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

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2006 AMC 10B Problem 12

Attempt Problem 12 of the 2006 AMC 10B 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 2006 AMC 10B solutions, or check the answer key.

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12.

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

$0$

$\dfrac{3}{4}$

$1$

$2$

$\dfrac{9}{4}$

Answer: E

Concepts:substitution system of equations

Difficulty rating: 1140

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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