Question

What is one of the solutions to the following system of equations?

Answer 1

$y^2+x^2=53 \\ y-x=5 \\ \\ \hbox{solve the 1st equation for y:} \\ y-x=5 \\ y=5+x \\ \\ \hbox{substitute 5+x for y in the 1st equation:} \\ (5+x)^2+x^2=53 \\ 25+10x+x^2+x^2=53 \\ 2x^2+10x+25-53=0 \\ 2x^2+10x-28=0 \\ 2x^2+14x-4x-28=0 \\ 2x(x+7)-4(x+7)=0 \\ (2x-4)(x+7)=0 \\ 2x-4=0 \ \lor \ x+7=0 \\ 2x=4 \ \lor \ x=-7 \\ x=2 \ \lor \ x=-7 \\ \\ y=5+x \\ y=5+2 \ \lor \ y=5-7 \\ y=7 \ \lor \ y=-2 \\ \\ (x,y)=(2,7) \ or \ (x,y)=(-7,-2)$

The answer is B.