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Manager  G
Joined: 29 May 2019
Posts: 121
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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(x−y)^2 + 2y^2=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

We need sum of both (x−y)^2 and 2y^2 to be 18.
As both are squares 18 is sum of two positive integers.

If we consider y to be 1.
then 2y^2 = 2
therefore, (x−y)^2 must be 16
x-y = 4
y is 1 so x will be 5
( 5, 1 )

We can take y as
1,
$$\sqrt{2}$$
$$\sqrt{3}$$
$$\sqrt{4}$$
$$\sqrt{5}$$
$$\sqrt{6}$$
$$\sqrt{7}$$
$$\sqrt{8}$$

y can not be $$\sqrt{9}$$ as y = 3 then x will also be 3.

Its not an ordered pair.

So total values will be 8.

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MAKE IT HAPPEN Senior Manager  P
Joined: 18 May 2019
Posts: 453
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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(x-y)^2+2y^2=18
We are to find integer values of x and y that satisfy the above equation.
(x-y)>4 will produce LHS>RHS
So our combination integer values of x and y which can satisfy this equation must be such that (x-y)<=4.

When y=-1, (x+1)^2+2=18
x^2+2x-15=0
(x-3)(x+5)=0
Hence x=-5 or x=3
So we have (x,y) pairs of (-5,-1) and (3,-1)

When y=1, (x-1)^2+2=18
x^2-2x-15=0
(x+5)(x-3)=0
Hence x=-5 or x=3
We have (x,y) integer pairs of (-5,1) and (3,1).

Every other integral combinations of (x,y) does not result in a LHS which equals 18 (RHS).

Therefore there are 4 ordered pairs (x,y) that are solutions of the equation.

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Intern  B
Joined: 02 Jun 2013
Posts: 44
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Here (x−y)^2+2^2=18

So we can try this with number picking method too. Both elements are squared , so they have to be positive.

Now lets say y =0 then after putting this value in above equation we dont get the integer value of x.

Y= 1 , then x= 5,-3 so , here got two pair ( 5,1) (-3,1)
y= 2, then x cant be an integer
y= 3, then then x can be 3 so here the pair is (3,3) now we can have (-3,-3) too

So there are 4 pairs . Ans - B
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Manager  S
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Start populating with y=0 ---> x integer
y=1 ---> x not integer
y=2---->x not integer
y=3------> x integer.
therefore 1,3 satisfies also -1,-3 will satisfy
total 4 solutions
GMAT Club Legend  V
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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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solving for the given eqn ; we get
x^2+3y^2-2xy=18
test with y=1 we get x=5,-3 then at x=5 y=3,7
total pairs ; IMO B ; 4

(x−y)2+2y2=18(x−y)2+2y2=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

A. 2
B. 4
C. 6
D. 8
E. 10
Intern  B
Joined: 22 Jun 2019
Posts: 41
Concentration: Operations, Organizational Behavior
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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(x - y)^2 + 2 y^2 = 18
x^2 - 2 x y + 3 y^2 - 18 = 0

plug in numbers
x = -5, y = -1
x = -3, y = -3
x = -3, y = 1
x = 3, y = -1
x = 3, y = 3
Senior Manager  P
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Given equation is (x−y)^2+2y^2=18 , and x & y are integers

Instead of trying to solve this equation , we can do trail and error method

lets say (x−y)^2+2y^2=18

18 - could be written as 16+2 , 2*9,6*3 , 18*1

But Only 16+2 and 2*9 can will us solutions where x and y are integers

==> case 1 : where x = y = 3 , (3−3)^2+2*3^2=18
Case 2 : where x = y = -3 , (-3+3)^2+2(-3)^2=18
Case 3 : where x = 5 , y=1 , (5-1)^2+2*1^2=18
Case 4 : where x = 3 , y= -1 , (3+1)^2+2(-1)^2=18

Total 4 cases ( 3,3) ,(-3,-3) , (5,1) ,(4,-1)--Option B
Manager  G
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GMAT 1: 730 Q51 V36 Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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There are 6 integer solutions to this equation.
x = -5, y = -1
x = 5, y = 1
x = 3, y = -1
x = -3, y = 1
x = -3, y = -3
x = 3, y = 3
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Senior Manager  P
Joined: 20 Mar 2018
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Concentration: Finance, Real Estate
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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(x−y)^(2)+2y^(2)=18
(x-y)^(2) + 2y^(2)= (9•2)
(x-y)^(2) + 2y^(2)=3^2•2

(x-y)^2 =3^2 , 2y^2=2
x-y= +/-3, y= +/-1

Four instances (1) +and + ,(2) -and + ,(3)+ and - (4)- and -

Instance 1 : x-y =3 ,y=1 —> x=4
.: (4,1)
Instance 2: x-y=-3,y=1 —> x=-2
.: (-2,1)
Instance 3: x-y=3,y=-1 —> x=2
.: (2,-1)
Instance 4: x-y=-3,y=-1 —> x=-4
.: (-4,-1)

And yes x and y are integers since we have an integer on the RHS of the equation and sum of squares on the LHS

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Intern  B
Joined: 09 Jul 2019
Posts: 38
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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I manually counted and tried different numbers and got only four pairs of integers, therefore B
Intern  B
Joined: 11 Jun 2014
Posts: 26
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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(x-y)^2+2y^2=18
putting values as follow
x=3,y=3
x=5,y=1
x=-3,y=1
x=3,y=-1
x=-5,y=-1
x=-3,y=-3
Senior Manager  P
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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(x−y)^2+2y^2=18

For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

I've tried to put different numbers to this equation and there are 4 (x,y) pairs possible.

A. 2
B. 4
C. 6
D. 8
E. 10

B is the answer. _________________
My SC approach flowchart

(no one is ideal, please correct if you see any mistakes or gaps in my explanation, it will be helpful for both of us, thank you)

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Intern  B
Joined: 10 Aug 2017
Posts: 29
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?

A. 2
B. 4
D. 8
E. 10

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Manager  S
Joined: 30 Nov 2016
Posts: 102
Location: India
Concentration: Finance, Strategy
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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$$(x-y)^2$$=18-$$2y^2$$
the only values of y which gives us a integer value for x are 1,-1
so the corresponding pair(x,y) would be:(5,1),(-5,-1)
Manager  S
Joined: 10 Nov 2014
Posts: 63
Location: United States
Concentration: Marketing, Strategy
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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Square + even no = 18 - therefore the square should also be even or 0.

The only possibilities are
0 + 18
4 + 14
16 + 2

First scenario, 0 + 18
(x-y)^2 = 0 and 2y^2=18
x=y and y=+/-3
(x,y) = (3,3) or (-3,-3)

Second scenario, 4 + 14
2y^2 =14, y^2 =7 which is not possible since y is an integer

Third scenario, 16 + 2
(x-y)^2=16 and 2y^2=2
x-y=+/-4 and y=+/-1

if y=1 and x-y=4 , we get y=1 and x=5
if y=1 and x-y=-4 we get y=1 and x=-3
if y=-1 and x-y=4 , we get y=-1 and x=3
if y=-1 and x-y =-4 , we get y=-1 and x=-5

(3,3)
(3,-3)
(5,1)
(3,-1)
(3,-1)
(-5,-1)

6 pairs
Ans choice C Re: For how many ordered pairs (x , y) that are solutions of the equation   [#permalink] 17 Jul 2019, 21:39

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