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Use the identity (x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2. We are given x+y=8x+y=8 and x2+y2=50x^2+y^2=50. Square the sum: (x+y)2=82=64(x+y)^2 = 8^2 = 64. Substitute into the identity: 64=(x2+y2)+2xy=50+2xy64 = (x^2 + y^2) + 2xy = 50 + 2xy. Hence 2xy=64−50=142xy = 64 - 50 = 14, so xy=7xy = 7. Choice B is correct. This is a standard GRE maneuver: when a system gives you sums of first and second powers, the symmetric identity lets you avoid solving for xx and yy individually. A frequent pitfall is to try to factor or to set up a quadratic immediately; while possible, it’s slower. If you did proceed that way, you’d let tt be a root of t2−(x+y)t+xy=0t^2 - (x+y)t + xy = 0 with x+y=8x+y=8 and unknown xyxy, then use the second equation to solve for xyxy—ultimately arriving at the same result. Keep a list of core algebraic identities (sum of squares, difference of squares, binomial expansions); they streamline problems that otherwise appear to require more computation.