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# help very confused struggeled for like 2 hours

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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

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nine...... right?

May 31, 2019
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i think so. I checked with my friend

ProffesorNobody  May 31, 2019
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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   May 31, 2019
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THANK YOU SO MUCHHH !!!!!!! U AO NICE THANKSSSSSS U BEST

May 31, 2019