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For how many ordered pairs (x , y) that are solutions of the equation

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 16 Jul 2019, 13:15
(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.

Answer: D

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 16 Jul 2019, 13:39
(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.

The answer is B.

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 16 Jul 2019, 15:44
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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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 16 Jul 2019, 16:13
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
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For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 16 Jul 2019, 17:34
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
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 16 Jul 2019, 18:46
(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
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 16 Jul 2019, 20:12
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
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 17 Jul 2019, 00:41
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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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 17 Jul 2019, 01:40
(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

Answer B

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 17 Jul 2019, 02:06
I manually counted and tried different numbers and got only four pairs of integers, therefore B
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 17 Jul 2019, 03:43
Answer C
(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
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 17 Jul 2019, 05:08
(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. :heart
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 17 Jul 2019, 05:27
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 (Answer)
D. 8
E. 10

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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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New post 17 Jul 2019, 05:38
\((x-y)^2\)=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)
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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

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