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For how many ordered pairs (x , y) that are solutions of the equation
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Updated on: 17 Jul 2019, 00:04
Trying to plug some integers into x and y at the very moment is not the way we should approach this problem. At least we should try to modify the equation to ease our calculation and thereby reduce the time spent. \((x−y)^2+2y^2=18\) \((x−y)^2=18  2y^2\) now the task is much easier. As we see \((x−y)^2\) must be a nonnegative integer and a perfect square becuase both \(x\) and \(y\) are integers. Accordingly, \(18  2y^2\) must be equal to that perfect square (examples of perfect squares: 4, 9, 16, 25, 36, etc). Making \(18  2y^2\) a perfect square is an easy task by plugging some numbers in: If \(y=0\), then \(18  2*0^2=18\) is not a perfect square. Thus \(0\) is out. If \(y=1\), then \(18  2*1^2=16\) is a perfect square. Thus \(y\) can be \(1\). Substituting \(y\) with \(1\) in the original equation we get that \(x=5\) If \(y=1\), then \(18  2*(1)^2=16\) is a perfect square. Thus \(y\) can be \(1\). Substituting \(y\) with \(1\) in the original equation we get that \(x=3\) If \(y=2\), then \(18  2*2^2=10\) is not a perfect square. Thus \(2\) is out. If \(y=2\), then \(18  2*(2)^2=10\) is not a perfect square. Thus \(2\) is out. If \(y=3\), then \(18  2*3^2=0\) is a perfect square. Thus \(y\) can be \(3\). Substituting \(y\) with \(3\) in the original equation we get that \(x=3\) If \(y=3\), then \(18  2*(3)^2=0\) is a perfect square. Thus \(y\) can be \(3\). Substituting y with \(3\) in the original equation we get that \(x=3\) Any integer greater than \(3\) or lower than \(3\) will make \(18  2y^2\) negative. Thus we have \(4\) pairs of \(x\) and \(y\): \((5, 1), (3, 1), (3, 3)\), and \((3, 3)\) Hence B
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Originally posted by JonShukhrat on 16 Jul 2019, 08:59.
Last edited by JonShukhrat on 17 Jul 2019, 00:04, 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, 09:07
Sum of 2 squares is equal to 18 so options are 9 +9 16+2



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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, 09:10
Sum of 18 by addition of 2 (one square and other twice of square ) can be done in 2 ways 0+2*9 , here y can be 3 and x can be 3 16+2*1 , x can be 5 y will be 1 so 4 options (3,3)(3,3)(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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16 Jul 2019, 09:10
(xy)^2 + 2*y^2 = 18 => It means (x,y) should be those values which yield squares. => for 18 the possible values for the above equation are: 0,2,16,18 => such as first term can be 0 and second term can be 18 (2*9). If we try to choose values for (x,y) then there are 6 possible values which will satisfy the above equation: (3,3) , (3,3),(5,1),(5,1),(3,1),(3,1) As we can only these values will satisfy the equation.
IMO the answer is C.
Please hit kudos if you like the solution.



