AMC 12A 2023 (Problem 10)Positive real numbers xxx and yyy satisfy y3=x2y^3=x^2y3=x2 and (y−x)2=4y2(y-x)^2=4y^2(y−x)2=4y2. What is x+yx+yx+y?(A) 12\text{(A)}\;12(A)12(B) 18\text{(B)}\;18(B)18(C) 24\text{(C)}\;24(C)24(D) 36\text{(D)}\;36(D)36(E) 42\text{(E)}\;42(E)42Hints (5)Hint 1y−x=±2yy-x=\pm 2yy−x=±2yHint 2Case 1: y−x=2y ⇒ y=−xy-x=2y\ \Rightarrow\ y=-xy−x=2y ⇒ y=−xHint 3In Case 1: y3=x2 ⇒ (−x)3=x2 ⇒ −x3=x2 ⇒ x=0,−1y^3=x^2\ \Rightarrow\ (-x)^3=x^2\ \Rightarrow\ -x^3=x^2\ \Rightarrow\ x=0,-1y3=x2 ⇒ (−x)3=x2 ⇒ −x3=x2 ⇒ x=0,−1. But x>0x>0x>0 and y>0y>0y>0, so Case 1 gives no solution.Hint 4Case 2: y−x=−2y ⇒ x=3yy-x=-2y\ \Rightarrow\ x=3yy−x=−2y ⇒ x=3yHint 5In Case 2: y3=(3y)2 ⇒ y3=9y2 ⇒ y=0,9y^3=(3y)^2\ \Rightarrow\ y^3=9y^2\ \Rightarrow\ y=0,9y3=(3y)2 ⇒ y3=9y2 ⇒ y=0,9. Since y>0y>0y>0, y=9y=9y=9 and therefore x=27x=27x=27.Final Answer(D) 36