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Re: For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 22:09
(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 (xy) 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, xy=4 and y=1 ................possible (x,y) pairs: (3,1),(3,1),(5,1),(5,1). second case, xy=0 and y = 3...........possible (x,y) pairs: (0,3),(0,3).
These are the only possibilities for given expression. (total 6)
ANSWER : C



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Re: For how many ordered pairs (x , y) that are solutions of the equation
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17 Jul 2019, 00:29
\((x−y)^{2}\)+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
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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
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17 Jul 2019, 01:40
(x−y)^(2)+2y^(2)=18 (xy)^(2) + 2y^(2)= (9•2) (xy)^(2) + 2y^(2)=3^2•2
(xy)^2 =3^2 , 2y^2=2 xy= +/3, y= +/1
Four instances (1) +and + ,(2) and + ,(3)+ and  (4) and 
Instance 1 : xy =3 ,y=1 —> x=4 .: (4,1) Instance 2: xy=3,y=1 —> x=2 .: (2,1) Instance 3: xy=3,y=1 —> x=2 .: (2,1) Instance 4: xy=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
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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
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17 Jul 2019, 03:43
Answer C (xy)^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
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17 Jul 2019, 04:22
\((x−y)^2\)+\(2y^2\)=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
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17 Jul 2019, 04:25
Solution: \((x  y)^2\) + \(2y^2\) = 18 Let's take xy = a & y=b We get, \(a^2\) + \(2b^2\) = 18...(i) Substituting values in equation (i), we can see that, if a=0 & xy =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 xy = 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
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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. 2B. 4C. 6 D. 8 E. 10B is the answer.
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Re: For how many ordered pairs (x , y) that are solutions of the equation
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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 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
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17 Jul 2019, 05:38
\((xy)^2\)=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
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17 Jul 2019, 06:09
\((x−y)^2+2y^2=18\)
Let xy = 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 xy=4, y=1 or xy=4, y=1 or xy=4, y=1 or xy=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 xy=0, y=3 or xy=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
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17 Jul 2019, 06:54
(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 : "(xy)^2" and "2y^2" could be positive integer or 0(zero) to satisfy the equation. We have only two options: 1) (xy)^2=16 and 2y^2=2> 16+2=18 2) (xy)^2=0 and 2y^2=18> 0+18=18.
1) 2y^2=2 > y=1 >(x1)^2=16 (x1)=4 > x=5 (x1)=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 >(x3)^2=0 (x3)=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
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17 Jul 2019, 07:27
\((x−y)^2+2y^2=18\)
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
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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 (xy)^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 (xy)^2=16 and 2y^2=2 xy=+/4 and y=+/1
if y=1 and xy=4 , we get y=1 and x=5 if y=1 and xy=4 we get y=1 and x=3 if y=1 and xy=4 , we get y=1 and x=3 if y=1 and xy =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
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