MathCounts 2025 State — Sprint Round (Problem 30)Expressions & IdentitiesHardIf xxx and yyy are real numbers such that (4−x)(4+y)=2(4-x)(4+y)=2(4−x)(4+y)=2 and (4+x)(4−y)=3(4+x)(4-y)=3(4+x)(4−y)=3, what is the value of (x2−1)(y2−1)(x^2-1)(y^2-1)(x2−1)(y2−1)? Express your answer as a common fraction.Related TopicsCoreEquation Solving TechniquesMinorAlgebraic IdentitiesCheck AnswerYour answer:CheckHints (9)Hint 1(4−x)(4+y)=2(4-x)(4+y)=2(4−x)(4+y)=2⟹16+4y−4x−xy=2\Longrightarrow 16+4y-4x-xy=2⟹16+4y−4x−xy=2⟹xy+4x−4y=14\Longrightarrow xy+4x-4y=14⟹xy+4x−4y=14Hint 2(4+x)(4−y)=3(4+x)(4-y)=3(4+x)(4−y)=3⟹16−4y+4x−xy=3\Longrightarrow 16-4y+4x-xy=3⟹16−4y+4x−xy=3⟹xy−4x+4y=13\Longrightarrow xy-4x+4y=13⟹xy−4x+4y=13Hint 3Find xyxyxy.Hint 4By Approaches to Similar Equations and Hints 1 and 2, add the two equations:xy+4x−4y+xy−4x+4y=14+13xy+4x-4y+xy-4x+4y=14+13xy+4x−4y+xy−4x+4y=14+132xy=272xy=272xy=27xy=272xy=\frac{27}{2}xy=227Hint 5Find x−yx-yx−y.Hint 6By Hints 1 and 4,xy+4x−4y=14xy+4x-4y=14xy+4x−4y=14272+4(x−y)=14\frac{27}{2}+4(x-y)=14227+4(x−y)=144(x−y)=14−272=124(x-y)=14-\frac{27}{2}=\frac124(x−y)=14−227=21x−y=18x-y=\frac18x−y=81Hint 7(x2−1)(y2−1)=x2y2−(x2+y2)+1(x^2-1)(y^2-1)=x^2y^2-(x^2+y^2)+1(x2−1)(y2−1)=x2y2−(x2+y2)+1Hint 8By Square of a difference,x2+y2=(x−y)2+2xyx^2+y^2=(x-y)^2+2xyx2+y2=(x−y)2+2xyHint 9By Hints 4, 6, 7, and 8,(x2−1)(y2−1)=x2y2−(x2+y2)+1(x^2-1)(y^2-1)=x^2y^2-(x^2+y^2)+1(x2−1)(y2−1)=x2y2−(x2+y2)+1=(272)2−(18)2−2(272)+1=\left(\frac{27}{2}\right)^2-\left(\frac18\right)^2-2\left(\frac{27}{2}\right)+1=(227)2−(81)2−2(227)+1=7294−164−27+1=\frac{729}{4}-\frac{1}{64}-27+1=4729−641−27+1=7294−164−26=\frac{729}{4}-\frac{1}{64}-26=4729−641−26=11664−1−166464=\frac{11664-1-1664}{64}=6411664−1−1664=999964=\frac{9999}{64}=649999Final Answer999964\frac{9999}{64}649999Related Problems (1)AMC 10A/12A 2024 (Problem 23/17)