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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, 09:16
(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?
1. Let's transform and draw inference:
(x−y)^2+2y^2=18 (x−y)^2=182y^2 (x−y)^2=2(3y)(3+y)
As (x−y)^2 => 0 in any cases: 2(3y)(3+y) => 0 as well then 3<=y<=3
2. Let's find all possible variants: We know that x and y are integers, so: y = 3, 2 ...2,3 Let's input: (x−y)^2=2(3y)(3+y) 1.y = 3 ; x = 3 2.y = 3 ; x = 3 3.y = 2 ; x not int 4.y = 2 ; x not int 5.y = 1 ; x = 5 or 3 (x−1)^2 = 16 x1 = 4 x = 5 or 3 6.y = 1; x = 3 or 5 (x+1)^2 = 16 x+1 = 4 x = 3 or 5 7.y = 0 ; x not int
So # of unique pairs: 6
ANSWER C
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: 17 Jul 2019, 07:37
the given equation can be rewritten as: \((xy)^2 = 2(y3)(y+3)\) the LHS must be \(≥ 0\), if \(LHS = 0\), then \(x = y\), and y can be either \(3\) or \(3\), so the possible pairs are \((3,3)\) or \((3,3)\) > 2 pointsif \(LHS > 0\), then RHS must >0 which can happen when \((y3)<0\) and \((y+3)>0\) or vise versa if (y3)<0, then y<3 and when (y+3)>0, then y >3, so the possible common values in the interval \(3>y>3\) is \(2,1,0,1,2\) By trying \(2\) or \(2\), the equation will be \((x2)^2 = 10\), so x can't be an integer By trying \(0\), the equation will be \(x^2 = 18\), so x can't be an integer By trying \(1\) or \(1\), the equation will be \((x1)^2 = 16\), so \(x1 = 4\), and x can be \(5\) or \(3\), and the possible pairs will be \((5,1), (3,1), (5,1), (3,1)\) > 4 pointsso we have a total of \(6\) points where (x,y) pairs are integers C
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Originally posted by MahmoudFawzy on 16 Jul 2019, 09:16.
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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, 09:17
(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?
Solution: (x−y)^2+2y^2=18 = 0 + 2*3^2 = 4^2 + 2*1^2 case1: (x−y)^2+2y^2 = 18 = 0 + 2*3^2 will give us integer solution of x & y so xy =0 & 2y^2=2*3^2 => y = +/ 3 so solution1: x=y =3 & solution2: x=y =3 case2: (x−y)^2+2y^2=18 = 4^2 + 2*1^2 so (xy)^2 = 4^2 & 2y^2 = 2*1^2 => xy= +/4 & y = +/ 1 so solution3: y = +1, x y = +4 => y = +1, x = +5 solution4: y = +1, x y = 4 => y = +1, x = 3 solution5: y = 1, x y = +4 => y = 1, x = +3 solution6: y = 1, x y = 4 => y = 1, x = 5 So total 6 integer solutions of the equation
A. 2 B. 4 C. 6 > correct 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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(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 here its given x and y are integers for for that (x−y)^2 has to be positive and also perfect square , and for that 2y^2 has to be less than 18. ( as (x−y)^2=18  2y^2) so y takes values 3,2,1,0,1,2,3 ( for 2y^2 to be less than 18) out of which 2, 2 and 0 dont give (x−y)^2 a perfect square. so remaining 4 numbers y can take for which x are if y=1 , then (x−1)^2 =16 gives x as 5 or 3 so 2 pairs if y = 1 , then (x+1)^2 =16 gives x as 3 and 5 so 2 more pairs if y =3, , then (x−3)^2 =0 gives x = 3 one pair if y = 3, then (x+3)^2 =0 gives x =3 one more pair on total 6 pairs ie ans C
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Originally posted by ccheryn on 16 Jul 2019, 09:20.
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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, 09:28
IMO C
(x−y)^2 + 2y^2=18
(xy)^2 = 18 2y^2
For Integral solution, left hand side should be a perfect square.
Perfect squares less than / equal to 18= {16, 9 , 4, 1 , 0}
Now, {16, 9 , 4, 1, 0 } = 18 2y^2 y= integer, y= +1, 18 2y^2= 16 [ Perfect square ] y= +3, 18 2y^2= 0 [ Perfect square ] .... Satisfies condition
Case 1 : y= +1, (xy)^2=18 2y^2= 16 (xy)= +4 x= +4 +y [/color]
(x,y)= (3,1), (5,1), (3,1), (5,1)
Case 2 : y= +3, (xy)^2=18 2y^2= 0 (xy)= 0 x=y
(x,y)= (3,3), (3,3)
Total ordered pair= 4+2=6



