Rational Inequality

[harpazo]

Solve the rational inequality below.

(x + 4)/(x - 2) ≤ 1

Solution:

(x + 4)/(x - 2) - 1 ≤ 0

After simplifying the left side, I end up with

6/(x - 2) ≤ 0

I then set denominator to zero and solve for x. Doing so, x = 2.

I place 2 on the real number line.

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

Select numbers from each interval including 2 to evaluate using 6/(x - 2) ≤ 0.

When x = 0, I get -3 ≤ 0. This is true.

My answer is (-infinity, 2).

Correct?

 

[Otis]

My answer is (-infinity, 2).

Correct?

Yes.

Here are some comments (for other readers, too).

(x + 4)/(x - 2) ≤ 1

This inequality says the left-hand side is less than or equal to 1.

Yet, it never equals 1, and we can realize that by thinking about what it means for a ratio to equal 1: A ratio equals 1 whenever numerator and denominator are the same non-zero number. In this exercise, we note that expressions x+4 and x-2 must represent different numbers, so their ratio cannot equal 1.

... Select numbers from each interval including 2 to evaluate ...

We cannot evaluate for x = 2. (If you're not convinced, try it.)

? \(\quad\) Writing x ≠ 2 could be the very first step, in this exercise.

... 6/(x - 2) ≤ 0 ...

At this point, we can get the solution without defining intervals and using test values.

The left-hand side is less than zero when it's negative, and it is negative when the numerator and denominator have opposite signs. We see that x-2 must be the negative number because 6 is the positive one.

x - 2 < 0

x < 2

?

 

[harpazo]

Yes.

Here are some comments (for other readers, too).


This inequality says the left-hand side is less than or equal to 1.

Yet, it never equals 1, and we can realize that by thinking about what it means for a ratio to equal 1: A ratio equals 1 whenever numerator and denominator are the same non-zero number. In this exercise, we note that expressions x+4 and x-2 must represent different numbers, so their ratio cannot equal 1.


We cannot evaluate for x = 2. (If you're not convinced, try it.)

? \(\quad\) Writing x ≠ 2 could be the very first step, in this exercise.


At this point, we can get the solution without defining intervals and using test values.

The left-hand side is less than zero when it's negative, and it is negative when the numerator and denominator have opposite signs. We see that x-2 must be the negative number because 6 is the positive one.

x - 2 < 0

x < 2

?

Good study notes.

 

[topsquark]

Try to look at it this way: We need to find critical points so we can "map out" what the solution is going to tell us. There are two good sets of points to use: One set is the values of x that make the numerator zero, and the other is the values of x that make the denominator zero.

So you have two critical points here: One at x = -4 and one at x = 2. That means we need to check on values in the intervals \(\displaystyle ( -\infty, -4)\), \(\displaystyle (-4, 2)\) and \(\displaystyle (2, \infty )\). And now check the values to see where the corresponding + and - are.

-Dan

 

[Otis]

Good study notes.

Only if you use them.

\(\;\)

 

[harpazo]

Try to look at it this way: We need to find critical points so we can "map out" what the solution is going to tell us. There are two good sets of points to use: One set is the values of x that make the numerator zero, and the other is the values of x that make the denominator zero.

So you have two critical points here: One at x = -4 and one at x = 2. That means we need to check on values in the intervals \(\displaystyle ( -\infty, -4)\), \(\displaystyle (-4, 2)\) and \(\displaystyle (2, \infty )\). And now check the values to see where the corresponding + and - are.

-Dan

The textbook answer is (-00, -1) U (1, 00).

 

[MarkFL]

The textbook answer is (-00, -1) U (1, 00).

That is incorrect. We have:

[MATH]\frac{x+4}{x-2}\le1[/MATH]
[MATH]\frac{x+4}{x-2}-1\le0[/MATH]
[MATH]\frac{x+4-(x-2)}{x-2}\le0[/MATH]
[MATH]\frac{6}{x-2}\le0[/MATH]
[MATH]\frac{1}{x-2}\le0[/MATH]
We have one critical value of \(x=2\), and testing to the left at \(x=0\) we get a true statement, thus our solution is:

[MATH](-\infty,2)[/MATH]
Even though our inequality is weak, the interval must be open because of division by zero.

 

[mmm4444bot]

The textbook answer is (-00, -1) U (1, 00).

No it is not. That is the book's answer for the exercise in your other thread.

You are not paying attention.

\(\;\)