Solve 1/3(x-6)<x+8

$$\frac { 1 } { 3 } ( x - 6 ) < x + 8$$

Solve for x

$$x>-15$$

Solution Steps

Use the distributive property to multiply \frac{1}{3} by x-6.

$$\frac{1}{3}x+\frac{1}{3}\left(-6\right)<x+8$$

Multiply \frac{1}{3} and -6 to get \frac{-6}{3}.

$$\frac{1}{3}x+\frac{-6}{3}<x+8$$

Divide -6 by 3 to get -2.

$$\frac{1}{3}x-2<x+8$$

Subtract x from both sides.

$$\frac{1}{3}x-2-x<8$$

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

$$-\frac{2}{3}x-2<8$$

Add 2 to both sides.

$$-\frac{2}{3}x<8+2$$

Add 8 and 2 to get 10.

$$-\frac{2}{3}x<10$$

Multiply both sides by -\frac{3}{2}, the reciprocal of -\frac{2}{3}. Since -\frac{2}{3} is negative, the inequality direction is changed.

$$x>10\left(-\frac{3}{2}\right)$$

Express 10\left(-\frac{3}{2}\right) as a single fraction.

$$x>\frac{10\left(-3\right)}{2}$$

Multiply 10 and -3 to get -30.

$$x>\frac{-30}{2}$$

Divide -30 by 2 to get -15.

$$x>-15$$