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If x + y = 2 and x^2 + y^2 = 2, what is the value of xy ?
A. –2
B. –1
C. 0
D. 1
E. 2
_________________
A. –2
B. –1
C. 0
D. 1
E. 2
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hello,
Here is the answer, hope it is clear
Here is the answer, hope it is clear
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Posts: 50
Bunuel
(x+y)^2 = x^2+y^2+2xy
4=2+2xy
xy=1
D
General Discussion
Bunuel
It is based on algebraic formula..
(X+y)^2=x^2+y^2+2xy..
Substitute the values given..
2^2=2+2xy...
2xy=4-2=2.... xy=1
D
_________________
Bunuel
x^2 + y^2 = 2 - Given
we can write the above equation as \((x+y)^2\) - 2xy = 2
\((2)^2\)-2xy = 2
=> 4 - 2xy = 2 => -2xy = -2 => xy = 1.
Option D.
Given
(x+Y) = 2 and X^2 + Y^2 = 2
(x+Y) = 2
Squaring both sides.
X^2 + Y^2 + 2xy = 4
2 + 2xy = 4
Solving for xy = 1
Option D
(x+Y) = 2 and X^2 + Y^2 = 2
(x+Y) = 2
Squaring both sides.
X^2 + Y^2 + 2xy = 4
2 + 2xy = 4
Solving for xy = 1
Option D
x+y = 2 or x = 2-y
x^2+Y^2=2
(2-Y)^2+Y^2=2
y^2-2Y+1=0
Y=1
X=2-y
X=1
Ans 1 D
x^2+Y^2=2
(2-Y)^2+Y^2=2
y^2-2Y+1=0
Y=1
X=2-y
X=1
Ans 1 D
Shrivathsan
Nice! Simple question.... How is it possible to see this under 2 min? LOL!?
Bunuel
Guys what is the relationship between x + y = 2 and x^2 + y^2 = 2 in question itself
? 
dave13
Hi dave13
In the question itself, there is no relationship between the 2 equations except
that they are talking about the same variable x and y.
In questions like these, we are basically given two simultaneous linear equations
and are excepted to find the value of x and y solving these equations.
Some basic formulae you need to remember are \((x+y)^2 | (x-y)^2 | x^2 - y^2\)
Hope this helps!
Two equations are given to form a basic algebra equation.
(x+y)^2= x^2+y^2+2xy
Hence value of xy=1
Sent from my iPhone using GMAT Club Forum
(x+y)^2= x^2+y^2+2xy
Hence value of xy=1
Sent from my iPhone using GMAT Club Forum
pushpitkc
Hello pushpitkc
thanks for your reply
ok i understand how from this x^2 + y^2 = 2, we do this (x+y) (x+y) = x^2+xy+yx+y^2 = x^2+y^2+2xy = 2 ok now what ? and what am i to do with x + y = 2 ? there are no roots, powers etc here
Hey dave13
We are given in the question stem that \(x+y = 2\) and \(x^2 + y^2 = 2\)
After you have figured that \((x+y)^2 = x^2 + y^2 + 2xy\)
Substituting these values, we get\(2^2 = 2 + 2xy\) -> \(4 - 2 = 2xy\) -> \(xy = \frac{2}{2} = 1\)(Option D)
Hope this helps!
We are given in the question stem that \(x+y = 2\) and \(x^2 + y^2 = 2\)
After you have figured that \((x+y)^2 = x^2 + y^2 + 2xy\)
Substituting these values, we get\(2^2 = 2 + 2xy\) -> \(4 - 2 = 2xy\) -> \(xy = \frac{2}{2} = 1\)(Option D)
Hope this helps!
pushpitkc
pushpitkc many thanks for explanation. yes it helps just partially
ok from this \(x^2 + y^2 = 2\) we got this \(x^2 + y^2 + 2xy =2\) ok I understand this part so far now you say we substitute values - which values are you talking about? and in which equation do you substitute ? how do you get this \(2^2 = 2 + 2xy\) also how do you get in it \(2^2\)
I would appreciate your detailed explanation for dummies
thanks!
dave13
Hey dave13
So we have an equation \((x+y)^2 = x^2 + y^2 + 2xy\) -> (1)
The question stem has given us the following details:
\(x+y = 2\)
\(x^2 + y^2 = 2\) -> (2)
Because \(x+y = 2\), value of \((x+y)^2 = 2^2 = 4\) -> (3)
Substituting values from (2) and (3) in equation (1)
\((x+y)^2\) = \(x^2 + y^2\)\(+ 2xy\)
4 = 2 + 2xy
2xy = 4-2 = 2
Therefore, xy = \(\frac{2}{2} = 1\)
Hope it is clear now.
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Bunuel
(x+y)^2= x^2+y^2+2xy
4 = 2 + 2xy
xy = 1
IMO D
Posted from my mobile device
x+y=2
x^2+y^2=2
replace values of xy from options , that give answer =2 as per equation.
x+y=2 => 1+1=2
x^2+y^2 =>1^2+1^2=2
xy =>1*1=2
So, 1 is the answer
Option D
x^2+y^2=2
replace values of xy from options , that give answer =2 as per equation.
x+y=2 => 1+1=2
x^2+y^2 =>1^2+1^2=2
xy =>1*1=2
So, 1 is the answer
Option D
If x and y were any integer other than 1, then the sum of the two squares would not be 2. To verify that both are 1 even more, we can see that x+y=2.
Bunuel
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