}{49\times 68}$$

Do the multiplications in the fraction \frac{34\times 147}{49\times 68}.

$$x=\frac{4998}{3332}$$

Reduce the fraction \frac{4998}{3332} to lowest terms by extracting and canceling out 1666.

$$x=\frac{3}{2}$$