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Find all ordered pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 32 + 2xy$.
For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(2,4),(-3,9)" (without the quotation marks).

 Jan 10, 2024

Best Answer 

 #1
avatar+36919 
+1

x+y = 10     =====>  y = 10 -x       sub this definition of y into the second equation

 

x^2 + (10-x)^2  - 2 x(10-x) -32 = 0      now solve for the value(s) of x 

x^2 + x^2 - 20x + 100  - 20x + 2x^2 -32 = 0 

4 x^2 -40x + 68 = 0            Quadratic Formula shows x =  7.828    and    2.172 

using these values in the line equation shows            y =   2.172     and  7.828

 Jan 11, 2024
 #1
avatar+36919 
+1
Best Answer

x+y = 10     =====>  y = 10 -x       sub this definition of y into the second equation

 

x^2 + (10-x)^2  - 2 x(10-x) -32 = 0      now solve for the value(s) of x 

x^2 + x^2 - 20x + 100  - 20x + 2x^2 -32 = 0 

4 x^2 -40x + 68 = 0            Quadratic Formula shows x =  7.828    and    2.172 

using these values in the line equation shows            y =   2.172     and  7.828

ElectricPavlov Jan 11, 2024
 #2
avatar+36919 
+1

  Here is perhaps a simler , exact - answer way:

 

x^2 + y^2 - 2xy = 32 

(x-y)^2 = 32 

x -y = +-  4 sqrt 2 

 

y =   x - 4 sqrt 2    and     x +  4sqrt 2 

 

then use the line quation:

x + y = 10 

x + x-  4 sqrt 2 = 10 

2x = 10 +4 sqrt 2

x = 5 +2 sqrt 2        and the other is x = 5 - 2 sqrt 2    Then the corresponding 'y' s are 

y=  5 - 2 sqrt 2        and                          5 + 2 sqrt2                                                                   

 

  Same answers....just maybe easier...and exact. laugh  

 Jan 11, 2024

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