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For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 07:00
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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? A. 2 B. 4 C. 6 D. 8 E. 10
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For how many ordered pairs (x , y) that are solutions of the equation
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Updated on: 16 Jul 2019, 09:12
(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? A. 2 B. 4 C. 6 D. 8 E. 10 Let us take y^2=0 => (xy)^2 = 18 NOT FEASIBLE since it is not a perfect square. (No integer solution) Let us take y^2= 1 => 2y^2 = 2 & (xy)^2=16 (1). FEASIBLE since it is a perfect square (integer solutions) Let us take y^2= 4 => 2y^2=8 => (xy)^2 = 10 NOT FEASIBLE since it is NOT a perfect square (No integer solutions) Let us take y^2=9 => 2y^2 =18 => (xy)^2 =0 (2) FEASIBLE since it is a perfect square (integer solutions) Take equation (1) 2y^2 = 2=> y^2 = 1 => y=1 or y=1 (xy)^2 = 16 => xy = 4 or xy = 4 If y=1 and xy=4 => x =5, y=1 (5,1) (i) If y=1 and xy=4 => x=3, y=1 (3,1). (ii) If y=1 and xy=4 => x=3 y=1 (3,1). (iii) If y=1 and xy=4 => x= 5 y=1 (5,1) (iv) Equation (1) provides 4 ordered pairs Take equation (2) (xy)^2=0 => x=y 2y^2=18 => y=3 or y=3 If y=3 => x=3 (3,3) If y=3 => x=3 (3,3) Equation (2) provides 2 ordered pairs Total six (6) ordered pairs are solution IMO C
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Kinshook Chaturvedi Email: kinshook.chaturvedi@gmail.com
Originally posted by Kinshook on 16 Jul 2019, 07:29.
Last edited by Kinshook on 16 Jul 2019, 09:12, edited 1 time in total.




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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, 07:33
C
Clearly y can't be greater than 3. In order for x to be integer, y can only take +/ 1 and +/ 3
For y =1, we get x = 5 and 3  two solutions For y = 1, we get x = 3 and 5  two solutions For y = 3, we get x = 3 For y = 3, we get x = 3
Total 6 solutions.



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For how many ordered pairs (x , y) that are solutions of the equation
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Updated on: 16 Jul 2019, 21:32
(x−y)^2+2y^2=18  Includes both +ve and ve Integers
(x−y)^2 can have Perfect squares values of 0,1,4,9,16 only. If (x−y)^2 is odd then 2*y^2 must be odd  which is not possible. So (x−y)^2 can take only 0,4 or 16.
If (x−y)^2 is 4 then 2*y^^2=14; y^2=7 not possible. Hence (xy)^2 can be 16 which gives y=1 or 1. If Y=1 then X=5 or 3 and Y=1 then X= 3 or 5. When (xy)^2 is 0, 2y^=18, y=3,3. Then x=3,3 Hence 4 ordered pairs are possible (5,1),(3,1),(3,1) , (5,1), (3,3) and (3,3).
IMO C
Originally posted by Lampard42 on 16 Jul 2019, 07:38.
Last edited by Lampard42 on 16 Jul 2019, 21:32, edited 1 time in total.



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For how many ordered pairs (x , y) that are solutions of the equation
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Updated on: 16 Jul 2019, 08:18
(xy)^2 and 2y^2 are always non negative
So let's see what are the cases we can get a perfect square + 2*perfect square = 18 (We take perfect squares since we need x and y to be integers)
Perfect squares: 0, 1, 4, 9, 16  (1) 2*Perfect squares: 0, 2, 8, 18  (2)
So only cases when an element from list 1 and an element from list 2 add up to 18 is (0+18) and (16+2)
(0+18)  Possible when (x,y) = (3,3) and (3,3)  2 cases (16+2)  Possible when (x,y) = (5,1), (5,1), (3,1) and (3,1)  4 cases
Therefore we have 6 cases in total
Answer is (C)
Originally posted by firas92 on 16 Jul 2019, 07:41.
Last edited by firas92 on 16 Jul 2019, 08:18, edited 1 time in total.



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For how many ordered pairs (x , y) that are solutions of the equation
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Updated on: 16 Jul 2019, 12:02
(xy)^2+2y^2=18 As \(18=0^2+2*3^2\) and \(18= 4^2+2*1^2\) Hence there are 2 cases possible Case 1 when \((xy)^2\)=0 and \(y^2\)=9 \((xy)^2=0\) or xy=0.....(1) \(y^2=9\) y= 3 or 3 We have 2 distinct integral values of y and 1 integral value of xy. Hence number of solutions possible= 2*1=2 Case 2  When \((xy)^2=16\) and \(y^2=1\) \((xy)^2=16\) (xy)=4 or 4 and \(y^2=1\) y=1 or 1 We have 2 distinct integral values of xy and 2 distinct integral values of y Number of solutions possible= 2*2=4Total number of solutions possible= 2+4=6 IMO C[There are couple of reasons for why we don't need to find the solutions of (x, y) 1. If (xy) and y are integers, x will always be an integer. 2. Values of y are different in both cases, hence there is no overlap possible. 3 it will save some precious time. ]
Originally posted by nick1816 on 16 Jul 2019, 09:33.
Last edited by nick1816 on 16 Jul 2019, 12:02, edited 1 time in total.



