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

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

Solve for x

$$x = \frac{17}{7} = 2\frac{3}{7} \approx 2.428571429$$

Steps for Solving Linear Equation

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

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

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

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

Cancel out 4 in both numerator and denominator.

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

Fraction \frac{-1}{3} can be rewritten as -\frac{1}{3} by extracting the negative sign.

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

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

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

Do the multiplications in the fraction \frac{-\left(-2\right)}{4\times 3}.

$$\frac{1}{3}\left(\frac{1}{2}x-\frac{1}{3}x+\frac{2}{12}\right)=1-\frac{1}{3}x$$

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

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

Combine \frac{1}{2}x and -\frac{1}{3}x to get \frac{1}{6}x.

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

Use the distributive property to multiply \frac{1}{3} by \frac{1}{6}x+\frac{1}{6}.

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

Multiply \frac{1}{3} times \frac{1}{6} by multiplying numerator times numerator and denominator times denominator.

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

Do the multiplications in the fraction \frac{1\times 1}{3\times 6}.

$$\frac{1}{18}x+\frac{1}{3}\times \frac{1}{6}=1-\frac{1}{3}x$$

Multiply \frac{1}{3} times \frac{1}{6} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{18}x+\frac{1\times 1}{3\times 6}=1-\frac{1}{3}x$$

Do the multiplications in the fraction \frac{1\times 1}{3\times 6}.

$$\frac{1}{18}x+\frac{1}{18}=1-\frac{1}{3}x$$

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

$$\frac{1}{18}x+\frac{1}{18}+\frac{1}{3}x=1$$

Combine \frac{1}{18}x and \frac{1}{3}x to get \frac{7}{18}x.

$$\frac{7}{18}x+\frac{1}{18}=1$$

Subtract \frac{1}{18} from both sides.

$$\frac{7}{18}x=1-\frac{1}{18}$$

Convert 1 to fraction \frac{18}{18}.

$$\frac{7}{18}x=\frac{18}{18}-\frac{1}{18}$$

Since \frac{18}{18} and \frac{1}{18} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7}{18}x=\frac{18-1}{18}$$

Subtract 1 from 18 to get 17.

$$\frac{7}{18}x=\frac{17}{18}$$

Multiply both sides by \frac{18}{7}, the reciprocal of \frac{7}{18}.

$$x=\frac{17}{18}\times \frac{18}{7}$$

Multiply \frac{17}{18} times \frac{18}{7} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{17\times 18}{18\times 7}$$

Cancel out 18 in both numerator and denominator.

$$x=\frac{17}{7}$$