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Re: Is x² - y > 1/y² where y is non-zero?
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19 Oct 2018, 03:10
This is the solution that I found. I do not find this explanation satisfactory.
Can any of the experts please explain the solution better? Thanks in advance
Let's start by simplifying the original by multiplying both sides by y² (we can safely do so because we know that y² is positive): Is \(x^2y^3\) - \(y^3\) > 1? (1) x < y If x and y are big positive numbers, \(y^3\) will dominate x²y² and the left side will be negative; if x and y are big negative numbers, x²y² and (-\(y^3\)) will both be positive and the left side will be a big positive number: insufficient. (2) x² < y² Tells us that |x| < |y|, but nothing about the signs of x and y. If we pick really big values with different signs we can generate both "yes" and "no" answers. Together: combining doesn't help us at all, since we have the same generalizations from both: choose E.
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Re: Is x² - y > 1/y² where y is non-zero?
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31 Jan 2019, 12:21
PriyankaPalit7 wrote:
Is x² - y > 1/y² where y is non-zero?
(1) x < y (2) x² < y²
\(x^2\) - y > 1/\(y^2\)
\(x^2\) - y - 1/\(y^2\) > 0
Lets try 2 first (2) x² < y² or y^2 > x^2, 9 > 4, y can be -3 or 3 and x can be -2 or 2 1/4 > 1/16, y can be 1/2 or -1/2 and x can be 1/4 or -1/4
4 +3 - 1/9 > 0, This answers a yes to the question 1/16 - 1/2 - 1/4 < 0, This answers a no to the question
from 1) x < y or y>x, here y & x can be again a fraction, integer or a number 1/2 > 1/4 or 3 > 2 1/16 - 1/2 - 1/4 < 0, This answers a no to the question 4 -3 - 1/9 > 0, This answers a yes to the question
The same test cases can be used again to show that after combining the 2, we wont get an unique answer
E
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Quote which i can relate to. Many of life's failures happen with people who do not realize how close they were to success when they gave up.
What \(x^2<y^2\) means is \(|x|<|y|\).. X<y and |x|<|y| when combined mean y>x and y>0.. By looking at the equation, \(x^2-y>\frac{1}{y^2}\), we should realise that \(\frac{1}{y^2}\) will be positive. We have to see if x^2-y can be both negative and positive. 1) x=2 and y=3 So \(2^2-3>\frac{1}{3^2}....4-3>\frac{1}{9}...1>\frac{1}{9}....\)yes 2) x=1 and y=4 \(1^2-4>\frac{1}{4^2}.....1-4>\frac{1}{16}.....-3>\frac{1}{16}...\).no
Is x² - y > 1/y² where y is non-zero?
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01 Feb 2019, 08:47
1
Here's how I solved, feel free to critique!
Is \(x^{2}\) - y > \(\frac{1}{y^{2}}\)? If \(y\neq{0}\).
I simplified the expression as follows:
Because \(y^{2}\) is always positive I multiplied both sides by \(y^{2}\) to get \(x^{2}y^{2}\) - \(y^{3}\) > 1.
From here I factored out \(y^{2}\) to get \(y^{2}\)(\(x^{2}\) - y) > 1.
The question is now:
Is \(y^{2}\)(\(x^{2}\) - y) > 1? If Y \(x\neq{0}\).
Statement 1: x < y
I picked x as -1 and y as 1. Plugging these values into the question the result is 0 > 1.
I then picked x as -3 and y as -2. Plugging these values into the question the result is 44 > 1.
Clearly insufficient.
Statement 2: \(x^{3}\) < \(y^{3}\)
I picked x as -1 and y as 1. Since they are raised to an odd power the signs remain the same for x and y (1 and -1, respectively). Plugging these values into the question the result is 0 > 1.
I then picked x as -3 and y as -2. Since they are raised to an odd power the signs remain the same for x and y (-3 and -2, respectively). Plugging these values into the question the result is 44 > 1.
Clearly insufficient.
Statement 1&2
Clearly insufficient.
Therefore, the answer is E.
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(1) x < y: case [I.] and [II.] are possible; insufic. (2) x² < y²: case [I.] and [II.] are possible; insufic. (1&2): case [I.] and [II.] are possible; insufic.
Answer (E)
Originally posted by exc4libur on 08 Oct 2019, 08:22.
Last edited by exc4libur on 09 Oct 2019, 07:20, edited 1 time in total.
(1) x < y: case [I.] and [II.] are possible; insufic. (2) x² < y²: case [I.] and [II.] are possible; insufic. (1&2): case [I.] and [II.] are possible; insufic.
Answer (E)
You are still using number plugging (which is pretty much the way to go in this question). Also, how come the two examples work for both statements? When x = -3 and y = -4, x > y, not x < y
x < y and x^2 < y^2 when x and y are both positive. Try x = 1, y = 2 x = 1/4, y = 1/2
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Re: Is x² - y > 1/y² where y is non-zero?
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09 Oct 2019, 07:21
VeritasKarishma wrote:
You are still using number plugging (which is pretty much the way to go in this question). Also, how come the two examples work for both statements? When x = -3 and y = -4, x > y, not x < y
x < y and x^2 < y^2 when x and y are both positive. Try x = 1, y = 2 x = 1/4, y = 1/2
There I fixed it. Thanks But can you solve this algebraically? Or is it a waste of time?
Re: Is x² - y > 1/y² where y is non-zero?
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10 Oct 2019, 03:34
exc4libur wrote:
VeritasKarishma wrote:
You are still using number plugging (which is pretty much the way to go in this question). Also, how come the two examples work for both statements? When x = -3 and y = -4, x > y, not x < y
x < y and x^2 < y^2 when x and y are both positive. Try x = 1, y = 2 x = 1/4, y = 1/2
There I fixed it. Thanks But can you solve this algebraically? Or is it a waste of time?
I wouldn't even try algebra here. x < y and x^2 < y^2 should remind you of number properties. For what kind of numbers does this hold? For non negative numbers. We know that numbers behave differently 'between 0 and 1' and '1 onwards'. So I will try both type of numbers.
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51% (02:15) correct 
