2005 AMC 12A Problems/Problem 2: Difference between revisions

Revision as of 18:07, 17 May 2010

The following problem is from both the 2005 AMC 10A #2 and 2005 AMC 10A #3, so both problems redirect to this page.

The equations $2x + 7 = 3$ and $bx - 10 = - 2$ have the same solution. What is the value of $b$ ?

$(\mathrm {A}) \ -8 \qquad (\mathrm {B}) \ -4 \qquad (\mathrm {C})\ 2 \qquad (\mathrm {D}) \ 4 \qquad (\mathrm {E})\ 8$

$2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = -4\ \mathrm{(B)}$

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Revision as of 21:04, 21 June 2009 view source 5849206328x (talk \| contribs) editor 1,190 edits mNo edit summary ← Older edit		Revision as of 18:07, 17 May 2010 view source YottaByte (talk \| contribs) 39 edits m Said 12A, had me worried I had been doing AMC 12 problems. Newer edit →
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	{{Duplicate\|[[2005 AMC ~~12A~~ Problems\|2005 AMC ~~12A~~ #2]] and [[2005 AMC 10A Problems\|2005 AMC 10A #3]]}}		{{Duplicate\|[[2005 AMC 10A Problems\|2005 AMC 10A #2]] and [[2005 AMC 10A Problems\|2005 AMC 10A #3]]}}

	== Problem ==		== Problem ==