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

Question

(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

Kinshook · Accepted Answer

(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 => (x-y)^2 = 18 NOT FEASIBLE since it is not a perfect square. (No integer solution)
Let us take y^2= 1 => 2y^2 = 2 & (x-y)^2=16 (1). FEASIBLE since it is a perfect square (integer solutions)
Let us take y^2= 4 => 2y^2=8 => (x-y)^2 = 10 NOT FEASIBLE since it is NOT a perfect square (No integer solutions)
Let us take y^2=9 => 2y^2 =18 => (x-y)^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
(x-y)^2 = 16 => x-y = 4 or x-y = -4
If y=1 and x-y=4 => x =5, y=1 (5,1) (i)
If y=1 and x-y=-4 => x=-3, y=1 (-3,1). (ii)
If y=-1 and x-y=4 => x=3 y=-1 (3,-1). (iii)
If y=-1 and x-y=-4 => x= -5 y=-1 (-5,-1) (iv)
Equation (1) provides 4 ordered pairs

Take equation (2)
(x-y)^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

gmatenthureturns · Answer

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.

Lampard42 · Answer

(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 (x-y)^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 (x-y)^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

firas92 · Answer

(x-y)^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)

nick1816 · Answer

(x-y)^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  (x-y)^2 =0 and  y^2 =9
 (x-y)^2=0 
or x-y=0.....(1)

y^2=9 
y= 3 or -3

We have 2 distinct integral values of y and 1 integral value of x-y. 
 Hence number of solutions possible= 2*1=2

Case 2  - When  (x-y)^2=16  and  y^2=1 
 (x-y)^2=16 
(x-y)=4 or -4

and  y^2=1 
y=1 or -1

We have 2 distinct integral values of x-y and 2 distinct integral values of y
 Number of solutions possible= 2*2=4

Total number of solutions possible= 2+4=6

IMO C

somesh86 · Answer

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

manish1708 · Answer

(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?

(x-y)^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 (x-1)^2 = 16 --> x-1 = (+/-)4 --> x = 5 or -3. Hence two pairs  (5,1)  and  (-3,1) 
If y = 2, then solving we get (x-2)^2 = 10...not a perfect square.
If y = 3, then solving we get (x-3)^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

Sayon · Answer

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. ->

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. ->    
So (9, 0) is a ordered pair. ->

y = 1  
 (x – 1)^2  + 2 = 18
 (x – 1)^2  = 16
x = 5 OR x = -3
So (-3, 1) is a ordered pair. ->    
So (5, 1) is a ordered pair. ->

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. ->

From  ,  ,  ,  ,   and   we have  6 ordered pairs .

Answer C

AntrikshR · Answer

Given: (x-y)^2 +2y^2 = 18
To Find: No. of pairs of (x,y) for which x and y will be integers.

(x-y)^2 +2y^2 = 18 can be written as (x-y)^2 = 18 - 2y^2
=> (x-y)^2 = 2(9-y^2)
=> (x-y)^2 = 2 (3+y)(3-y)
Now, since L.H.S is a perfect square, y can take values between   only.
(Any value greater than 3 or less than -3 will make R.H.S negative and L.H.S can't be negative as it's a perfect square.)
i.e. -3,-2,-1,0,1,2,3
For y = +3 or -3, R.H.S will be 0
=> x = +3 or -3
=> 2 pairs (3,3) and (-3,-3)
For y = +2 or -2,R.H.S will be 10
Now, 10 is not a perfect square so we will not get integral solutions for x.
For y = +1 or -1,R.H.S will be 16
Now, 16 is a perfect square
=> (x+1)^2 = 16 and (x-1)^2 = 16
=> x+1 = +-4 and x-1 = +-4
we will get 4 pairs by the above 2 quadratics i.e. (5,1),(-3,1),(3,-1),(-5,-1)
For y = 0,R.H.S will be 18
Now, 18 is not a perfect square so we will not get integral solutions for x.
=> total 6 integral solutions i.e. (3,3),(-3,-3),(5,1),(-3,1),(3,-1),(-5,-1)
Option C.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards