## Algebra 2 (1st Edition)

$(1,0),(0,1)$
Plugging in the first equation into the second one we get: $1+4x+4y-5=0\\4x+4y-4=0\\x+y-1=0$ This is the same as the third equation. (i.e. we only need to consider the first two equations because the third will be satisfied if the first two are satisfied). From that we get: $x=1-y$, plugging this into the first one we get: $(1-y)^2+y^2=1\\y^2-2y+1+y^2=0\\2y^2-2y=0\\y^2-y=0\\y(y-1)=0.$Thus $y=0$ or $y=1$. Then our solutions are: $(1,0),(0,1)$