How to find x and y for this? x^2 + (1-y)^2 + (x-y)^2 = 1/3

NinjaTxas

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How to find x and y for this? x^2 + (1-y)^2 + (x-y)^2 = 1/3

Here is the imagine of this equation wiritten in a more readable way: http://prntscr.com/hin4zf

It was in a math competition in Lithuania for 10th grade students. I couldn't figure out the answer.
 
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How to find x and y for this? x^2 + (1-y)^2 + (x-y)^2 = 1/3

Here is the imagine of this equation wiritten in a more readable way: http://prntscr.com/hin4zf

It was in a math competition in Lithuania for 10th grade students. I couldn't figure out the answer.
Easiest way to do this is to plot it!!!
 
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It seems wrong to ask you to "find x and y" for a single equation; is that what the problem actually said? It is important to quote the entire problem exactly (or translate it carefully, if it is in Lithuanian).

I would expand the left side (eliminate parentheses) and solve for y (which will require completing the square or using the quadratic formula). Then I might initially try to graph it. But what you will find is that y is a real number for only one value of x, which makes the question suddenly make sense. It actually has an answer.
 
It seems wrong to ask you to "find x and y" for a single equation; is that what the problem actually said? It is important to quote the entire problem exactly (or translate it carefully, if it is in Lithuanian).

I would expand the left side (eliminate parentheses) and solve for y (which will require completing the square or using the quadratic formula). Then I might initially try to graph it. But what you will find is that y is a real number for only one value of x, which makes the question suddenly make sense. It actually has an answer.

Thank you for your answer. I study maths in Lithuanian, so I don't see English exercises often. It was written in Lithuanian. I can translate the task from LT to EN. It sound like that "Which what x and y values this is a correct equality?"

So what you're saying is that I should make a "function" (That is how we call it in LT, I don't know how it is called in English).I find what y is equal to, then I choose 2 or more y's and calculate the x's. Then I make a graph and see where it intersects. I seem to get the concept, but I get stuck at trying to solve for y. I can't part x from y. Even if I get that "y=..." in that "..." there is y too, and it also looks really wrong.

Here is where I go out of ideas.
http://prntscr.com/hj4hgh
http://prntscr.com/hj4hku
 
Plot means graph.

Subhotosh may have misread the exercise. This equation is not easy to plot. :cool:
I know... but I did not misread it.

While trying to plot it - I discovered the solution suggested by Peterson.

I was hoping OP will discover that too...
 
Thank you for your answer. I study maths in Lithuanian, so I don't see English exercises often. It was written in Lithuanian. I can translate the task from LT to EN. It sound like that "Which what x and y values this is a correct equality?"

My point was not that the question was written badly, but that it is surprising! Normally one would not solve one equation for both variables, but in this case it can be done, as you discover when you do the work of trying to graph the equation (as Khan implied and I stated explicitly).

So what you're saying is that I should make a "function" (That is how we call it in LT, I don't know how it is called in English).I find what y is equal to, then I choose 2 or more y's and calculate the x's. Then I make a graph and see where it intersects. I seem to get the concept, but I get stuck at trying to solve for y. I can't part x from y. Even if I get that "y=..." in that "..." there is y too, and it also looks really wrong.

Here is where I go out of ideas.
http://prntscr.com/hj4hgh
http://prntscr.com/hj4hku

The word "function" is correct in English; "solving for y" is the same as "expressing y as a function of x".

But I don't understand what you are saying you describe here, "choose 2 or more y's and calculate the x's".

The work in the first link is correct until you started trying to solve for y. You have, in effect,

2x^2 + 2y^2 - 2xy - 2y + 2/3 = 0

But you did not write it as a quadratic in y, which is why you still had y in the result. I would rearrange the equation as

2y^2 - (2x + 2)y + (2x^2 + 2/3) = 0

so that a = 2, b = 2x + 2, and c = 2x^2 + 2/3. Then you will be using the quadratic formula and looking at the discriminant, in particular.
 
While trying to plot it - I discovered the solution suggested by Peterson.
I misunderstood your claim that the "easiest way to solve [this exercise] is to plot [the equation]". I had not interpreted your instruction as meaning 'trying to plot'. Please excuse me. :cool:
 
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