1955 AMC 12 Problem 46

Attempt Problem 46 of the 1955 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1955 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

46.

The graphs of 2x+3y6=0,2x+3y-6=0, 4x3y6=0,4x-3y-6=0, x=2,x=2, and y=23y=\dfrac23 intersect in:

66 points

11 point

22 points

no points

an unlimited number of points

Answer: B
Concepts:linear equationscommon intersectioncoordinate geometry
Difficulty rating: 1340
Small Hint:

Solve the first two linear equations simultaneously

Big Hint:

Check whether that solution also satisfies each of the last two displayed equations

Solution:

Adding the first two equations gives 6x12=0,6x-12=0, so x=2.x=2. Substitution gives 3y=2,3y=2, hence y=23.y=\frac{2}{3}. This point also lies on the last two given lines. Therefore all four graphs have the single common point (2,23).(2,\frac{2}{3}).

Thus, the correct answer is B.

← Problem 45#45
Full Exam

Problem 46 in Other Years

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12