Difficult System of Equation Prob: Given xy(x+y)=48 and xy+x+y=14. Calculate x^2+y^2

Janlloyd

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The problem is as follows:

Let x and y be integers such that xy(x+y)=48 and xy+x+y=14. Calculate x^2+y^2.

Answer Choices:
(A) 13
(B) 16
(C) 20
(D) 40
(E) 68

So far I have come up that xy=48/(x+y) and I substitute that into the other equation making 48/(x+y)+x+y=14. After that I feel completely stuck,
If I multiply x+y to everything else I still end up with two variables leaving me nowhere. Still I do kind of know the answer. I looked at the multiple
choices and saw that 20 could be separated into 16 and 4 having roots of 4 and 2. 4 and 2 do indeed fit into both equations, but this method is
likely not going to help in other problems. If anyone can explain it would help me a great deal!
 
The problem is as follows:

Let x and y be integers such that xy(x+y)=48 and xy+x+y=14. Calculate x^2+y^2.

Answer Choices:
(A) 13
(B) 16
(C) 20
(D) 40
(E) 68

So far I have come up that xy=48/(x+y) and I substitute that into the other equation making 48/(x+y)+x+y=14. After that I feel completely stuck,
If I multiply x+y to everything else I still end up with two variables leaving me nowhere. Still I do kind of know the answer. I looked at the multiple
choices and saw that 20 could be separated into 16 and 4 having roots of 4 and 2. 4 and 2 do indeed fit into both equations, but this method is
likely not going to help in other problems. If anyone can explain it would help me a great deal!

14 - (x+y) = xy = 48/(x+y) ................. Let x+y = u

48/u = 14 - u

u2 - 14u + 48 = 0

u = 6, 8 → x+y = 6 or 8 → xy = 8, 6

x2 + y2 = (x+y)2 - 2xy ...................continue
 
The problem is as follows:

Let x and y be integers such that xy(x+y)=48 and xy+x+y=14. Calculate x^2+y^2.

Answer Choices:
(A) 13
(B) 16
(C) 20
(D) 40
(E) 68

So far I have come up that xy=48/(x+y) and I substitute that into the other equation making 48/(x+y)+x+y=14. After that I feel completely stuck,
If I multiply x+y to everything else I still end up with two variables leaving me nowhere. Still I do kind of know the answer. I looked at the multiple
choices and saw that 20 could be separated into 16 and 4 having roots of 4 and 2. 4 and 2 do indeed fit into both equations, but this method is
likely not going to help in other problems. If anyone can explain it would help me a great deal!
My 1st thought would be to let xy=u and x+y=v.
Then uv=48 and u+v=14. Want two numbers that multiply out to 48 and add up to 14. How about 6 and 8? What can you do from here?
 
My 1st thought would be to let xy=u and x+y=v.
Then uv=48 and u+v=14. Want two numbers that multiply out to 48 and add up to 14. How about 6 and 8? What can you do from here?

Interestingly, you can express x^2 + y^2 directly in terms of u and v: (x^2 + 2xy + y^2) - 2xy = v^2 - 2u. But since either u or v can be 6, this gives two possible answers.

It looks like you have to actually solve for x and y (or at least partly solve) in order to answer the question. Only one choice of assignments for u and v yields integer values for x and y.
 
Interestingly, you can express x^2 + y^2 directly in terms of u and v: (x^2 + 2xy + y^2) - 2xy = v^2 - 2u. But since either u or v can be 6, this gives two possible answers.

It looks like you have to actually solve for x and y (or at least partly solve) in order to answer the question. Only one choice of assignments for u and v yields integer values for x and y.
OK, I see what you are saying about getting two solutions. Thank you. You are correct, that this is very interesting.

Edit: I see now that I am very guilty of sloppy thinking as I never verified that both x and y are integers! I'll go sit in the corner for a bit.
 
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This is a fairly well-known fact, provable by expanding the right hand side:

(x+y)2 - 2xy = (x2 + 2xy + y2) - 2xy = x2 + y2.

Ahhh I see, thank you very much, this problem makes a lot more sense now!
I'm guessing the key to this problem was to substitute x+y with another variable

Also sorry for asking a lot of questions, I'm trying to compete in a math competition and my teacher does not offer much help
so thank you very much for your help.
 
My 1st thought would be to let xy=u and x+y=v.
Then uv=48 and u+v=14. Want two numbers that multiply out to 48 and add up to 14. How about 6 and 8? What can you do from here?

48/u=14-u2
u2-14u+48=0
u=8,6

xy=8
x+y=6; x=6-y

(6-y)y=8
y2-6y+8=0
(y-2)(y-4)=0
y=2,4
 
48/u=14-u2
u2-14u+48=0
u=8,6

xy=8
x+y=6; x=6-y

(6-y)y=8
y2-6y+8=0
(y-2)(y-4)=0
y=2,4

Correct so far; if you had chosen to make xy = 6, you would have found non-integer values.

Now find x, and then x2 + y2; or else use my trick to get the answer directly from u and v without needing specific values for x and y.
 
While there, work on this:
xy(x + y) = 819
xy + x + y = 63
Yes, x and y are positive integers :rolleyes:
My corner is very dark. I can't even use a calculator or get any online help while there.
 
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