Question : prove that (0,0), (1,-1),(-1,1) are solutions of the system of non linear equations equations given as
\[x=y\left(1-\left(x^2+y^2\right)\right) , y=x\left(1-\left(x_2+y^2\right)\right) \]
Solution :
Given system of non linear equations
\[x=y\left(1-\left(x^2+y^2\right)\right) \] (1) \[y=x\left(1-\left(x_2+y^2\right)\right) \] (2) From eqn (1) \[\frac{x}{y}=1-\left(x^2+y^2\right)\] From eqn (2) \[\frac{y}{x}=1-\left(x^2+y^2\right)\] Equating these equations \[\frac{x}{y}=\frac{y}{x}\] \[x^2-y^2=0\] x=y, x=-y Substituting x=y into eqn (1) \[x=x\left(1-\left(x^2+y^2\right)\right)\] \[x-x\left(1-\left(x^2+y^2\right)\right)=0\] \[x\left(1-1+\left(x^2+x^2\right)\right)=0\] 2x=0 x=0 Since x=y y=0 (0,0) Is a solution of the given system of non linear equations. Now substituting x=-y into eqn (1) \[x=-x\left(1-\left(x^2+x^2\right)\right)\] \[x+x\left(1-\left(x^2+x^2\right)\right)=0\] \[x\left(1+1-2x^2\right)=0\] \[x=0,\ 2-2x^2=0\] \[x^2=1\] \[x=\pm1\] Since y=-x (1,-1), (-1,1) Are also solutions of the system of given non linear equations. Therefore (0,0), (1,-1), (-1,1) are solutions of the system of non linear equations equations.