+1768 Find all ordered pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 32 + 2xy$.
For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(2,4),(-3,9)" (without the quotation marks).
+37147 x+y = 10 =====> y = 10 -x sub this definition of y into the second equation
x^2 + (10-x)^2 - 2 x(10-x) -32 = 0 now solve for the value(s) of x
x^2 + x^2 - 20x + 100 - 20x + 2x^2 -32 = 0
4 x^2 -40x + 68 = 0 Quadratic Formula shows x = 7.828 and 2.172
using these values in the line equation shows y = 2.172 and 7.828
+37147 x+y = 10 =====> y = 10 -x sub this definition of y into the second equation
x^2 + (10-x)^2 - 2 x(10-x) -32 = 0 now solve for the value(s) of x
x^2 + x^2 - 20x + 100 - 20x + 2x^2 -32 = 0
4 x^2 -40x + 68 = 0 Quadratic Formula shows x = 7.828 and 2.172
using these values in the line equation shows y = 2.172 and 7.828
+37147 Here is perhaps a simler , exact - answer way:
x^2 + y^2 - 2xy = 32
(x-y)^2 = 32
x -y = +- 4 sqrt 2
y = x - 4 sqrt 2 and x + 4sqrt 2
then use the line quation:
x + y = 10
x + x- 4 sqrt 2 = 10
2x = 10 +4 sqrt 2
x = 5 +2 sqrt 2 and the other is x = 5 - 2 sqrt 2 Then the corresponding 'y' s are
y= 5 - 2 sqrt 2 and 5 + 2 sqrt2
Same answers....just maybe easier...and exact. 
