Solve 2x+4/3x-2<1

$$\frac { 2 x + 4 } { 3 x - 2 } < 1$$

Solve for x

$$x\in \left(-\infty,\frac{2}{3}\right)\cup \left(6,\infty\right)$$

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.

$$2x+4<3x-2$$

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

$$2x-3x<-4-2$$

Combine like terms.

$$-x<-6$$

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

$$x>6$$

Consider condition x>\frac{2}{3} specified above. The result remains the same.

$$x>6$$

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.

$$2x+4>3x-2$$

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

$$2x-3x>-4-2$$

Combine like terms.

$$-x>-6$$

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

$$x<6$$

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

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

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,\frac{2}{3}\right)\cup \left(6,\infty\right)$$