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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).

 Jun 13, 2021
 #1
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\(x+y=10\)

\(x^2+y^2+2xy=100\)

\(2xy=100-52\)

\(2xy = 48\)

\(xy=24\)

 

\(x+{24\over x}=10\)

\(x^2-10x+24=0\)

\(x=6,4\)

 

1st Case : when x = 6 ⇒y = 4 

2nd Case : when x = 4 ⇒y = 6 

 

∴ All pairs of solutions are (6,4) and (4,6) 

 Jun 14, 2021

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