Solve 2x-3/3x-4leq2

$$\frac { 2 x - 3 } { 3 x - 4 } \leq 2$$

Solve for x

$$x\in (-\infty,\frac{5}{4}]\cup (\frac{4}{3},\infty)$$

Solution Steps

Denominator 3x-4 cannot be zero since division by zero is not defined. There are two cases.

$$3x-4>0 3x-4<0$$

Consider the case when 3x-4 is positive. Move -4 to the right hand side.

$$3x>4$$

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

$$x>\frac{4}{3}$$

The initial inequality does not change the direction when multiplied by 3x-4 for 3x-4>0.

$$2x-3\leq 2\left(3x-4\right)$$

Multiply out the right hand side.

$$2x-3\leq 6x-8$$

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

$$2x-6x\leq 3-8$$

Combine like terms.

$$-4x\leq -5$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x\geq \frac{5}{4}$$

Consider condition x>\frac{4}{3} specified above.

$$x>\frac{4}{3}$$

Now consider the case when 3x-4 is negative. Move -4 to the right hand side.

$$3x<4$$

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

$$x<\frac{4}{3}$$

The initial inequality changes the direction when multiplied by 3x-4 for 3x-4<0.

$$2x-3\geq 2\left(3x-4\right)$$

Multiply out the right hand side.

$$2x-3\geq 6x-8$$

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

$$2x-6x\geq 3-8$$

Combine like terms.

$$-4x\geq -5$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x\leq \frac{5}{4}$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,\frac{5}{4}]\cup (\frac{4}{3},\infty)$$