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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, 10:42
Given,
(x − y)^2 + 2*y^2 = 18
This can be seen as 18 being the sum of 2 squares.
Case 1:
=> The first perfect square less than 18 is 16. Let us see if we can have integer values of x and y such that one of the squars is 16
if y = 1, we can try to get (x  y)^2 to be 16.
=> x = 5, y = 1 satisfies the equation. Now let us play with the signs of x and y and see which combinations satisfy the equation. By doing so we get the following 3 combinations of x and y.
1. (5 , 1) 2. (5 , 1) 3. (5 , 1)
Case 2:
=> The second perfect square less than 18 is 9. Let us see if we can have integer values of x and y such that one of the squars is 9
if y = 3, we can try to get (x  y)^2 to be 9.
=> x = 3, y = 3 satisfies the equation. Now let us play with the signs of x and y and see which combinations satisfy the equation. By doing so we get the following 3 combinations of x and y.
4. (3 , 3) 5. (3 , 3) 6. (3 , 3)
Case 3:
=> The third perfect square less than 18 is 4. Let us see if we can have integer values of x and y such that one of the squars is 4
By putting multiple values of y we see that we do not have an integer pair for x and y such that the equation satisfies where one of the squares in the LHS is the perfect square 4.
Thus we conclude that we have 6 unique integer pairs of x and y which satisfy the given equation.
Answer: C



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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, 10:46
(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?
(xy)^2 + 2y^2 = 18 In the above equation both square terms will be positive and less than 18 respectively.
By trail and error, If y = 0, then solving we get x^2 = 18.. not a perfect square. If y = 1, then solving we get (x1)^2 = 16 > x1 = (+/)4 > x = 5 or 3. Hence two pairs (5,1) and (3,1) If y = 2, then solving we get (x2)^2 = 10...not a perfect square. If y = 3, then solving we get (x3)^2 = 0 > x = 3. One pair (3,3) If y = 4, then 2y^2 becomes > 18 hence y cannot take any value > 3
Similarly y can take negative numbers 1, 2, 3... anything 4 and below we will have second term > 18 hence not possible.
If y = 1, then x+1 = (+/)4 > x = 3, 5. Hence two pairs (3, 1) and (5, 1) If y = 2, then (x+2)^2 = 10. Not a perfect square. If y = 3, then (x+3)^2 = 0 > x = 3. One pair (3, 3)
Hence total pairs (x, y) are (5,1); (3,1); (3,3); (3, 1); (5, 1); (3, 3) Total number of pairs = 6.
A. 2 B. 4 C. 6 D. 8 E. 10
Answer Choice: C



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For how many ordered pairs (x , y) that are solutions of the equation
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16 Jul 2019, 20:15
We need to find ordered pairs such that x and y both are integers. Let us look at the equation once and see what we can determine:
\((x – y)^2\) will always be positive > (a) \(y^2\) will always be positive > (b) Thus, 18 will be the result of the sum of squares of two positive numbers. > (c)
Looking at the equation and from (b) and (c) we find that y can take a range of value such that \(3 \leq{y} \leq{+3}\) Thus, we can find all ordered pairs in the given range of y
y = 3 \((x + 3)^2\) – 18 = 0 x = 3 So (3, 3) is a ordered pair. > [1]
y = 2 \((x + 2)^2\)+ 8 =18 \((x + 2)^2\) = 10 > 10 is not a square of any integer.
y = 1 \((x + 1)^2\) – 2 = 18 \((x + 1)^2\) = 20 > 20 is not a square of any integer.
y = 0 \(x^2\) = 18 x = 9 OR x = 9 So (9, 0) is a ordered pair. > [2] So (9, 0) is a ordered pair. > [3]
y = 1 \((x – 1)^2\) + 2 = 18 \((x – 1)^2\) = 16 x = 5 OR x = 3 So (3, 1) is a ordered pair. > [4] So (5, 1) is a ordered pair. > [5]
y = 2 \((x – 2)^2\) = 10 > 10 is not a square of any integer.
y = 3 \((x 3)^2\) + 18 = 18 x = 3 So (3, 3) is a ordered pair. > [6]
From [1], [2], [3], [4], [5] and [6] we have 6 ordered pairs.
Answer C




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