(1) y > 0
(2) x = y
Answer is C
My answer is B, but it is wrong.
My method:-
y^2 + 3 > y^2 (After plugging y for x)
Now, if I plug any value for y, whether it is positive or negative the y^2 + 3 > y^2 is true.
(-2)^2 + 3 > (-2)^2
4 + 3 > 4
7 > 4
Now putting positive value.
(2)^2 + 3 > (2)^2
4 + 3 > 4
7 > 4
I am not able to understand where is the problem. Even if i take fraction for y, then also i am concluding on the same terms.
Please help
Thanks & Regards
Vinni
Please help
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- kmittal82
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I think this is where you are wrong:
y^2 + 3 > y^2
The inequality sign would change if y is negative, so you can't really say the above is true, unless you know for a fact that y > 0
y^2 + 3 > y^2
The inequality sign would change if y is negative, so you can't really say the above is true, unless you know for a fact that y > 0
- neelgandham
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Is 1/x > y/((y^2)+3)
Is 1/y > y/((y^2)+3)
Is 1/y - y/((y^2)+3) > 0
Is (y^2)+3-(y^2)/(y*((y^2)+3)) > 0
Is 3/(y*((y^2)+3)) > 0 ?
(y^2)+3 > 0 , but for y can be > 0 or < 0, So we cannot, for certain, answer the question
So 3/(y*((y^2)+3)) > 0
Hence sufficient to answer the question.
Doesn't speak of y. *&**(1) y > 0
Is 1/x > y/((y^2)+3)(2) x = y
Is 1/y > y/((y^2)+3)
Is 1/y - y/((y^2)+3) > 0
Is (y^2)+3-(y^2)/(y*((y^2)+3)) > 0
Is 3/(y*((y^2)+3)) > 0 ?
(y^2)+3 > 0 , but for y can be > 0 or < 0, So we cannot, for certain, answer the question
We know that y > 0From 1 and 2
So 3/(y*((y^2)+3)) > 0
Hence sufficient to answer the question.
Anil Gandham
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- vinni.k
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Thanks, but how can an inequality sign changes. Rules are if you multiply or divide an inequality by a negative number , then inequality sign changes. Here i am not multiplying a value by negative number. I am just plugging a negative value inside the square. I am not convinced by your explanation.kmittal82 wrote:I think this is where you are wrong:
y^2 + 3 > y^2
The inequality sign would change if y is negative, so you can't really say the above is true, unless you know for a fact that y > 0
I liked the neelgandham's method, but I will really appreciate if someone can clear my doubt.
Thanks & Regards
Vinni
- krusta80
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kmittal82 is referring to when you cross-multiplied. You ALWAYS need to consider the possible signs of the denominators when cross multiplying inequalities, because in essence you are multiplying both sides by the denominators when you cross multiply.vinni.k wrote:Thanks, but how can an inequality sign changes. Rules are if you multiply or divide an inequality by a negative number , then inequality sign changes. Here i am not multiplying a value by negative number. I am just plugging a negative value inside the square. I am not convinced by your explanation.kmittal82 wrote:I think this is where you are wrong:
y^2 + 3 > y^2
The inequality sign would change if y is negative, so you can't really say the above is true, unless you know for a fact that y > 0
I liked the neelgandham's method, but I will really appreciate if someone can clear my doubt.
Thanks & Regards
Vinni
Now, let's look at the denominators...
y^2+3 will always be positive, but...
Since x can be negative or positive in part (2) alone, you need to consider both cases. So, in essence, x is the negative number that kmittal is referring to.
- kmittal82
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thanks for clarifying on my behalf krutsa, couldn't have explained it better myselfkrusta80 wrote:kmittal82 is referring to when you cross-multiplied. You ALWAYS need to consider the possible signs of the denominators when cross multiplying inequalities, because in essence you are multiplying both sides by the denominators when you cross multiply.vinni.k wrote:Thanks, but how can an inequality sign changes. Rules are if you multiply or divide an inequality by a negative number , then inequality sign changes. Here i am not multiplying a value by negative number. I am just plugging a negative value inside the square. I am not convinced by your explanation.kmittal82 wrote:I think this is where you are wrong:
y^2 + 3 > y^2
The inequality sign would change if y is negative, so you can't really say the above is true, unless you know for a fact that y > 0
I liked the neelgandham's method, but I will really appreciate if someone can clear my doubt.
Thanks & Regards
Vinni
Now, let's look at the denominators...
y^2+3 will always be positive, but...
Since x can be negative or positive in part (2) alone, you need to consider both cases. So, in essence, x is the negative number that kmittal is referring to.