Solve -5x+2/3geq-1/2x-1

$$- 5 x + \frac { 2 } { 3 } \geq - \frac { 1 } { 2 } x - 1$$

Solve for x

$$x\leq \frac{10}{27}$$

Solution Steps

Add \frac{1}{2}x to both sides.

$$-5x+\frac{2}{3}+\frac{1}{2}x\geq -1$$

Combine -5x and \frac{1}{2}x to get -\frac{9}{2}x.

$$-\frac{9}{2}x+\frac{2}{3}\geq -1$$

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

$$-\frac{9}{2}x\geq -1-\frac{2}{3}$$

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

$$-\frac{9}{2}x\geq -\frac{3}{3}-\frac{2}{3}$$

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

$$-\frac{9}{2}x\geq \frac{-3-2}{3}$$

Subtract 2 from -3 to get -5.

$$-\frac{9}{2}x\geq -\frac{5}{3}$$

Multiply both sides by -\frac{2}{9}, the reciprocal of -\frac{9}{2}. Since -\frac{9}{2} is negative, the inequality direction is changed.

$$x\leq -\frac{5}{3}\left(-\frac{2}{9}\right)$$

Multiply -\frac{5}{3} times -\frac{2}{9} by multiplying numerator times numerator and denominator times denominator.

$$x\leq \frac{-5\left(-2\right)}{3\times 9}$$

Do the multiplications in the fraction \frac{-5\left(-2\right)}{3\times 9}.

$$x\leq \frac{10}{27}$$