If x^2 = y^2, is true that x > 0 ? (1) x = 2y+1 (2) y <= -1

Hi!

I did a bit of a unnecessarily long way but got to the wrong answer and was hoping someone could tell me what my logical error is.

i squared both sides of statement 1 and got to \(x^2 = 4y^2 + 4y +1\)

then I replaced \(x^2\) with \(y^2\) and got the same as everyone else that y = -1 or y = -(1/3) NOT sufficient

Using statement B i eliminated y=-(1/3) but where I got it wrong is that I thought that since y = -1 then x could be equal to +/- 1 so I chose E

Am i missing something?

Thank you in advance or any responses!

If x^2 = y^2, is true that x>0?

(1) x=2y+1

(2) y<= -1

Hi Bunuel & Karishma, I will appreciate if you can comment:

1- How can we see this question and decide whether we need to pick numbers or not?

2- In other words, combining the two statements here is fairly easy, but how do we get it to a 50-50 between C and E in 5 to 10 seconds?

3 - In other words, how do we ensure that we don't waste time plugging in numbers in statement 1?

Thanks. What are the implications of |y|<|x|, is it "y<x or y>-x" or something different?

Question time:

for #2.) we are given y<=-1. This is not sufficient because of the following:

|x| = |y|

x=y
OR
x=-y

x=-y
OR
x=-(-y) x=y

Correct?

Another thing that has bothered me is this. If x=-y, and for example, y=5, then would x=(-5)?

As always, thanks for the help!

In the following statement,

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

what if, y=-x=-1 ?
Then x=1>0 and hence (1)+(2) = insufficient

Don't we assume x<0 when we infer x=-y from x^2=y^2 (and when x>0 x=y)? Isn't this the normal procedure when we have |x|=something. I got x=-y and x=y from the statement, but then got confused by the assumptions made to open the abs value. Thanks.

Hi chetan2u, Bunuel,
I did understand the solution that you have shared , but what did i do wrong in my approach. I am getting answer E . I understand correct answer is C
here is how i tried to solve it
So we are given that x^2 =y^2
means we have |x|=|y|
So we have four cases from it ( understand that it finally transoms to two cases but just listed the cases as i did in my original attempt)
\(x\geq{0}\) , y\(\geq{0}\) we have x=y
\(x\geq{0}\) , y< 0 we have x=-y
x<0 , y\(\geq{0}\) we have -x=y
x<0 , y<0 we have x=y

Now statement A says :
x= 2y+1
squaring on both sides
\(x^2\)= (2y+1)^2
now since we are given that\(x^2\)=\(y^2\)replace \(x^2\) with\(y^2\)
we get
y=-1 or/and y=-\(\frac{1}{3}\)
now we can have
x=y=-1 or x=1 and y=-1
x=y=-\(\frac{1}{3}\) or x=\(\frac{1}{3}\) and y=-\(\frac{1}{3}\)

Statement 2:
\(y\leq{1}\)
we will have either x>0 or x<0
because
\(x\geq{0}\) , y< 0
or
x<0 , y<0


Now combing 1 and 2 we have
y=-1
so two cases are applicable ( i am referring to 4 cases that show relation between x and y)
x<0 , y<0 we have x=y
we will get x=y=-1
and for case this we get
\(x\geq{0}\) , y< 0 we have x=-y
x=1 and y=-1

I did not understand what Kris01 meant to consider true for all statements. I considered both statements and came to conclusion that y=-1

Where did i go wrong .
Please help me understand my mistake .
Probus

If x^2 = y^2, is true that x>0?

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunnel,

I need a clarification here, when i put statement 1 in X^2 = Y^2 i get the below

3Y^2= -4Y-1,

and since LHS is positive so the RHS needs to be positive so in order to RHS to be positive Y should be "-ve" and when Y is negative, Now

considering statement 1 we can get to know that X is negative.....

So my point is again statement 2 needed?

Please clarify me if i am wrong?

If x^2 = y^2, is true that x>0?

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

Can someone tell me if my reasoning is sound?

x^2=y^2 and thus, x=y or x=-y

1.)

I.) x=2y+1

y=2y+1
-y=1
y=-1

y=2(-1)+1
x=-1

II.) x=2y+1

-y=2y+1
-3y=1
y=-1/3

x=2(-1/3)+1
x=1/3

Insufficient because we have two values for x. (however, using the same reasoning to arrive at c. cant we say that both answers we arrive at show that x ≠ 0 and thus, a is sufficient?)

2. y≤-1 Not sufficent

1+2 2. says that y≤-1. In #1, the only case where y≤-1 is y=2y+1

-y=1
y=-1

y=2(-1)+1
x=-1

Here we get an answer of x=-1 which is obviously ≠ to 0.

If x^2 = y^2, is true that x>0?

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel

Why cant answer be B

From x^2=y^2
x and y can take the following values:
1 and -1 or
-1 and 1
as x and y different so not considering same integers like both positive or both negative

from 1st statement none of the values satisfy the given equation
from 2nd statement, y is negative which makes x as positive. Where am I wrong?