Confusing Question on Compass Test

T

tylerlopez

New member
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Nov 9, 2013
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Hi everyone, I recently took the Compass test and one particular question stumped me,
it said... for all X>1

question.jpg

Thank you for the help!
 
JeffM said:
What was the question?
Click to expand...

The question was my picture.

I'd like to add my answer was 1-2xy^2/x^2+3x-1, I don't know whether or not I was correct as the compass did not tell me. However, I do not believe I was.
 
tylerlopez said:
The question was my picture.

I'd like to add my answer was 1-2xy^2/x^2+3x-1, I don't know whether or not I was correct as the compass did not tell me. However, I do not believe I was.
Click to expand...
I'm sorry. I cannot answer an expression. I can answer some questions, but I have no clue what the answer to an expression is. Now if there had been instructions such as "simplify" or "evaluate when x = 2 and y = 7," there would be an answer.
 
JeffM said:
I'm sorry. I cannot answer an expression. I can answer some questions, but I have no clue what the answer to an expression is. Now if there had been instructions such as "simplify" or "evaluate when x = 2 and y = 7," there would be an answer.
Click to expand...

Oh I'm sorry, there weren't instructions about simplifying but I know that's what the particular problem wanted.

Sorry for the misconception.
The simplifying is what I had troubles with.
 
tylerlopez said:
Oh I'm sorry, there weren't instructions about simplifying but I know that's what the particular problem wanted.

Sorry for the misconception.
The simplifying is what I had troubles with.
Click to expand...
\(\displaystyle \dfrac{2x - 4x^2y}{2x^3 + 6x^2 - 2x} = \dfrac{2x(1 - 2xy^2)}{2x(x^2 + 3x - 1)} = \dfrac{1 - xy^2}{x^2 + 3x - 1}.\)

Now that is not the answer that you gave because you forgot to use grouping symbols, but I suspect that you meant the correct answer.

However, I still am unsure that you really understood the question that was being asked because x > 1 does not seem particularly relevant to simplification. (x > 0 would be relevant.)
 
JeffM said:
\(\displaystyle \dfrac{2x - 4x^2y}{2x^3 + 6x^2 - 2x} = \dfrac{2x(1 - 2xy^2)}{2x(x^2 + 3x - 1)} = \dfrac{1 - xy^2}{x^2 + 3x - 1}.\)

Now that is not the answer that you gave because you forgot to use grouping symbols, but I suspect that you meant the correct answer.

However, I still am unsure that you really understood the question that was being asked because x > 1 does not seem particularly relevant to simplification. (x > 0 would be relevant.)
Click to expand...
It may have said then X =....?
I don't understand the concept of x > 1 and x > 0 and x(does not equal)0. I saw several of those on the compass test and wasn't certain if that was some sort of clarification or if it was relevant to the question.
Would you happen to know, what is the term for when there is a preceding message such as x>1 and x>0 so that I may look for a lesson on that topic?
 
tylerlopez said:
It may have said then X =....?
I don't understand the concept of x > 1 and x > 0 and x(does not equal)0. I saw several of those on the compass test and wasn't certain if that was some sort of clarification or if it was relevant to the question.
Would you happen to know, what is the term for when there is a preceding message such as x>1 and x>0 so that I may look for a lesson on that topic?
Click to expand...
Without having the context of specific problem, I cannot tell you the precise meaning of a qualification such as x > 1. Generically, however,
x > 1 is as an instruction to ignore values of x unless they exceed 1. If you understand the concept of domain with respect to functions, such an instruction defines the "domain" of the problem.

With respect to rational functions or indeed any function that involves division, the domain necessarily excludes values of the variable that make the denominator equal to zero.
 
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