Intermediate Algebra

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High-SchoolMath

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Hey, so this is a subquestion to a problem in a high-school algebra exam. I couldn't unfortunately solve it and I can't seem to find any answers online so it would be appreciated if anyone can take their time to review this:

We know that the common solutions of two inequalities belong to [-3,-1).
So if two numbers r_1 and r_2 are also solutions to the above inequalities then prove that r_1-r_2 belongs to (-2,2)
 
Are you actually sitting in the exam, right now?

What did you conclude when you considered those two differences?
 
I think you need to give us more information.

[MATH]r_1 = 0 \text { and } r_2 = -\ 3[/MATH]
is, as I understand the question that you wrote, a feasible solution.

[MATH]0 - (-\ 3) = 3 > 2.[/MATH]
So clearly trying to show that the difference between any two numbers in the interval [- 3, 1) has an absolute value < 2 is impossible.

[MATH] -\ 3 \le x < 1.[/MATH]
[MATH]-\ 3 \le y < 1 \implies -\ 1 < -\ y \le 3 \implies -\ 3 + (-\ 1) < x + (-\ y) < 1 + 3 \implies[/MATH]
[MATH]-\ 4 < x - y < 4.[/MATH]
Now the language of your question is very odd. We are given information that common solutions to two inequalities are in a given interval. Do we know what those inequalities are? Then we are told that two numbers are also solutions to the "above" inequalities. What are the "above" inequalities?
 
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I don't remember one of them but I don't it is important to know what they are since we know the interval of their common solutions and the question says: two numbers r_1 and r_2 are solutions to the above inequalities, so they also belong to [-3,-1) and we have to prove that r_1-r_2 belongs to (-2,2). We don't know the value of r_1 and r_2.
 
High-SchoolMath said:
We know that the common solutions of two inequalities belong to [-3,-1).
So if two numbers r_1 and r_2 are also solutions to the above inequalities then prove that r_1-r_2 belongs to (-2,2)
Click to expand...
As I read this, we can ignore the information about common solutions and what the inequalities are specifically.

We are told that two numbers [MATH]r_1[/MATH] and [MATH]r_2[/MATH] are both in the set [MATH][-3, -1)[/MATH], and we want to prove that [MATH]r_1 - r_2[/MATH] is in the set [MATH](-2, 2)[/MATH].

That is, given that [MATH]-3\le r_1\lt -1[/MATH] and [MATH]-3\le r_2\lt -1[/MATH], prove that [MATH]-2\lt r_1 - r_2\lt 2[/MATH].

First try to show that [MATH]r_1 - r_2\lt 2[/MATH]. You know that [MATH]r_1\lt -1[/MATH] and that [MATH]-3\le r_2[/MATH], so that [MATH]r_2\ge -3[/MATH]. Can you combine the known inequalities to obtain the one we want?
 
I got to that point in the exam ,but couldn't express a relationship of those inequalities in order to prove that they belong to (-2,2). So, essentially I am asking the same thing as Dr.Peterson :/ . Also, by combining them, how are we going to get rid of the bigger or equal to -3 ( -3<= r_1 ) ?
 
High-SchoolMath said:
I got to that point in the exam ,but couldn't express a relationship of those inequalities in order to prove that they belong to (-2,2). So, essentially I am asking the same thing as Dr.Peterson :/ . Also, by combining them, how are we going to get rid of the bigger or equal to -3 ( -3<= r_1 ) ?
Click to expand...
Did you even read my post?

"Those inequalities"??????????

It is not even clear which inequalities you are talking about.

I have already demonstrated that. if we are just talking about the two inequalities with common solutions in [-3, 1),it is possible for solutions in that interval to have an absolute value > 2. Either the problem on your test was erroneous, or you are leaving something critical out. What were the additional inequalities specified "above"?
 
the inequalities of r_1 and r_2 since they belong to [-3,-1) they can be written as -3<=r_1<-1 and the same applies for r_2
 
JeffM said:
I have already demonstrated that. if we are just talking about the two inequalities with common solutions in [-3, 1),it is possible for solutions in that interval to have an absolute value > 2.
Click to expand...
But you've misread it. It says they are in [-3, -1).

As for the inequalities, as I said, we don't need to know what they started with; the interval tells us the numbers are at least -3 and less than 1. My understanding is that "above" means the ones they are supposing led to the interval. The word "also" is questionable, but I don't think it kills the problem.
 
High-SchoolMath said:
I got to that point in the exam ,but couldn't express a relationship of those inequalities in order to prove that they belong to (-2,2). So, essentially I am asking the same thing as Dr.Peterson :/ . Also, by combining them, how are we going to get rid of the bigger or equal to -3 ( -3<= r_1 ) ?
Click to expand...
Dr.Peterson said:
First try to show that [MATH]r_1 - r_2\lt 2[/MATH]. You know that [MATH]r_1\lt -1[/MATH] and that [MATH]-3\le r_2[/MATH], so that [MATH]r_2\ge -3[/MATH]. Can you combine the known inequalities to obtain the one we want?
Click to expand...
Please note that I narrowed things down for you, isolating certain individual inequalities to make it easier to see. I am being much more specific than you.

Can you combine [MATH]r_1\lt -1[/MATH] and [MATH]r_2\ge -3[/MATH] to get [MATH]r_1 - r_2\lt 2[/MATH]? It isn't hard. Hint: multiply one of them by -1 and add.

Then try to get the other side of the desired compound inequality.
 
Dr.Peterson said:
But you've misread it. It says they are in [-3, -1).

As for the inequalities, as I said, we don't need to know what they started with; the interval tells us the numbers are at least -3 and less than 1. My understanding is that "above" means the ones they are supposing led to the interval. The word "also" is questionable, but I don't think it kills the problem.
Click to expand...
Dr.Peterson said:
But you've misread it. It says they are in [-3, -1).

As for the inequalities, as I said, we don't need to know what they started with; the interval tells us the numbers are at least -3 and less than 1. My understanding is that "above" means the ones they are supposing led to the interval. The word "also" is questionable, but I don't think it kills the problem.
Click to expand...
Ahh yes so I did. [-3, - 1).

Then I really do not understand what the OP is asking. The answer is obvious and was given by Harry long ago.
 
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