Solve 2x-2/3(x-2)=1/3[x-1/2(3x+1)]

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

Solve for x

$$x=-1$$

Steps for Solving Linear Equation

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

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

Express -\frac{2}{3}\left(-2\right) as a single fraction.

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

Multiply -2 and -2 to get 4.

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

Combine 2x and -\frac{2}{3}x to get \frac{4}{3}x.

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

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

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

Express -\frac{1}{2}\times 3 as a single fraction.

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

Fraction \frac{-3}{2} can be rewritten as -\frac{3}{2} by extracting the negative sign.

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

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

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

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

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

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

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

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

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

Fraction \frac{-1}{6} can be rewritten as -\frac{1}{6} by extracting the negative sign.

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

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

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

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

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

Fraction \frac{-1}{6} can be rewritten as -\frac{1}{6} by extracting the negative sign.

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

Add \frac{1}{6}x to both sides.

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

Combine \frac{4}{3}x and \frac{1}{6}x to get \frac{3}{2}x.

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

Subtract \frac{4}{3} from both sides.

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

Least common multiple of 6 and 3 is 6. Convert -\frac{1}{6} and \frac{4}{3} to fractions with denominator 6.

$$\frac{3}{2}x=-\frac{1}{6}-\frac{8}{6}$$

Since -\frac{1}{6} and \frac{8}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3}{2}x=\frac{-1-8}{6}$$

Subtract 8 from -1 to get -9.

$$\frac{3}{2}x=\frac{-9}{6}$$

Reduce the fraction \frac{-9}{6} to lowest terms by extracting and canceling out 3.

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

Multiply both sides by \frac{2}{3}, the reciprocal of \frac{3}{2}.

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

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

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

Cancel out 2 in both numerator and denominator.

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

Divide -3 by 3 to get -1.

$$x=-1$$