(1.1) When we multiply both sides of an inequality by a negative number, the inequality flips oven and instead of "≥" becomes "≤"
(1.2) When we multiply both sides of an inequality by a possitive number, the inequality sign remains unaltered.
(1.3) We need to consider 2 cases because the term (x + 1) could be possitive or negative. If it's possitive the sing remains unaltered, but if (x + 1) < 0, then the sign must change.
(1.4) To solve this, first let's leave 0 in the right hand side of teh inequality:
[tex]\begin{gathered} \frac{x-5}{x+1}\ge2 \\ \frac{x-5}{x+1}-2\ge0 \end{gathered}[/tex]
Now add the terms:
[tex]\frac{x-5}{x+1}-\frac{2(x+1)}{x+1}=\frac{x-5-2x-2}{x+1}=\frac{-x-7}{x-1}\ge0[/tex]
Now we need to calculate the critical points, where the numerator and denominator are 0:
numerator:
-x - 7 = 0
x = -7
Denominator:
x + 1 = 0
x = -1
Now, we have three intervals:
Now we need to test in which intervals teh inequality is true.
For the interval (-∞, -7) Lets take -10:
[tex]\begin{gathered} x=-10 \\ \frac{-10-5}{-10+1}=\frac{-15}{-9}=\frac{5}{3}\ge2 \end{gathered}[/tex]
Which is false, 5/3 ≈ 1.666667 which is less than 2.
Next let's test the interval [-7, -1). Let's grab x = -5
[tex]\begin{gathered} x=-5 \\ \frac{-5-5}{-5+1}=\frac{-10}{-4}=\frac{5}{2}=2.5\ge2 \end{gathered}[/tex]
Which is TRUE
Finally, for (-1, ∞) Let's take x = 0:
[tex]\frac{0-5}{0+1}=-\frac{5}{1}=-5[/tex]
And (-5) is less than 2.
The interval of solutions is:
[tex]\lbrack-7,-1)[/tex]
Or:
[tex]-7\le x<-1[/tex]
