Solve x+1/2x-3>2

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

Solve for x

$$x\in \left(\frac{3}{2},\frac{7}{3}\right)$$

Solution Steps

Denominator 2x-3 cannot be zero since division by zero is not defined. There are two cases.

$$2x-3>0 2x-3<0$$

Consider the case when 2x-3 is positive. Move -3 to the right hand side.

$$2x>3$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

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

The initial inequality does not change the direction when multiplied by 2x-3 for 2x-3>0.

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

Multiply out the right hand side.

$$x+1>4x-6$$

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

$$x-4x>-1-6$$

Combine like terms.

$$-3x>-7$$

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

$$x<\frac{7}{3}$$

Consider condition x>\frac{3}{2} specified above.

$$x\in \left(\frac{3}{2},\frac{7}{3}\right)$$

Now consider the case when 2x-3 is negative. Move -3 to the right hand side.

$$2x<3$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

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

The initial inequality changes the direction when multiplied by 2x-3 for 2x-3<0.

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

Multiply out the right hand side.

$$x+1<4x-6$$

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

$$x-4x<-1-6$$

Combine like terms.

$$-3x<-7$$

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

$$x>\frac{7}{3}$$

Consider condition x<\frac{3}{2} specified above.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{3}{2},\frac{7}{3}\right)$$