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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, 09:38
by plugging the numbers, pairs of x,y would be 3, 3 3, 3 0, 4 0, 4 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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16 Jul 2019, 09:51
Keep calm and sharpen your pencils: Algebra and inequalities are coming
1) Deduct \(2y^2\) from both sides to get: \((xy)^2=182y^2\)
2) Factor out 2 from the righthand side: \((xy)^2=2*(9y^2)\)
3) Note that the righthand side cant be less than 0, because the lefthand side is the square and the square>=0
4) So \((9y^2)>=0\) \((3y)*(3+y)>=0\) \(3<=y<=3\) We have following integer values for y: 3, 2, 1, 0, 1, 2, 3
5) insert each of the y value and check which yields integer value of x y=3; \((x+3)^2=2*(99)\); x=3 y=2; \((x+2)^2=2*(94)\); x is not an integer y=1; \((x+1)^2=2*(91)\); \((x+1)^2=16\); x may be 3 or 5, so here two x and y pairs ar possible y=0; \((x0)^2=2*(90)\); x is not an integer y=1; \((x1)^2=2*(91)\); \((x1)^2=16\); x=5 or x=3; two pairs y=2; \((x2)^2=2*(94)\); x is not an integer y=3; \((x3)^2=2*(99)\); x=3
So we have got 6 pairs of x and y
IMO Ans: 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, 09:54
Answer is B. but Im still not sure if its right.



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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, 09:57
IMO C
(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?
Notice that, one part of the equation is square of an integer (xy)^2 and other part has to be even 2y^2 After plugging in few integers for e.g. x=3, y=3, we get (xy)^2=0 and 2y^2=18, next x=1,y=17 is not in the above mentioned form so we can check for the next set of values. So, we can clearly find the pattern and see that, There are only 6 pairs possible for the given equations and they are as below: 1. (3,3) 2. (3,3) 3. (3,1) 4. (3,1) 5. (5,1) 6. (5,1)
One thing to notice here is that for each pair for e.g. (3, 3), there is another pair exists in which we just need to flip the signs, as (3, 3). Same goes others pairs.



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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, 10:01
2*y^2 is always even so , we need to have (xy)^2 to be also always be even. Also the LHS is always positive. Now, 2*y^2 given its integer and given the value to choose for y from 3,2,1,0,1,2,3.else 2*y^2 becomes more than 18. y=0 ,x we end up with x^2 and no integer has 18 as their square. y=2/2, (xy)^2 needs to be 10 ,not possible with integers
y=1, x can be 5 y=1 x can be 3 y=3 then x can be 3 y can be 3 then x can be 3 so we have 4 such cases , hence B



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For how many ordered pairs (x , y) that are solutions of the equation
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Quote: \((x−y)^2+2y^2=18\)
For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers? First of all, let us find out how we can write the equation in a different way to make it easier to solve it. \((x  y) ^ 2 = 18  2y ^ 2\) \((x  y) ^ 2 = 2 * (3 ^ 2  y ^ 2)\) Obviously, first solution is when \(y^2 = 3^2\) and the right part of the equation is 0. y ^ 2 = 9 when y = +3 or when y = 3. Thus, in the first case x = 3 and in the second case x = 3. As the left part of the equation is in square, it is always \((x  y) ^ 2 >= 0\). Thus, \(2 * (3^2  y^2) >= 0\). As \(2 * (3^2  y^2) >= 0\), then \(y^2 =< 3^2\) or \(y^2 =< 9\). Let us consider now the option when y = +1 or y = 1. With y=1: \(2*(9  1^2)=2*(91)=2*8=16\) \((xy)^2=16\) => \((x1)^2=16\) \(x=5\) and \(x=3\) With y = 1: \(2*(9  (1)^2)=2*(91)=2*8=16\) \((xy)^2=16\) => \((x(1)^2=16\) \((x+1)^2=16\) \(x=3\) and \(x=5\) Let us now consider when y=+2 or y=2: \(2*(92^2)=2*(9(2)^2)=2*(94)=2*5=10\) Then, \((xy)^2=10\) but \(\sqrt{10}\) is not an integer, thus this pair of integers does not interest us. Till now, we came up with four pairs of integers: 1) x = 5 and y = 1 2) x = 3 and y = 1 3) x = 3 and y = 1 4) x = 5 and y = 1 5) x = 3 and y = 3 6) x = 3 and y= 3 Thus, the answer is 6 options. Answer: C
Originally posted by RusskiyLev on 16 Jul 2019, 10:06.
Last edited by RusskiyLev on 16 Jul 2019, 10:35, 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, 10:07
\((x−y)^2+2y^2=18\) Solving this equation we get, \(x^2+y^22xy + 2y^2=18\) = \(x^2 + 3 y^2  2xy = 18\)
By the method of substitution, 1. Let y=1, we get \(x^2 + 3  2x =18.\) or \(x^2 2x 15=0.\) Solving this quadratic equation we get x=5 and 3. This gives us 2 ordered pairs  (5,1) and (3,1) .
2. Let y=1, we get \(x^2 + 3 + 2x =18\) or\(x^2 + 2x 15=0.\) Solving this quadratic equation we get x=5 and 3. This gives us 2 ordered pairs  (5,1) and (3,1) .
Similarly, substituting values for y=2 we get x=6 and 2. > (6,2) and (2,2) y=2 we dont get integer values for x. y=3 we get x=3. > (3,3) y=3 we get x=3. > (3,3)
Hence, giving aa total of 8 ordered pairs. The answer is 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, 10:13
taking value of y and finding corresponding values of x. they will be counted as a solution if we get an integer x for corresponding y chosen.
(xy)^2 + 2y^2 = 18
taking y =1 (x1)^2 = 16 x1 = +4 x = 5;3
taking y = 1 (x+1)^2 = 16 x = 3;5
taking y = 3 (x3)^2 = 0 x = 3;
taking y = 3 (x+3)^2 = 0 x = 3;
All the solutions together :
(5,1)(3,1)(3,1)(5,1)(0,3)(0,3)
Hence 6 solutions



