Solve x+1-3x+6/x-2leq0

$$\frac { x + 1 - 3 x + 6 } { x - 2 } \leq 0$$

Solve for x

$$x\in (-\infty,2)\cup [\frac{7}{2},\infty)$$

Solution Steps

Combine x and -3x to get -2x.

$$\frac{-2x+1+6}{x-2}\leq 0$$

Add 1 and 6 to get 7.

$$\frac{-2x+7}{x-2}\leq 0$$

For the quotient to be ≤0, one of the values 7-2x and x-2 has to be ≥0, the other has to be ≤0, and x-2 cannot be zero. Consider the case when 7-2x\geq 0 and x-2 is negative.

$$7-2x\geq 0 x-2<0$$

The solution satisfying both inequalities is x<2.

$$x<2$$

Consider the case when 7-2x\leq 0 and x-2 is positive.

$$7-2x\leq 0 x-2>0$$

The solution satisfying both inequalities is x\geq \frac{7}{2}.

$$x\geq \frac{7}{2}$$

The final solution is the union of the obtained solutions.

$$x<2\text{; }x\geq \frac{7}{2}$$