system of equation.

$\textbf{Solve for real system of equation}\\\\ $\mathbf{3.(x^2+y^2+z^2)-2(xy+yz+zx)=1}$\\\\ $\mathbf{x+y-z=-1}$

4 Answers

262

Aditya Bhutra ·2011-07-13 23:16:23

i could reach only till the point that
the soln will lie on -> (x+1)²+(y+1)²=1

then obviously there r infinite solutions

as we can see only 2 eqns r there

so 99% possibility we hav to get some sum of square = zero

but we r not getting that...

\sum_{cyc}(x^2-2xy+y^2)=1-(x^2+y^2+z^2)

Now from the second eqn, we get

1-(x^2+y^2+z^2)=2xy-2xz-2yz

From which we get

2(x-y)^2+2z^2=0

Thus the soln

x=y=\frac{-1}{2}

& z=0

1708

man111 singh ·2011-07-30 01:38:13

thanks soumik and shubhodip.

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