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HarveyKlaus
If |x| = |y| and xy = 0, which of the following must be true?

A \(xy^2>0\)
B. \(x^2y>0\)
C. \(x+y=0\)
D. \(x/(y+1)=2\)
E. \(1/x+1/y=1/2\)

Since |x| = |y| and xy = 0, both x and y must be zero.

Thus, we see that A and B are not true, because both answers are equal to zero.

C, however, must be true because 0 + 0 = 0.

Answer: C
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Bunuel niks18 chetan2u pushpitkc

Quote:
If |x| = |y| and xy = 0, which of the following must be true?


A \(xy^2>0\)

B. \(x^2y>0\)

C. \(x+y=0\)

D. \(\frac{x}{(y+1)}=2\)

E. \(\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\)

Can you explain solving modulus on both sides of equality if it was a PS problem.

|x| = |y|

Do I definitely need to know if x or y is positive or negative?

I approached in below manner:

If product of two numbers x and y is zero, then EITHER of x or y is zero, or
BOTH x and y are zero.

With this logic, A,B and E are out. (Since this is PS problem ALL of above
conditions must be satisfied.)

For D, if x = 0 then RHS = 2 is Not satisfied, since LHS = 0
If y = 0 , then |0| = 0 LHS = 0/1 or 0, RHS = 2. Not satisfied.

Let me know if my steps are correct.
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adkikani
Bunuel niks18 chetan2u pushpitkc

Quote:
If |x| = |y| and xy = 0, which of the following must be true?


A \(xy^2>0\)

B. \(x^2y>0\)

C. \(x+y=0\)

D. \(\frac{x}{(y+1)}=2\)

E. \(\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\)

Can you explain solving modulus on both sides of equality if it was a PS problem.

|x| = |y|

Do I definitely need to know if x or y is positive or negative?

I approached in below manner:

If product of two numbers x and y is zero, then EITHER of x or y is zero, or
BOTH x and y are zero.

With this logic, A,B and E are out. (Since this is PS problem ALL of above
conditions must be satisfied.)

For D, if x = 0 then RHS = 2 is Not satisfied, since LHS = 0
If y = 0 , then |0| = 0 LHS = 0/1 or 0, RHS = 2. Not satisfied.

Let me know if my steps are correct.

Hi adkikani

|x|=|y| implies magnitude of x and y are equal irrespective of their sign. For e.g. if x=2 and y=-2 then also |x|=|y|.

As you rightly mentioned xy =0 so either of the two has to be 0 but here as their magnitude are same so both x & y has to be 0

Only option C is valid in this case

Posted from my mobile device
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Can be solved using algebra as well -

Given -> |x| = |y| and xy = 0

squaring both sided \(x^2 = y^2 \)

=> \(x^2 - y^2 = 0 \)
=> (x+y)(x-y) = 0

From the given answer choices, we can say that x+y = 0.



HarveyKlaus
If |x| = |y| and xy = 0, which of the following must be true?


A \(xy^2>0\)

B. \(x^2y>0\)

C. \(x+y=0\)

D. \(\frac{x}{(y+1)}=2\)

E. \(\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\)
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