Solve the following inequalities for x:

$\frac{x+3}{x-4}≥0$

$(−∞,−3]∪(4,∞)$

The correct answer is Option (1) → $(−∞,−3]∪(4,∞)$

Given $\frac{x+3}{x-4}≥0$. First, we note that $x ≠ 4$.

Since $(x-4)^2 > 0$ for all $x ∈ R, x ≠4,$

$\frac{x+3}{x-4}≥0⇒(x+3)(x-4) ≥0$  (multiplying by $(x-4)^2$)

$⇒ (x-(-3)) (x −4) ≥ 0$   ...(1)

Mark the numbers -3 and 4 on the number line.

By the method of intervals, the inequality (1) is satisfied when $x ≥ 4$ or $x < -3$ but $x ≠ 4$.

∴ The solution set is $(−∞,−3]∪(4,∞)$.