Answer:
[tex]\begin{bmatrix}\mathrm{Solution:}\:&\:-6\le \:x<3\quad \mathrm{or}\quad \:x\ge \:5\:\\ \:\mathrm{Interval\:Notation:}&\:[-6,\:3)\cup \:[5,\:\infty \:)\end{bmatrix}[/tex]
The number line graph is also attached below.
Step-by-step explanation:
Given the inequality
[tex]\frac{\left(x-5\right)\left(x+6\right)}{\left(x-3\right)}\ge 0[/tex]
[tex]\:\frac{x^2+x-30}{\left(x-3\right)}\ge \:0[/tex]
Let's find the critical points of the inequality.
[tex]\frac{x^2+x-30}{\left(x-3\right)}=0[/tex]
[tex]x^2+x-30=0[/tex] (Multiply both sides by x-3)
[tex](x-5)(x+6)=0[/tex] (Factor left side of equation)
[tex]x-5=0[/tex] or [tex]x+6=0[/tex] (Set factors equal to 0)
[tex]x=5[/tex] or [tex]x=-6[/tex]
Check possible critical points.
x = 5 (Works in original equation)
x = −6 (Works in original equation)
Critical points:
x = 5 or x = −6 (Makes both sides equal)
x = 3 (Makes left denominator equal to 0)
Check intervals in between critical points. (Test values in the intervals to see if they work.)
x ≤ −6 (Doesn't work in original inequality)
−6 ≤ x < 3 (Works in original inequality)
3 < x ≤ 5 (Doesn't work in original inequality)
x ≥ 5 (Works in original inequality)
so
[tex]-6\le \:x<3\quad \mathrm{or}\quad \:x\ge \:5[/tex]
Therefore,
[tex]\begin{bmatrix}\mathrm{Solution:}\:&\:-6\le \:x<3\quad \mathrm{or}\quad \:x\ge \:5\:\\ \:\mathrm{Interval\:Notation:}&\:[-6,\:3)\cup \:[5,\:\infty \:)\end{bmatrix}[/tex]
The number line graph is also attached below.
