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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, 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 xy = 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
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16 Jul 2019, 13:39
(xy)^2+2y^2=18 We are to find integer values of x and y that satisfy the above equation. (xy)>4 will produce LHS>RHS So our combination integer values of x and y which can satisfy this equation must be such that (xy)<=4.
When y=1, (x+1)^2+2=18 x^2+2x15=0 (x3)(x+5)=0 Hence x=5 or x=3 So we have (x,y) pairs of (5,1) and (3,1)
When y=1, (x1)^2+2=18 x^22x15=0 (x+5)(x3)=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
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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
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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
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16 Jul 2019, 17:34
solving for the given eqn ; we get x^2+3y^22xy=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
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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
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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 , (51)^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
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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, 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, 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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