Solve 2/3+5xleq-4/3+x

$$\frac { 2 } { 3 } + 5 x \leq - \frac { 4 } { 3 } + x$$

Solve for x

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

Solution Steps

Subtract x from both sides.

$$\frac{2}{3}+5x-x\leq -\frac{4}{3}$$

Combine 5x and -x to get 4x.

$$\frac{2}{3}+4x\leq -\frac{4}{3}$$

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

$$4x\leq -\frac{4}{3}-\frac{2}{3}$$

Since -\frac{4}{3} and \frac{2}{3} have the same denominator, subtract them by subtracting their numerators.

$$4x\leq \frac{-4-2}{3}$$

Subtract 2 from -4 to get -6.

$$4x\leq \frac{-6}{3}$$

Divide -6 by 3 to get -2.

$$4x\leq -2$$

Divide both sides by 4. Since 4 is positive, the inequality direction remains the same.

$$x\leq \frac{-2}{4}$$

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

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