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Manager  S
Joined: 24 Jan 2019
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Location: India
Concentration: Strategy, Finance
GMAT 1: 730 Q51 V38 GPA: 4
Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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1
(x−y)^2+2(y^2)=18

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

Both terms have minimum value of zero. if x, y are integers then (x-y) is also an integer.

Here, we need to divide 18 in such a way that it is of the form : (square of some integer) + 2* (square of other integer). Then we need to check for the values of x, y.

18 = 16 + 2*(1) = 0 + 2*(9)

first case, |x-y|=4 and |y|=1 ................possible (x,y) pairs: (3,-1),(-3,1),(5,1),(-5,-1).
second case, |x-y|=0 and |y| = 3...........possible (x,y) pairs: (0,-3),(0,3).

These are the only possibilities for given expression. (total 6)

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

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1
$$(x−y)^{2}$$+2$$y^{2}$$=18

As we can see y can vary between -3 and +3.

Case 1: @y=+/-3, x is also +/-3.
So we have 2 solutions: (3,3) and (-3,-3)

Case 2:@y=+/-2 , no integer value of x satisfies the equation.

Case 3:@y=+1, x=5 or x=-3
So we have 2 solutions (-3,1) (5,1)

Case 4: @y=-1, x=3, x=-5
Sp we have 2 solutions (-5,-1) (3,-1)

Case 5: @y=0, no integer value of x satisfies the equation.

Therefore solutions are,
(3,3)
(-3,3)
(-3,1)
(5,1)
(-5,-1)
(3,-1)
Total 6 solutions.

Ans: C

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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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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
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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
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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
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
Manager  S
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GMAT 1: 670 Q49 V32 Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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1
$$(x−y)^2$$+$$2y^2$$=18

Possible values for $$y^2$$ can be from 1 to 9.
But to make y an integer, $$y^2$$ can only be 1, 4 or 9.
If $$y^2$$=1 =>y=1 or -1
If $$y^2$$=4 =>y=2 or -2
If $$y^2$$=9 =>y=3 or -3

Now, if y=1,
$$(x−y)^2$$+$$2y^2$$=18
$$(x−1)^2$$+2=18
$$(x−1)^2$$ =16
So, x=5 or x=-3

Thus we get two points : (5,1), (-3,1)

Similarly if y=-1,
$$(x−y)^2$$+$$2y^2$$=18
We will get two values of x=3 & -5

Thus we get two more points : (-5,-1), (3,-1)

If y=-2 or 2, X will not be an integer.

If y=-3 or 3, we will get two values of x : -3 & 3 respectively.

Thus we get two more points : (-3,-3), (3,-3)

So, ans should be 6.
Option (C)
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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1
Solution:
$$(x - y)^2$$ + $$2y^2$$ = 18

Let's take x-y = a
& y=b
We get,
$$a^2$$ + $$2b^2$$ = 18...(i)
Substituting values in equation (i),

we can see that, if a=0 & x-y =0
b = 3 then & y=3. Therefore x= 3
(3,3) satisfying
(-3,-3) satisfying

Substituting values in equation (i)
a=4,-4,
b=1,-1

x-y = 4 & y=1
only (5,1) , (-5,-1) ,(3,-1) ,(-3,1) are satisfying,

So in total we have integer 6 pairs which are the solutions for x & y

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

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1
$$(x−y)^2+2y^2=18$$

Let x-y = z

$$z^2+2y^2=18$$

Both z^2 and y^2 are perfect squares. Following equations are possible with perfect square values:

16+2.(1) = 18 ---> eqn 1
0+2.(9) = 18 ---> eqn 2

For eqn 1, z=4, y=1 or z=-4, y=-1 or z=4, y=-1 or z=-4,y=1
x-y=4, y=1 or x-y=-4, y=-1 or x-y=4, y=-1 or x-y=-4,y=1
x=5,y=1 or x=-5,y=-1 or x=3,y=-1 or x=-3,y=1
There are 4 ordered pairs

For eqn 2, z=0, y=3 or z=0, y=-3
x-y=0, y=3 or x-y=0, y=-3
x=3,y=3 or x=-3,y=-3
There are 2 ordered pairs

In total, there are 6 ordered pairs.

Option (C).
Senior Manager  G
Joined: 25 Jul 2018
Posts: 273
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?

Both two parts of the equation cannot be negative :
"(x-y)^2" and "2y^2" could be positive integer or 0(zero) to satisfy the equation.
We have only two options:
1) (x-y)^2=16 and 2y^2=2---> 16+2=18
2) (x-y)^2=0 and 2y^2=18---> 0+18=18.

1) 2y^2=2 --> y=1 --->(x-1)^2=16
(x-1)=4 --> x=5
(x-1)=-4 --> x=-3 (That means we have 2 pairs for y=1: (5;1) (-3;1)

y=-1 --> (x+1)^2=16
(x+1)=4 --> x=3
(x+1)=-4 --> x=-5 (That means we have 2 pairs for y=-1: (3;-1) (-5;-1)

2) 2y^2=18 y=3 --->(x-3)^2=0
(x-3)=0 --> x=3 (That means we have 1 pair for y=3: (3;3)

y=-3 --> (x+3)^2=0
(x+3)=0 --> x=-3 (That means we have 1 pair for y=-3: (-3;-3)
In total, 4 pairs for Case1 + 2 pairs for Case2 =6

The answer choice is C.
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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1
$$(x−y)^2+2y^2=18$$

Possible values
Case 1: 0^2 + 2*3^2 = 18
--> (x, y) = (3, 3), (-3, -3)

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

So, 6 pairs are possible

IMO Option C

Pls Hit Kudos if you like the solution
Manager  S
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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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