How to solve the system of equations?

newtagi

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Feb 19, 2016
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I need help to solve this system of equations. I got the answer by checking) There are two answers. (1;3;0) and (-1;-3;0)

x^2 + y^2 - z(x+y) = 10
y^2 + z^2 - x(y+z) = 6
z^2 + x^2 - y(z+x) = -2
 
I need help to solve this system of equations:

x^2 + y^2 - z(x+y) = 10
y^2 + z^2 - x(y+z) = 6
z^2 + x^2 - y(z+x) = -2

I got the answer by checking. There are two answers. (1, 3, 0) and (-1, -3, 0)
What do you mean by "getting the answers by checking"?

By the way, Wolfram Alpha confirms that your answers are the only solutions. ;)
 
I need help to solve this system of equations. I got the answer by checking) There are two answers. (1;3;0) and (-1;-3;0)

x^2 + y^2 - z(x+y) = 10
y^2 + z^2 - x(y+z) = 6
z^2 + x^2 - y(z+x) = -2
Knowing the answer helps. Since z=0 in both solutions, the third equation must be
z^2 + x^2 - y(z+x) = z^2 - 2 = -2
or
x^2 - y(z+x) = 0.

So start by treating the first equation as a quadratic in x2 and get x in terms of z and y. Then compute z+x and x2, again in terms of y and z, and substitute them into the third equation and see what you get. You should get z=0 which will change the first two equations to
x^2 + y^2 = 10
y^2 + x y = 6
and your equation for x into a solution involving only y. Use the second equation to also solve for x in terms of y and equate the solutions for x. This gives an equation only in y. Solve for y. Use the y value to obtain x.
 
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