Solve 5x+8/x+5geq2

$$\frac{ 5x+8 }{ x+5 } \geq 2$$

Solve for x

$$x\in (-\infty,-5)\cup [\frac{2}{3},\infty)$$

Solution Steps

Denominator x+5 cannot be zero since division by zero is not defined. There are two cases.

$$x+5>0 x+5<0$$

Consider the case when x+5 is positive. Move 5 to the right hand side.

$$x>-5$$

The initial inequality does not change the direction when multiplied by x+5 for x+5>0.

$$5x+8\geq 2\left(x+5\right)$$

Multiply out the right hand side.

$$5x+8\geq 2x+10$$

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

$$5x-2x\geq -8+10$$

Combine like terms.

$$3x\geq 2$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

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

Now consider the case when x+5 is negative. Move 5 to the right hand side.

$$x<-5$$

The initial inequality changes the direction when multiplied by x+5 for x+5<0.

$$5x+8\leq 2\left(x+5\right)$$

Multiply out the right hand side.

$$5x+8\leq 2x+10$$

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

$$5x-2x\leq -8+10$$

Combine like terms.

$$3x\leq 2$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

$$x\leq \frac{2}{3}$$

Consider condition x<-5 specified above.

$$x<-5$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-5)\cup [\frac{2}{3},\infty)$$