Solve 1/2(x+1)leq2/3x-1

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

Solve for x

$$x\geq 9$$

Solution Steps

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

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

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

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

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

$$-\frac{1}{6}x+\frac{1}{2}\leq -1$$

Subtract \frac{1}{2} from both sides.

$$-\frac{1}{6}x\leq -1-\frac{1}{2}$$

Convert -1 to fraction -\frac{2}{2}.

$$-\frac{1}{6}x\leq -\frac{2}{2}-\frac{1}{2}$$

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

$$-\frac{1}{6}x\leq \frac{-2-1}{2}$$

Subtract 1 from -2 to get -3.

$$-\frac{1}{6}x\leq -\frac{3}{2}$$

Multiply both sides by -6, the reciprocal of -\frac{1}{6}. Since -\frac{1}{6} is negative, the inequality direction is changed.

$$x\geq -\frac{3}{2}\left(-6\right)$$

Express -\frac{3}{2}\left(-6\right) as a single fraction.

$$x\geq \frac{-3\left(-6\right)}{2}$$

Multiply -3 and -6 to get 18.

$$x\geq \frac{18}{2}$$

Divide 18 by 2 to get 9.

$$x\geq 9$$