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

Given : x^2 = y^2 => x=+/-y
DS : x>0

Option 1 : x = 2y+1
x^2 = [(x-1)/2]^2
=> 4x^2 = x^2 +1 -2x
=> 3x^2 +2x -1 = 0
=> (x+1) (3x-1) = 0
=> x= -1, 1/3

NOT SUFFICIENT

Option 2 :
y <= -1
So there can be many values of x . x<=-1 or x>=1
NOT SUFFICIENT

Combined :
x= -1,1/3
so y = 1, -1, 1/3, -1/3
But y <=-1
So x = -1 is possible

SUFFICIENT

Answer C

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

From (1) we have only two possibilities:
a. \(y=x=-1\)
b. \(y=-x=-\frac{1}{3}\)

y = -1 and x = 1 does not satisfy the first statement.

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.

You could use number plugging to check whether you are right.

x^2 = y^2:

The above could be true for example if x = y = -1 or if x = 1 and y = -1. These examples answer no to both of your question. Generally x^2 = y^2 means that |x| = |y|, so x and y are at the same distance from 0 on the number line.

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

|x| is the distance from x to 0 on the number line, so |y| < |x| means that x if further from 0 than y is.

----x----y----0--------------
----x---------0----y---------
---------y----0---------x----
--------------0----y----x----

Thanks Bunuel. It looks like I did not capture all cases with my inequalities. For example, when y>0, I have y is always less than x, which is not true based on case 2. When faced with this condition on DS, is the best way to draw out the number line like this? How does this compare to getting a condition that says |x-5|>2 in terms of how you would get the constraints.

10. Absolute Value

• Theory
  Math: Absolute value (Modulus)
  Absolute Value: Tips and hints
  Properties of Absolute Values on the GMAT
  Absolute modulus : A better understanding
  GMAT Question Patterns: Abolute Value
  The Reason Behind Absolute Value Questions on the GMAT
  
  • Questions
    Inequality and absolute value questions from my collection
    The E-GMAT Question Series on ABSOLUTE VALUE
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For more check Ultimate GMAT Quantitative Megathread

Hope it helps.

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

Hi...

You have gone wrong in not considering equation 1 while taking value of X..
1) X=2y+1
You got two values of y as -1/3 and -1
Now substitute y as -1/3 , X=2*(-1/3)+1=-1/3
has -1, x=2*(-1)+1=-2+1=-1
So only two values are left -1/3 and -1 for X..
|x|=|y| may give as 1 as a value but statement I negates it
2) y<=-1
Nothing about X

Combined it tells you that y=-1 and so X=-1
Sufficient

C

Bunuel
Can I write the highlighted part like the bold part?
----y----x----0--------------
----x---------0----y---------
---------y----0---------x----
--------------0----x----y----

Lets solve the question
1)
x = 2y +1

we can't know if x>0 if we don't know the value of y.
Insufficient

2) y<=-1

Alone, this obviously tells us nothing about x! Insufficient. Answer is either C or E.

1+2)

Let's plug a few values of y that are <= -1 into the equation from situation #1 to see how they affect x:

x=2y+1
x=2(-1)+1 = -1
x=2(-2)+1 = -3
x=2(-3)+1 = -5

C is the answer.

No. |y| < |x| means that x if further from 0 than y is. Is x further from 0 than y is in your examples?

I get my answer. Thank you so much...

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

(1) x = 2y+1
\((2y+1)^2 = y^2\)
\((2y+1)^2 - y^2 =0\)
(y+1)(3y+1) =0
y = -1 or -1/3
Since x = 2y+1
If y=-1; x = -1
If y=-1/3; x = 1/3
(x,y) = {(-1,-1),(1/3,-1/3)}
x = -1 or 1/3
NOT SUFFICIENT

(2) y <= -1
Still y^2 = +ve
x = +y or -y
NOT SUFFICIENT

(1)+(2)
(1) x = 2y+1
\((2y+1)^2 = y^2\)
\((2y+1)^2 - y^2 =0\)
(y+1)(3y+1) =0
y = -1 or -1/3
Since x = 2y+1
If y=-1; x = -1
If y=-1/3; x = 1/3
(x,y) = {(-1,-1),(1/3,-1/3)}
x = -1 or 1/3
(2) y <= -1
y=-1
x=-1<0
SUFFICIENT

IMO C

Can we solve this using graphical approach? Could someone post a solution as well?