Solve the system `|x^(2)-2x|+y=1,x^(2)+|y|=1`

Answer

The pairs `(0,1),(1,0),((1-sqrt(5))/2,(1-sqrt(5))/2)` are solutions of the original system of equations.

Explanation

Let `yge0`, then `|y|=y`
and then given system reduces to
`|x^(2)-2x|+y=1`….i
ad `x^(2)+y=1`………ii
From Eqs (i) and (ii) we get
`x^(2)-|x^(2)-2x|`
`impliesx^(2)=|x||x-2|`
Now `x lt 0, 0lexlt2,xge2`
`x^(2)-x(x-2),x^(2)=-x(x-2)`
`x^(2)=x(x-2)`
`:.x=0`
`impliesx(x+x-2)=0`
`:.x=0`
fail `:.x=0,1` fail
`impliesx=0,1` then `y=1,0`
`:.` Solutions are (0,1) and (1,0).
If `ylt0` then `|y|=-y` and then given system reduces to
`|x^(2)-2x|+y=1` ......iii
or `x^(2)-y=1`....iv
From Eqs (iii) and (iv) we get
`|x^(2)-2x|+x^(2)=2`
`implies|x||x-2|+x^(2)=2`
Now `xlt0, 0lexlt2, xge2`
`x(x-2)+x^(2)=2`
`-x(x-2)+x^(2)=2`
`x(x-2)+x^(2)=2`
`implies2x^(2)-2x-2=0implies2x=2`
`impliesx^(2)-x-1=0`
`impliesx^(2)-x-1=0`
`:.x=1`
`:.x=(1+-sqrt(5))/2`
`:.x=(1+1sqrt(5))/2` fail
`impliesx+(1-sqrt(5))/2 [ :' x lt0]`
`impliesx=(1-sqrt(5))/2,1` then `y=(1-sqrt(5))/2,0`
`:.` Solutions are `((1-sqrt(5))/2,(1-sqrt(5))/2)` and (1,0)
Hence all pairs `(0,1),(1,0)` and `((1-sqrt(5))/2,(1-sqrt(5))/2)` are solutions of the original system of equations.