AMC 10A/12A 2026 (Problem 7/5)How many ordered triples of positive integers (x,y,z)(x,y,z)(x,y,z) satisfy both the following equations?x+y+z=31x+y+z=31x+y+z=31x+3y+3z=41x+3y+3z=41x+3y+3z=41(A) 4\text{(A)}\;4(A)4(B) 5\text{(B)}\;5(B)5(C) 6\text{(C)}\;6(C)6(D) 7\text{(D)}\;7(D)7(E) 8\text{(E)}\;8(E)8Related TopicsCoreEquation Solving TechniquesCheck AnswerYour answer:ABCDECheckHints (4)Hint 1x+y+z=31x+y+z=31x+y+z=31z=31−x−yz=31-x-yz=31−x−yHint 2x+3y+3z=41x+3y+3z=41x+3y+3z=41By Hint 1,x+3y+3(31−x−y)=41x+3y+3(31-x-y)=41x+3y+3(31−x−y)=41−2x=41−93=−52-2x=41-93=-52−2x=41−93=−52x=26x=26x=26Hint 3By Hints 1 and 2,z=31−x−y=31−26−y=5−yz=31-x-y=31-26-y=5-yz=31−x−y=31−26−y=5−yHint 4(x,y,z)=(26,y,5−y)(x,y,z)=(26,y,5-y)(x,y,z)=(26,y,5−y)y≥1y\ge 1y≥15−y≥1 ⟹ 4≥y5-y\ge 1\implies 4\ge y5−y≥1⟹4≥yy=1,2,3,4y=1,2,3,4y=1,2,3,4Ans=4\text{Ans}=4Ans=4Final Answer(A) 4Related Problems (2)AMC 10B 2024 (Problem 25)MATHCOUNTS State Target 2025 (Problem 8)