Solve 23x-2x/3x-2leq-2

$$\frac { 23 x - 2 x } { 3 x - 2 } \leq - 2$$

Solve for x

$$x\in [\frac{4}{27},\frac{2}{3})$$

Solution Steps

Combine 23x and -2x to get 21x.

$$\frac{21x}{3x-2}\leq -2$$

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.

$$21x\leq -2\left(3x-2\right)$$

Multiply out the right hand side.

$$21x\leq -6x+4$$

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

$$21x+6x\leq 4$$

Combine like terms.

$$27x\leq 4$$

Divide both sides by 27. Since 27 is positive, the inequality direction remains the same.

$$x\leq \frac{4}{27}$$

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

$$x\in \emptyset$$

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.

$$21x\geq -2\left(3x-2\right)$$

Multiply out the right hand side.

$$21x\geq -6x+4$$

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

$$21x+6x\geq 4$$

Combine like terms.

$$27x\geq 4$$

Divide both sides by 27. Since 27 is positive, the inequality direction remains the same.

$$x\geq \frac{4}{27}$$

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

$$x\in [\frac{4}{27},\frac{2}{3})$$

The final solution is the union of the obtained solutions.

$$x\in [\frac{4}{27},\frac{2}{3})$$