Find all pairs $x+y=10$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 56$.
For example, to enter the solutions and , you would enter "(2,4),(-3,9)" (without the quotation marks).
I reposted this since the last response was incorrect
btw i accidentally left this part out its "to enter the solutions (2,4) and (-3,9), you-"
+23245 x + y = 10 ---> (x + y)2 = (10)2 ---> x2 + 2xy + y2 = 100
Combining this with x2 + y2 = 56 ---> x2 + 2xy + y2 = 100
x2 + y2 = 56
subtracting: 2xy = 44
xy = 22
Since x + y = 10 ---> y = 10 - x.
Combining these two equations: xy = 22 ---> x(10 - x) = 22
10x - x2 = 22
x2 - 10x + 22 = 0
Using the quadratic equation: x = 5 + sqrt(3) ---> y = 10 - ( 5 + sqrt(3) ) = 5 - sqrt(3)
Also, x = 5 - sqrt(3) and y = 5 + sqrt(3)
