Solve the following inequalities for x:

$\frac{x + 3}{x-2}≤ 2$

$(−∞,2)∪[7,∞)$

The correct answer is Option (2) → $(−∞,2)∪[7,∞)$

Given $\frac{x + 3}{x-2}≤ 2$. First we note that $x≠2$

$⇒\frac{x + 3}{x-2}-2≤0$

$⇒\frac{x + 3-2(x-2)}{x-2}-2≤0$

$⇒\frac{-x + 7}{x-2}≤0⇒\frac{x - 7}{x-2}≥0$  (Since $(x-2)^2>0,∀x∈R,x≠2)$

$⇒(x-7)(x-2)≥0$   ...(1)  (Multiplying by $(x-2)^2$)

Mark the numbers 2 and 7 on the number line. By the method of intervals, the inequality (1) is satisfied when $x ≤2$ or $x≥ 7$ but $x ≠ 2$

∴ The solution set is $(−∞,2)∪[7,∞)$.