Solve 3x-1/2x-1+1leq0

$$\frac { 3 x - 1 } { 2 x - 1 } + 1 \leq 0$$

Solve for x

$$x\in [\frac{2}{5},\frac{1}{2})$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2x-1}{2x-1}.

$$\frac{3x-1}{2x-1}+\frac{2x-1}{2x-1}\leq 0$$

Since \frac{3x-1}{2x-1} and \frac{2x-1}{2x-1} have the same denominator, add them by adding their numerators.

$$\frac{3x-1+2x-1}{2x-1}\leq 0$$

Combine like terms in 3x-1+2x-1.

$$\frac{5x-2}{2x-1}\leq 0$$

For the quotient to be ≤0, one of the values 5x-2 and 2x-1 has to be ≥0, the other has to be ≤0, and 2x-1 cannot be zero. Consider the case when 5x-2\geq 0 and 2x-1 is negative.

$$5x-2\geq 0 2x-1<0$$

The solution satisfying both inequalities is x\in \left[\frac{2}{5},\frac{1}{2}\right).

$$x\in [\frac{2}{5},\frac{1}{2})$$

Consider the case when 5x-2\leq 0 and 2x-1 is positive.

$$5x-2\leq 0 2x-1>0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in [\frac{2}{5},\frac{1}{2})$$