Solve -3(1/9-x)=x-(4x+1/3)+6x

$$- 3 ( \frac { 1 } { 9 } - x ) = x - ( 4 x + \frac { 1 } { 3 } ) + 6 x$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Solution Steps

Use the distributive property to multiply -3 by \frac{1}{9}-x.

$$-3\times \frac{1}{9}+3x=x-\left(4x+\frac{1}{3}\right)+6x$$

Multiply -3 and \frac{1}{9} to get \frac{-3}{9}.

$$\frac{-3}{9}+3x=x-\left(4x+\frac{1}{3}\right)+6x$$

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

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

To find the opposite of 4x+\frac{1}{3}, find the opposite of each term.

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

Combine x and -4x to get -3x.

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

Combine -3x and 6x to get 3x.

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

Subtract 3x from both sides.

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

Combine 3x and -3x to get 0.

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

Compare -\frac{1}{3} and -\frac{1}{3}.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

Use the distributive property to multiply -3 by \frac{1}{9}-x.

$$-3\times \frac{1}{9}+3x=x-\left(4x+\frac{1}{3}\right)+6x$$

Multiply -3 and \frac{1}{9} to get \frac{-3}{9}.

$$\frac{-3}{9}+3x=x-\left(4x+\frac{1}{3}\right)+6x$$

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

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

To find the opposite of 4x+\frac{1}{3}, find the opposite of each term.

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

Combine x and -4x to get -3x.

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

Combine -3x and 6x to get 3x.

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

Subtract 3x from both sides.

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

Combine 3x and -3x to get 0.

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

Compare -\frac{1}{3} and -\frac{1}{3}.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$