Solve 2x-5/x+3leq5

$$\frac{ 2x-5 }{ x+3 } \leq 5$$

Solve for x

$$x\in (-\infty,-\frac{20}{3}]\cup (-3,\infty)$$

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.

$$2x-5\leq 5\left(x+3\right)$$

Multiply out the right hand side.

$$2x-5\leq 5x+15$$

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

$$2x-5x\leq 5+15$$

Combine like terms.

$$-3x\leq 20$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

$$x\geq -\frac{20}{3}$$

Consider condition x>-3 specified above.

$$x>-3$$

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.

$$2x-5\geq 5\left(x+3\right)$$

Multiply out the right hand side.

$$2x-5\geq 5x+15$$

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

$$2x-5x\geq 5+15$$

Combine like terms.

$$-3x\geq 20$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

$$x\leq -\frac{20}{3}$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-\frac{20}{3}]\cup (-3,\infty)$$