Is x² - y 1/y² where y is non-zero?

Question

Is x² - y > 1/y² where y is non-zero?

(1) x < y
(2) x² < y²

PriyankaPalit7 · Answer

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.

arpitkansal · Answer

chetan2u  please help with this question.
I guessed the answer right but not sure about the approach.

KanishkM · Answer

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

chetan2u · Answer

Hi arpitkansal , What x^2x 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 Thus no definite answer.. E

rahul16singh28 · Answer

Take (x,y) = (0,1) & (0,-1).

We get E.

chetan2u · Answer

Hi Rahul, be careful with numbers you pick up..
(o, -1) does not fit into statement 1, which is x<y.. as 0>-1, meaning x>y

emockus · Answer

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.

ocelot22 · Answer

(1) x < y Let x = -3, y =-2. Then we have (-3)^2 --2 > 1/(-2)^2 --> is 9 -2 > 1/4? Yes Let x =-1/2 let y =-1/4 is (-1/2)^2 --1/4 > 1/(-1/4)^2 ? is 1/4 - 1/4 >4 NO NS (2) x^2 1/(-3)^2 ? is 4+3 > 1/9? Yes let x = 3 y =4 is (3)^2 -4 > 1/4^2? is 9-4 > 1/16 No NS (1) and (2) x < y and |x|<|y| so values like x = -4, y =-3 won't work Let x=0 y =1 is 0^2 -1 >1/1^2 No Let x =3, y=4 is 3^2 -4 > 1/4^2 ? is 9-4 >1/4^2 Yes NS Answer is E.

exc4libur · Answer

Hey VeritasKarishma Bunuel , could this be solved algebraically? x^2-y>1/y^2…x^2y^2-y^3>1? …x=2,y=3:(4)(9)-(27)>1?…36-27>1?…answer=yes …x=1,y=4:(1)(16)-(64)>1?…16-64>1?…answer=no (1) x < y: case and are possible; insufic. (2) x² < y²: case and are possible; insufic. (1&2): case and are possible; insufic. Answer (E)

KarishmaB · Answer

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

exc4libur · Answer

There I fixed it. Thanks

But can you solve this algebraically? Or is it a waste of time?

KarishmaB · Answer

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.

Is x² - y > 1/y² where y is non-zero?

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