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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).

 Jun 8, 2024
 #1
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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))

 Jun 9, 2024

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