Solve E=(3/5+8/3+3/9)*(5/4-frac{1}{4})]div(9+frac{2}{3}-frac{3}{4})

$$E = ( \frac { 3 } { 5 } + \frac { 8 } { 3 } + \frac { 3 } { 9 } ) \cdot ( \frac { 5 } { 4 } - \frac { 1 } { 4 } ) ] \div ( 9 + \frac { 2 } { 3 } - \frac { 3 } { 4 } )$$

Solve for E

$$E = \frac{18}{5} = 3\frac{3}{5} = 3.6$$

Steps for Solving Linear Equation

Least common multiple of 5 and 3 is 15. Convert \frac{3}{5} and \frac{8}{3} to fractions with denominator 15.

$$E=\left(\frac{9}{15}+\frac{40}{15}+\frac{3}{9}\right)\left(\frac{5}{4}-\frac{1}{4}\right)$$

Since \frac{9}{15} and \frac{40}{15} have the same denominator, add them by adding their numerators.

$$E=\left(\frac{9+40}{15}+\frac{3}{9}\right)\left(\frac{5}{4}-\frac{1}{4}\right)$$

Add 9 and 40 to get 49.

$$E=\left(\frac{49}{15}+\frac{3}{9}\right)\left(\frac{5}{4}-\frac{1}{4}\right)$$

Reduce the fraction \frac{3}{9} to lowest terms by extracting and canceling out 3.

$$E=\left(\frac{49}{15}+\frac{1}{3}\right)\left(\frac{5}{4}-\frac{1}{4}\right)$$

Least common multiple of 15 and 3 is 15. Convert \frac{49}{15} and \frac{1}{3} to fractions with denominator 15.

$$E=\left(\frac{49}{15}+\frac{5}{15}\right)\left(\frac{5}{4}-\frac{1}{4}\right)$$

Since \frac{49}{15} and \frac{5}{15} have the same denominator, add them by adding their numerators.

$$E=\frac{49+5}{15}\left(\frac{5}{4}-\frac{1}{4}\right)$$

Add 49 and 5 to get 54.

$$E=\frac{54}{15}\left(\frac{5}{4}-\frac{1}{4}\right)$$

Reduce the fraction \frac{54}{15} to lowest terms by extracting and canceling out 3.

$$E=\frac{18}{5}\left(\frac{5}{4}-\frac{1}{4}\right)$$

Since \frac{5}{4} and \frac{1}{4} have the same denominator, subtract them by subtracting their numerators.

$$E=\frac{18}{5}\times \frac{5-1}{4}$$

Subtract 1 from 5 to get 4.

$$E=\frac{18}{5}\times \frac{4}{4}$$

Divide 4 by 4 to get 1.

$$E=\frac{18}{5}\times 1$$

Multiply \frac{18}{5} and 1 to get \frac{18}{5}.

$$E=\frac{18}{5}$$

Assign E

$$E≔\frac{18}{5}$$