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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, 10:14
\((xy)^2\)=182\(y^2\), RHS should equal to square of a positive number from LHS If y=0, then RHS is 18, but there is no square of any integer that equals 18, thus y cannot be zero If y=1, then RHS is 16, this one works because \(4^2\)=16 > x=5 (\((51)^2\)=\(4^2\)) (5, 1) is a first set if y=2, then RHS is 10, does not work because no square of integer equals to 10 If y=3, then RHS is 0, it is possible because \(0^2\)=0, x must be 3 too to produce difference of 0. (3,3) is the second set No more positive numbers will work for y because then RHS will be less than zero and no square of anything produces a negative number. Let's try some negative integers (we will NOT get a negative result on RHS because y is in power of 2). If y=1, then RHS is 16, it works as we saw above. x=3 (3(1)=4), (3, (1)) is a third set If y=2, then RHS is 10, does not work as we saw above If y=(3), then RHS is 0, works as we saw from above, x in this case must be (3)>(3(3)=0, ((3), (3)) is a fourth set If y=(4), then RHS is a negative integer and it is possible to equal to square on RHS as square on RHS will always produce a nonnegative result. Thus, we have only 4 set of integers and 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, 10:16
For how many ordered pairs (x , y) that are solutions of the equation above are x and y both integers?
Thinking logically. We have two positive numbers and their squares ( in the other situation double square) cannot be more than 18. Thus if one of them if 4 then other no more then zero. Let's just plug some numbers keeping in mind our observation. It is easier to do with y. y=0, x is not an integer. y=1, x=3;5 y=1, x=5;3 y=2, x is not an integer. y=2 x is not an integer. y= 3 x=3 y=3, x=3 y=4 is impossible because square will be equal to the negative number.
Thus we have 6 pairs. IMO 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, 10:30
Given (xy)^2 + 2y^2 = 18 To find the pair solutions of x and y that are integers Let y = 1 therefore, (x1)^2 + 2(1)^2 = 18 (x1)^2 = 16 thus, x = +5 or 3 also, let y = 1 therefore, (x+1)^2 + 2(1)^2 = 18 we have (x+1)^2 = 16 thus, x = +3 or 5 simiarly, let y = 3 and y = 3 will yield x = 3 and x = 3 respectively Therefore the pairs for (x,y) include (3,3), (3,3), (3,1), (5,1), (5,1) and (3,1). Hence answer choice C.
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