Solve |1/3-1|-2x1/3

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

Evaluate

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

Solution Steps

Convert 1 to fraction \frac{3}{3}.

$$|\frac{1}{3}-\frac{3}{3}|-2x\times \frac{1}{3}$$

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

$$|\frac{1-3}{3}|-2x\times \frac{1}{3}$$

Subtract 3 from 1 to get -2.

$$|-\frac{2}{3}|-2x\times \frac{1}{3}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{2}{3} is \frac{2}{3}.

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

Multiply 2 and \frac{1}{3} to get \frac{2}{3}.

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

Factor

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

Solution Steps

Factor out \frac{1}{3}.

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

Consider -3\left(\frac{1}{3}-1\right)-2x. Multiply and combine like terms.

$$-2x+2$$

Consider -2x+2. Factor out 2.

$$2\left(-x+1\right)$$

Rewrite the complete factored expression.

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