Solve x-1/x-2geq3/4-2x

$$\frac{ x-1 }{ x-2 } \geq \frac{ 3 }{ 4-2x }$$

Solve for x

$$x\in (-\infty,-\frac{1}{2}]\cup (2,\infty)$$

Solution Steps

Subtract \frac{3}{4-2x} from both sides.

$$\frac{x-1}{x-2}-\frac{3}{4-2x}\geq 0$$

Factor 4-2x.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-2 and 2\left(-x+2\right) is 2\left(x-2\right). Multiply \frac{x-1}{x-2} times \frac{2}{2}. Multiply \frac{3}{2\left(-x+2\right)} times \frac{-1}{-1}.

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

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

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

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

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

Combine like terms in 2x-2+3.

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

Use the distributive property to multiply 2 by x-2.

$$\frac{2x+1}{2x-4}\geq 0$$

For the quotient to be ≥0, 2x+1 and 2x-4 have to be both ≤0 or both ≥0, and 2x-4 cannot be zero. Consider the case when 2x+1\leq 0 and 2x-4 is negative.

$$2x+1\leq 0 2x-4<0$$

The solution satisfying both inequalities is x\leq -\frac{1}{2}.

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

Consider the case when 2x+1\geq 0 and 2x-4 is positive.

$$2x+1\geq 0 2x-4>0$$

The solution satisfying both inequalities is x>2.

$$x>2$$

The final solution is the union of the obtained solutions.

$$x\leq -\frac{1}{2}\text{; }x>2$$