Solve 1/2(3/2x-2/3)=1/4(frac{2}{3}x+frac{1}{2})

$$\frac { 1 } { 2 } ( \frac { 3 } { 2 } x - \frac { 2 } { 3 } ) = \frac { 1 } { 4 } ( \frac { 2 } { 3 } x + \frac { 1 } { 2 } )$$

Solve for x

$$x=\frac{11}{14}\approx 0.785714286$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{1}{2} by \frac{3}{2}x-\frac{2}{3}.

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

Multiply \frac{1}{2} times \frac{3}{2} by multiplying numerator times numerator and denominator times denominator.

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

Do the multiplications in the fraction \frac{1\times 3}{2\times 2}.

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

Multiply \frac{1}{2} times -\frac{2}{3} by multiplying numerator times numerator and denominator times denominator.

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

Do the multiplications in the fraction \frac{1\left(-2\right)}{2\times 3}.

$$\frac{3}{4}x+\frac{-2}{6}=\frac{1}{4}\left(\frac{2}{3}x+\frac{1}{2}\right)$$

Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.

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

Use the distributive property to multiply \frac{1}{4} by \frac{2}{3}x+\frac{1}{2}.

$$\frac{3}{4}x-\frac{1}{3}=\frac{1}{4}\times \frac{2}{3}x+\frac{1}{4}\times \frac{1}{2}$$

Multiply \frac{1}{4} times \frac{2}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3}{4}x-\frac{1}{3}=\frac{1\times 2}{4\times 3}x+\frac{1}{4}\times \frac{1}{2}$$

Do the multiplications in the fraction \frac{1\times 2}{4\times 3}.

$$\frac{3}{4}x-\frac{1}{3}=\frac{2}{12}x+\frac{1}{4}\times \frac{1}{2}$$

Reduce the fraction \frac{2}{12} to lowest terms by extracting and canceling out 2.

$$\frac{3}{4}x-\frac{1}{3}=\frac{1}{6}x+\frac{1}{4}\times \frac{1}{2}$$

Multiply \frac{1}{4} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3}{4}x-\frac{1}{3}=\frac{1}{6}x+\frac{1\times 1}{4\times 2}$$

Do the multiplications in the fraction \frac{1\times 1}{4\times 2}.

$$\frac{3}{4}x-\frac{1}{3}=\frac{1}{6}x+\frac{1}{8}$$

Subtract \frac{1}{6}x from both sides.

$$\frac{3}{4}x-\frac{1}{3}-\frac{1}{6}x=\frac{1}{8}$$

Combine \frac{3}{4}x and -\frac{1}{6}x to get \frac{7}{12}x.

$$\frac{7}{12}x-\frac{1}{3}=\frac{1}{8}$$

Add \frac{1}{3} to both sides.

$$\frac{7}{12}x=\frac{1}{8}+\frac{1}{3}$$

Least common multiple of 8 and 3 is 24. Convert \frac{1}{8} and \frac{1}{3} to fractions with denominator 24.

$$\frac{7}{12}x=\frac{3}{24}+\frac{8}{24}$$

Since \frac{3}{24} and \frac{8}{24} have the same denominator, add them by adding their numerators.

$$\frac{7}{12}x=\frac{3+8}{24}$$

Add 3 and 8 to get 11.

$$\frac{7}{12}x=\frac{11}{24}$$

Multiply both sides by \frac{12}{7}, the reciprocal of \frac{7}{12}.

$$x=\frac{11}{24}\times \frac{12}{7}$$

Multiply \frac{11}{24} times \frac{12}{7} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{11\times 12}{24\times 7}$$

Do the multiplications in the fraction \frac{11\times 12}{24\times 7}.

$$x=\frac{132}{168}$$

Reduce the fraction \frac{132}{168} to lowest terms by extracting and canceling out 12.

$$x=\frac{11}{14}$$