Solve 1-3x/x-2-1geq0

$$\frac{ 1-3x }{ x-2 } -1 \geq 0$$

Solve for x

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

Solution Steps

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

$$\frac{1-3x}{x-2}-\frac{x-2}{x-2}\geq 0$$

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

$$\frac{1-3x-\left(x-2\right)}{x-2}\geq 0$$

Do the multiplications in 1-3x-\left(x-2\right).

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

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

$$\frac{3-4x}{x-2}\geq 0$$

For the quotient to be ≥0, 3-4x and x-2 have to be both ≤0 or both ≥0, and x-2 cannot be zero. Consider the case when 3-4x\leq 0 and x-2 is negative.

$$3-4x\leq 0 x-2<0$$

The solution satisfying both inequalities is x\in \left[\frac{3}{4},2\right).

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

Consider the case when 3-4x\geq 0 and x-2 is positive.

$$3-4x\geq 0 x-2>0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

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