Solve 2/3*(x+1)=1/6*(x-2)

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

Solve for x

$$x=-2$$

Steps for Solving Linear Equation

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

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

Use the distributive property to multiply \frac{1}{6} by x-2.

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

Multiply \frac{1}{6} and -2 to get \frac{-2}{6}.

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

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

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

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

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

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

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

Subtract \frac{2}{3} from both sides.

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

Since -\frac{1}{3} and \frac{2}{3} have the same denominator, subtract them by subtracting their numerators.

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

Subtract 2 from -1 to get -3.

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

Divide -3 by 3 to get -1.

$$\frac{1}{2}x=-1$$

Multiply both sides by 2, the reciprocal of \frac{1}{2}.

$$x=-2$$