Solve the simultaneous equations x^2 + y^2 =13 and x= y - 5.
(y-5)2 + y2 = 13y2-10y +25 + y2=13.2y2-10y+12=0.y2-5y+6=0.(y-3)(y-2)=0. So, y=3 or y=2. Then substituting these values of y back into x=y-5 gives, x=-2 when y=3 and x=-3 when y=2.
HD
(y-5)2 + y2 = 13y2-10y +25 + y2=13.2y2-10y+12=0.y2-5y+6=0.(y-3)(y-2)=0. So, y=3 or y=2. Then substituting these values of y back into x=y-5 gives, x=-2 when y=3 and x=-3 when y=2.
