Solve 0<1/2(x+2)-3

$$0 < \frac { 1 } { 2 ( x + 2 ) } - 3$$

Solve for x

$$x\in \left(-2,-\frac{11}{6}\right)$$

Solution Steps

Use the distributive property to multiply 2 by x+2.

$$0<\frac{1}{2x+4}-3$$

Factor 2x+4.

$$0<\frac{1}{2\left(x+2\right)}-3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{2\left(x+2\right)}{2\left(x+2\right)}.

$$0<\frac{1}{2\left(x+2\right)}-\frac{3\times 2\left(x+2\right)}{2\left(x+2\right)}$$

Since \frac{1}{2\left(x+2\right)} and \frac{3\times 2\left(x+2\right)}{2\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$0<\frac{1-3\times 2\left(x+2\right)}{2\left(x+2\right)}$$

Do the multiplications in 1-3\times 2\left(x+2\right).

$$0<\frac{1-6x-12}{2\left(x+2\right)}$$

Combine like terms in 1-6x-12.

$$0<\frac{-11-6x}{2\left(x+2\right)}$$

Use the distributive property to multiply 2 by x+2.

$$0<\frac{-11-6x}{2x+4}$$

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$\frac{-11-6x}{2x+4}>0$$

For the quotient to be positive, -6x-11 and 2x+4 have to be both negative or both positive. Consider the case when -6x-11 and 2x+4 are both negative.

$$-6x-11<0 2x+4<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when -6x-11 and 2x+4 are both positive.

$$2x+4>0 -6x-11>0$$

The solution satisfying both inequalities is x\in \left(-2,-\frac{11}{6}\right).

$$x\in \left(-2,-\frac{11}{6}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-2,-\frac{11}{6}\right)$$