Solve 2x+1/x-3>-3

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

Solve for x

$$x\in \left(-\infty,\frac{8}{5}\right)\cup \left(3,\infty\right)$$

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.

$$5x>8$$

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

$$x>\frac{8}{5}$$

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.

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

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.

$$5x<8$$

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

$$x<\frac{8}{5}$$

Consider condition x<3 specified above. The result remains the same.

$$x<\frac{8}{5}$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,\frac{8}{5}\right)\cup \left(3,\infty\right)$$