AMC 10 · 2019 · #4
Grade 6 algebraPick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #9 (Easier Problem): we don't need to handle every AP — pick the two simplest ones, say (a, b, c) = (1, 2, 3) and (1, 3, 5). These give two specific lines x + 2y = 3 and x + 3y = 5. Their common point must include the answer. Tool #3 (Eliminate) + Tool #6 (Guess and Check): plug each of the five answer choices into both lines; the only point that satisfies BOTH equations is the answer. No algebra needed.
Pick two simple APs: (1, 2, 3) gives x + 2y = 3, and (1, 3, 5) gives x + 3y = 5. The point must satisfy both.
Two concrete APs give two concrete lines — easier to test.
4.OA.C.5Solve An Easier Related ProblemPlug each choice into x + 2y = 3: only (-1, 2) works (-1 + 4 = 3); (0,1), (1,-2), (1,0), (1,2) all fail.
Plug each candidate into the first line; eliminate the ones that fail.
6.EE.A.2Eliminate PossibilitiesCheck (-1, 2) on the second line x + 3y = 5: -1 + 6 = 5 ✓, so it lies on both lines.
Double-check the surviving point on the second line.
6.EE.A.2Eliminate PossibilitiesTest more APs: (2, 5, 8) gives 2(-1)+5(2)=8 ✓, (3, 2, 1) gives 3(-1)+2(2)=1 ✓ — (-1, 2) works for every AP.
If a point survives on every AP we test, it's the universal common point.
5.G.A.1Guess And CheckThis AMC 10 problem only needs Grade 6 expression-evaluation you already know — pick two simple APs (1, 2, 3) and (1, 3, 5), plug each answer-choice point into both lines, and only (-1, 2) survives. Answer: (A)!