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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
+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
-1

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.......

trumpstinks  May 16, 2020
#3
+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
0

thanks!! I thought it was only integers ;( trumpstinks  May 16, 2020
#5
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