Solve x+1/3x-2>2

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

Solve for x

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

Solution Steps

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

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

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

$$3x>2$$

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

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

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

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

Multiply out the right hand side.

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

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

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

Combine like terms.

$$-5x>-5$$

Divide both sides by -5. Since -5 is negative, the inequality direction is changed.

$$x<1$$

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

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

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

$$3x<2$$

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

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

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

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

Multiply out the right hand side.

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

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

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

Combine like terms.

$$-5x<-5$$

Divide both sides by -5. Since -5 is negative, the inequality direction is changed.

$$x>1$$

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

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

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