Solve 3quad2x+1/x-2>-3

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

Solve for x

$$x\in \left(-\infty,\frac{1}{3}\right)\cup \left(2,\infty\right)$$

Solution Steps

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

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

Divide -3 by 3 to get -1.

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

Denominator x-2 cannot be zero since division by zero is not defined. There are two cases.

$$x-2>0 x-2<0$$

Consider the case when x-2 is positive. Move -2 to the right hand side.

$$x>2$$

The initial inequality does not change the direction when multiplied by x-2 for x-2>0.

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

Multiply out the right hand side.

$$2x+1>-x+2$$

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

$$2x+x>-1+2$$

Combine like terms.

$$3x>1$$

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

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

Consider condition x>2 specified above.

$$x>2$$

Now consider the case when x-2 is negative. Move -2 to the right hand side.

$$x<2$$

The initial inequality changes the direction when multiplied by x-2 for x-2<0.

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

Multiply out the right hand side.

$$2x+1<-x+2$$

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

$$2x+x<-1+2$$

Combine like terms.

$$3x<1$$

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

$$x<\frac{1}{3}$$

Consider condition x<2 specified above. The result remains the same.

$$x<\frac{1}{3}$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,\frac{1}{3}\right)\cup \left(2,\infty\right)$$