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

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

Solve for x

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

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.

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

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

$$2x-3x>-1-2$$

Combine like terms.

$$-x>-3$$

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

$$x<3$$

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

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

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.

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

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

$$2x-3x<-1-2$$

Combine like terms.

$$-x<-3$$

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

$$x>3$$

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

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

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