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

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

Solve for x

$$x\in [\frac{3}{5},\frac{4}{3})$$

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.

$$2x+1\leq -\left(3x-4\right)$$

Multiply out the right hand side.

$$2x+1\leq -3x+4$$

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

$$2x+3x\leq -1+4$$

Combine like terms.

$$5x\leq 3$$

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

$$x\leq \frac{3}{5}$$

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

$$x\in \emptyset$$

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.

$$2x+1\geq -\left(3x-4\right)$$

Multiply out the right hand side.

$$2x+1\geq -3x+4$$

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

$$2x+3x\geq -1+4$$

Combine like terms.

$$5x\geq 3$$

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

$$x\geq \frac{3}{5}$$

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

$$x\in [\frac{3}{5},\frac{4}{3})$$

The final solution is the union of the obtained solutions.

$$x\in [\frac{3}{5},\frac{4}{3})$$