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

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Manager
Joined: 24 Jan 2019
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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16 Jul 2019, 22:09
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)

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

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17 Jul 2019, 00:29
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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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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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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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

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

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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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17 Jul 2019, 03:43
(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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17 Jul 2019, 04:22
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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17 Jul 2019, 04:25
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

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

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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

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

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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
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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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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17 Jul 2019, 06:09
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).
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Re: For how many ordered pairs (x , y) that are solutions of the equation  [#permalink]

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17 Jul 2019, 06:54
1
(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

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

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17 Jul 2019, 07:27
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

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

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

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