Find all pairs x+y=10 of real numbers such that x + y = 10 and x^2 + y^2 = 52.
For example, to enter the solutions and (2,4) and (-3,9), you would enter "(2,4),(-3,9)" (without the quotation marks).
+526 \(x+y=10\)
\(x^2+y^2+2xy=100\)
\(2xy=100-52\)
\(2xy = 48\)
\(xy=24\)
⇒\(x+{24\over x}=10\)
\(x^2-10x+24=0\)
\(x=6,4\)
1st Case : when x = 6 ⇒y = 4
2nd Case : when x = 4 ⇒y = 6
∴ All pairs of solutions are (6,4) and (4,6)
