2006 AMC 10B Problem 12 — Solution | LIVE by Po-Shen Loh

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2006 AMC 10B Problem 12

Below is the professionally curated solution for Problem 12 of the 2006 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:substitution system of equations

Difficulty rating: 1140

12.

The lines $x=\tfrac14 y+a$ and $y=\tfrac14 x+b$ intersect at the point $(1,2).$ What is $a+b?$

$0$

$\dfrac{3}{4}$

$1$

$2$

$\dfrac{9}{4}$

Solution:

Substituting $(1,2)$ : from $1=\tfrac14(2)+a$ we get $a=\tfrac12,$ and from $2=\tfrac14(1)+b$ we get $b=\tfrac74.$

Then $a+b=\tfrac12+\tfrac74=\tfrac94.$

Thus, the correct answer is E.

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