Solve each of the following in equations and represent the solution

Given:

$\frac{x-7}{x-2} \geq 0, x \in R$

$\frac{x-7}{x-2} \geq 0$

Signs of $x-7$ :

$x-7=0 \rightarrow x=7$ (Adding 7 on both the sides)

$x-7>0 \rightarrow x>7$ (Adding 7 on both the sides)

$x-7<0 \rightarrow x<7$ (Adding 7 on both the sides)

Signs of $x-2$ :

$x-2=0 \rightarrow x=2$ (Adding 2 on both the sides)

$x-2>0 \rightarrow x>2$ (Adding 2 on both the sides)

$x-2<0 \rightarrow x<2$ (Adding 2 on both the sides)

Zeroes of denominator:

$x-2=0 \rightarrow$ at $x=2 \frac{x-7}{x-2}$ will be undefined.

Intervals that satisfy the required condition: ≥ 0

x < 2 or x = 7 or x >7

Therefore,

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