+1055 Find all ordered pairs x, y of real numbers such that x+y=10 and x^2+y^2=64.
For example, to enter the solutions (2, 4) and (-3, 9), you would enter "(2,4),(-3,9)" (without the quotation marks).
+106 To solve this, let's first isolate a.
\((a+b)^2 = 10^2\)
\(a^2+2ab+b^2 = 100\)
\(64 + 2ab = 100\)
\(2ab = 36\)
\(ab = 18\)
\(b = 10 - a\)
\(a(10-a) = 18\)
\(-a^2+10a-18 = 0\)
\(a^2-10a+18 = 0\)
Now, we can solve for a
\(a^2 - 10a + 25 = 7\)
\((a-5)^2 = 7\)
\(a-5 = \pm \sqrt{7}\)
\(a = 5 \pm \sqrt{7}\)
We can now finally solve for b by plugging in each value of a.
a = (5 + sqrt(7)):
\((5 + \sqrt{7}) + b = 10\)
\(b = 5 - \sqrt{7}\)
a = (5 - sqrt(7)):
\((5 - \sqrt7) + b = 10\)
\(b = 5 + \sqrt7\)
Therefore, the solutions are (5 + sqrt(7), 5 - sqrt(7)) and (5 - sqrt(7), 5 + sqrt(7))
