Solve 2x+1/x+3>3

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

Solve for x

$$x\in \left(-8,-3\right)$$

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.

$$2x+1>3\left(x+3\right)$$

Multiply out the right hand side.

$$2x+1>3x+9$$

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

$$2x-3x>-1+9$$

Combine like terms.

$$-x>8$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x<-8$$

Consider condition x>-3 specified above.

$$x\in \emptyset$$

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.

$$2x+1<3\left(x+3\right)$$

Multiply out the right hand side.

$$2x+1<3x+9$$

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

$$2x-3x<-1+9$$

Combine like terms.

$$-x<8$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x>-8$$

Consider condition x<-3 specified above.

$$x\in \left(-8,-3\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-8,-3\right)$$