The problem is: x + y = 50, x2 + y2 + (x - y)2 = 1400. [Hint: Subtract the squar
ID: 3107836 • Letter: T
Question
The problem is:x + y = 50, x2 + y2 + (x - y)2 = 1400.
[Hint: Subtract the square of the first equation from twice the second equation to get a quadratic in x - y.]
This is what I have so far:
2(x2 + y2 +(x - y)2 - (x + y)2 = 2500-1400
2x2 + 2y2 + 2(x - y)2 - (x + y)2 = 1100
2x2 + 2y2 + 2(x - y)(x - y) - (x + y)(x + y) = 1100
2x2 + 2y2 + 2(x2 - xy - xy + y2) - (x2 + xy + xy + y2) = 1100
2x2 + 2y2 - 2xy - 2xy - xy - xy = 1100
2x2 + 2y2 - 6xy = 1100
x2 + y2 - 3xy = 550
From here I don't know how to get it in a quadratic in (x - y).
Also, I know the answer is x=30 and y=20, but don't know how to get there.
Explanation / Answer
I would start out differently than the hint personally You have the equations x + y = 50 and x^2 + y^2 + (x - y)^2 = 1400. So the first equation can be y= 50-x Then substitute for the y's and get x^2 + (50-x)^2 + (x - (50-x))^2 = 1400 Then begin simplifying x^2 + (50-x)^2 + (2x - 50)^2 = 1400 x^2 + (2500-100x+x^2) + (4x^2 -200x +2500) = 1400 6x^2 + -300x +5000 = 1400 6x^2 + -300x + 3600 = 0 Then you can you the quadratic formula and find x=20 or 30 and then put that back into the other equation y=50-x and find that the answer to be x=30, y=20 and x=20, y=30
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