Solve 2x/x+3geq1

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

Solve for x

$$x\in (-\infty,-3)\cup [3,\infty)$$

Solution Steps

Denominator x+3 cannot be zero since division by zero is not defined. There are two cases.

$$x+3>0 x+3<0$$

Consider the case when x+3 is positive. Move 3 to the right hand side.

$$x>-3$$

The initial inequality does not change the direction when multiplied by x+3 for x+3>0.

$$2x\geq x+3$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$2x-x\geq 3$$

Combine like terms.

$$x\geq 3$$

Now consider the case when x+3 is negative. Move 3 to the right hand side.

$$x<-3$$

The initial inequality changes the direction when multiplied by x+3 for x+3<0.

$$2x\leq x+3$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$2x-x\leq 3$$

Combine like terms.

$$x\leq 3$$

Consider condition x<-3 specified above.

$$x<-3$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-3)\cup [3,\infty)$$