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Find all pairs of real numbers (x,y)$ such that $x + y = 6$ and $x^2 + y^2 = 28$. If you find more than one pair, then list your pairs in order by increasing $x$ value, separated by commas. For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(-3,9),(2,4)" (without the quotation marks).

 May 31, 2019
 #1
avatar+25 
-3

nine...... right?

 May 31, 2019
 #2
avatar+25 
-3

i think so. I checked with my friend

ProffesorNobody  May 31, 2019
 #3
avatar+102434 
+1

x + y  = 6     ⇒  y  = 6 - x       (1)

x^2 + y^2 = 28     (2)

 

Sub (1) into (2)  for y

 

x^2 + (6 - x)^2  = 28   simplify

 

x^2 + x^2 - 12x + 36  = 28

 

2x^2- 12x + 8  =  0       divide through by 2

 

x^2 - 6x + 4  = 0      complete the square on x

 

x^2 - 6x + 9  = - 4 + 9

 

(x - 3)^2  =  5        take both roots

 

x - 3  =   ± √5    add 3  to both sides

 

x =  3 ± √5

 

Because of the conjugate property of roots....

When x  =   3 +  √5, y =  3 -  √5

When x = 3 -  √5 , y  = 3 +  √5

 

 

cool cool cool

 May 31, 2019
 #4
avatar
0

THANK YOU SO MUCHHH !!!!!!! U AO NICE THANKSSSSSS U BEST

 May 31, 2019

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