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Thanks so much for the help previously CPhill, it helped me with some other problems. 

 

I'm stuck with this problem now however: 

 

Solve for x and y:  \(x^2+y^2=17 \)

                              \(y-1=x\)

 

There are two solutions: (x1, y1) and (x2, y2)

 

Evaluate x1+y1+x2+y2= ?

 Jan 6, 2016
 #1
avatar+129899 
+5

y - 1   = x    →  y = x + 1    (1)

 

x^2 + y^2 = 17   (2)

 

Subbing (1) into (2), we have

 

x^2 + ( x + 1)^2  = 17       simplify

 

x^2 + x^2 + 2x + 1  = 17

 

2x^2 + 2x - 16  = 0       divide through by 2

 

x^2 + x - 8  = 0

 

This will not factor.......using the quadratic formula, the two solutions for x are

 

[ -1 + sqrt(33)] / 2       and   [ -1 - sqrt(33)] / 2

 

And y = x + 1

 

So.....when x = [ -1 + sqrt(33)] / 2 , y = x + 1 =  [1 + sqrt(33)]/2

 

And, when x = [ -1 - sqrt(33)]/ 2 , y = x + 1 =   [1 - sqrt(33)] / 2

 

And the sum of all these solutions is

 

[ -1 + sqrt(33)] /2   +   [ -1 - sqrt(33)] / 2   +  [1 + sqrt(33)]/2 + [1 - sqrt(33)] / 2  = (suprisingly, 0  !!! )

 

You can see why this is so from the following graph: https://www.desmos.com/calculator/8axms04lcy

 

The solution coorrdinates just "cancel" each other out when we sum them !!!!!! 

 

 

cool cool cool

 Jan 6, 2016

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