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Find all pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 56$. For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(2,4),(-3,9)" (without the quotation marks).

 May 16, 2020
 #1
avatar+29978 
+2

Rearrange the first equation to get y = 10 - x.

 

Plug this into the second equation:  x^2 +(x - 10)^2 = 56

 

Expand this and rearrange to get a quadratic in x.  Solve for the two values of x. then substitute these back into y = 10 - x to find the corresponding two values of y.

 May 16, 2020
 #2
avatar+36 
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I am a bit confused. I did what you told me, and got x^2+x^2-20x+100 = 2x^2-20x+100=56. I divided my 2 to get x^2-10x+50=28, and subtracted by 25 on both sides to get x^2-10x-25=3. I factored the left side to give (x-5)(x-5) = 3. The two values of x I got from here were 8 and 6. So, the sets of x and y should be (8,2), and (6,4), but these do not satisfy x^2+y^2 = 56.......

 

Please help me! Thanks!

trumpstinks  May 16, 2020
 #3
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+1

x +y =10,

x^2 +y^2 =56, solve for x, y


Use substitution to get:
x = 5 - sqrt(3)  AND  y = 5 + sqrt(3), OR:
x = 5 + sqrt(3)  AND  y = 5 - sqrt(3)

 May 16, 2020
 #4
avatar+36 
+1

thanks!! I thought it was only integers ;(

indecision

trumpstinks  May 16, 2020
 #5
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0

As far as I can see, there is NO integer solution. 2, 8 or 3, 7 or 4, 6. None of these will balance the 2nd equation, unless it is 52, or 58 instead of 56.

Guest May 16, 2020
edited by Guest  May 16, 2020

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