[tex]y^2+x^2=53 \\
y-x=5 \\ \\
\hbox{solve the 1st equation for y:} \\
y-x=5 \\
y=5+x \\ \\
\hbox{substitute 5+x for y in the 1st equation:} \\
(5+x)^2+x^2=53 \\
25+10x+x^2+x^2=53 \\
2x^2+10x+25-53=0 \\
2x^2+10x-28=0 \\
2x^2+14x-4x-28=0 \\
2x(x+7)-4(x+7)=0 \\
(2x-4)(x+7)=0 \\
2x-4=0 \ \lor \ x+7=0 \\
2x=4 \ \lor \ x=-7 \\
x=2 \ \lor \ x=-7 \\ \\
y=5+x \\
y=5+2 \ \lor \ y=5-7 \\
y=7 \ \lor \ y=-2 \\ \\
(x,y)=(2,7) \ or \ (x,y)=(-7,-2)[/tex]
The answer is B.
