Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Problem Solving - Page 663: 43a

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Given: $x^2-y^2-8x+8=0\\y^2-x^2-8y+8=0$ Add the first equation to the second one: $$-8y-8x+16=0\\-y-x+2=0\\x=-y+2$$ Substitute $x$ in the first equation. $$(2-y)^2-y^2-8(2-y)+8=0\\4-4y+y^2-y^2-16+8y+8=0\\4y=4\\y=1$$ Find x: $x=-1+2=1$ The solution is $(1,1)$