Inequalities

[eladmanor101]

Hello, this is my first post to this forum and I would like to get some help regarding this inequality question (All that's needed to do is to solve for x):


This is my solution plus I tried to make it as clear as possible for you.
The only problem is that the solution in my book is different and I am not sure what I did wrong or if there is another way of approaching this inequality.

Correct solution (from my book) is "-3<x<=2"
(<= is smaller than or equal to)

I am looking forward for your responses, thanks!

 

[Dr.Peterson]

There was no benefit in adding 1; that doesn't prevent dividing by 0, and only makes the rest of the work harder. The way to prevent dividing by 0 is just to write down "x can't be -3". You should also have noted that x can't be 5, due to the factors you canceled.

Then when you multiplied by (x+3), you changed the addition on the left to multiplication! But the worst thing is that in multiplying by (x+3), you might be multiplying by a negative number, which changes the direction of the inequality. So you really should never do that at all.

Presumably your textbook has shown you a method to use; why not follow it?

The usual way is to treat (x-2)/(x+3) <= 0 more or less the same way you did your last step with (x-3)(x+3) <= 0. In fact, the solution of (x-2)/(x+3) <= 0 is almost the same as for (x-2)(x+3) <= 0; the only difference is that x+3 can't be zero.

 

[eladmanor101]

Thank you! Seems like I have made a few mistakes which I usually do not make. Maybe I wasn't concentrated.
About the dividing by 0, I meant to say dividing zero by a number which makes it zero so I thought maybe a solution will be missed.
I am studying other things from my class so the only thing I can rely on is the textbook and the examples shown, which could be simple or could not cover all of the methods/difficulties.

By the way, are you saying I should multiply everything by (x+3) and get x-2 <=0? If so, where does the x>-3 come from?

 

[Mr. Bland]

If I'm interpreting the image correctly, the problem is as follows:

[MATH]\frac{x^2 - 7x + 10}{x^2 -2x - 15} \leq 0[/MATH]​

And your work is as follows:

1. [MATH]\frac{(x - 5)(x - 2)}{(x - 5)(x + 3)} \leq 0[/MATH]
2. [MATH]\frac{x - 2}{x + 3} \leq 0[/MATH]
3. [MATH]\frac{x - 2}{x + 3} + \frac{1}{1} \leq \frac{1}{1}[/MATH]
4. [MATH](x - 2)(x + 3) \leq (x + 3)[/MATH]

Assuming I got that right, there is an error in the transition from step 3 to step 4: when multiplying by [MATH](x + 3)[/MATH], it is distributed across the addition but does not replace it. The adjusted version looks like this:

[MATH](x - 2) + (x + 3) \leq (x + 3)[/MATH]​

However, by eliminating the fraction, you've lost some of the information that pertains to the original inequality. The next step down this path leads here:

[MATH]x \leq 2[/MATH]​

While this is part of the answer, it is only part of it. Let's go back to step 2:

[MATH]\frac{x - 2}{x + 3} \leq 0[/MATH]​

From here, we can get to the correct answer by making a few observations:

1. If both the numerator and denominator are positive, then the result will be greater than zero. Therefore, we know the maximum value of the numerator is zero itself, giving us the partial answer [MATH]x \leq 2[/MATH].
2. We know that [MATH]x \ne -3[/MATH] because otherwise we would have a zero division.
3. If [MATH]x < -3[/MATH], then both the numerator and denominator will be negative, meaning the result will be greater than zero. Therefore, in conjunction with observation #2, we know that [MATH]x > -3[/MATH].

 

[eladmanor101]

Thanks for your answer, I understand now.
So basically, we need to use logic to get to the x > -3 part (Although normally there is a way to get to the solution using calculation maybe this question was an exception.

Going back to study, looks like I gotta practice this stuff!
(Again I really thank you for your very well written answers!)