Rational Inequality...2

[nycmathguy]

Section 2.7
Question 44

(x + 12)/(x + 2) ≥ 3

Simplifying the rational inequality just a bit, I get

(-2x + 6)/(x + 2) ≥ 0

Set denominator to zero and solve for x.

I get x = -2. This means x cannot be -2 here.
If x = -2, we get division by zero which is undefined.

Now, solving -2x + 6)/(x + 2) ≥ 0 for x, I get the following: x ≤ 3.

Plot x = -2 and x = 3 on the real number line and test each interval.

<-------(-2)---------(3)--------->

When x = -3, we get a false statement.
As stated above, x cannot be -2.
When x = 0, we get a true statement.
When x = 3, we get a true statement.
When x = 4, we get a false statement.

The only solution interval I see is [0,3].

Here it is on the real number line:

You say?

 

[MathLover1]

x cannot be -2, true, but x can be greater than -2
solution is:
-2<x<=3

(-2, 3]

 

[nycmathguy]

x cannot be -2, true, but x can be greater than -2
solution is:
-2<x<=3

(-2, 3]

View attachment 552

Can you break it down step by step? How did you end up with (-2, 3]?

While you show me how this is done, I will also watch a few video lessons. I need to solve at least 3 more rational inequality problems from the current section. I cannot go beyond Section 2.7 without knowing this very important precalculus topic. I need to feel confident with rational inequality. Otherwise, I stop right here.

P. S. Look for some Calculus 1 questions tonight.

 

[MathLover1]

(-2x + 6)/(x + 2) ≥ 0
if (-2x + 6) ≥ 0 => x<=3

if denominator (x + 2) ≥ 0 => x>=-2 ; since denominator cannot be = to 0, we use only x>-2

combine solutions: -2<x<=3

 

[nycmathguy]

(-2x + 6)/(x + 2) ≥ 0
if (-2x + 6) ≥ 0 => x<=3

if denominator (x + 2) ≥ 0 => x>=-2 ; since denominator cannot be = to 0, we use only x>-2

combine solutions: -2<x<=3

Thank you. I will do 46, 48, 50 and 52 for additional practice BUT only after watching a few video clips.