Solve 8/2-xgeqx/x-2+1

$$\frac{ 8 }{ 2-x } \geq \frac{ x }{ x-2 } +1$$

Solve for x

$$x\in [-3,2)$$

Solution Steps

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

$$\frac{8}{2-x}\geq \frac{x}{x-2}+\frac{x-2}{x-2}$$

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

$$\frac{8}{2-x}\geq \frac{x+x-2}{x-2}$$

Combine like terms in x+x-2.

$$\frac{8}{2-x}\geq \frac{2x-2}{x-2}$$

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

$$\frac{8}{2-x}-\frac{2x-2}{x-2}\geq 0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2-x and x-2 is x-2. Multiply \frac{8}{2-x} times \frac{-1}{-1}.

$$\frac{8\left(-1\right)}{x-2}-\frac{2x-2}{x-2}\geq 0$$

Since \frac{8\left(-1\right)}{x-2} and \frac{2x-2}{x-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8\left(-1\right)-\left(2x-2\right)}{x-2}\geq 0$$

Do the multiplications in 8\left(-1\right)-\left(2x-2\right).

$$\frac{-8-2x+2}{x-2}\geq 0$$

Combine like terms in -8-2x+2.

$$\frac{-6-2x}{x-2}\geq 0$$

For the quotient to be ≥0, -2x-6 and x-2 have to be both ≤0 or both ≥0, and x-2 cannot be zero. Consider the case when -2x-6\leq 0 and x-2 is negative.

$$-2x-6\leq 0 x-2<0$$

The solution satisfying both inequalities is x\in \left[-3,2\right).

$$x\in [-3,2)$$

Consider the case when -2x-6\geq 0 and x-2 is positive.

$$-2x-6\geq 0 x-2>0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in [-3,2)$$