Difference between revisions of "2005 AMC 12A Problems/Problem 2"

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{{Duplicate|[[2005 AMC 10A Problems|2005 AMC 10A #2]] and [[2005 AMC 10A Problems|2005 AMC 10A #3]]}}
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{{Duplicate|[[2005 AMC 12A Problems|2005 AMC 12A #2]] and [[2005 AMC 10A Problems|2005 AMC 10A #3]]}}
  
 
== Problem ==
 
== Problem ==
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<math>2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = -4\ \mathrm{(B)}</math>
 
<math>2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = -4\ \mathrm{(B)}</math>
  
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==Video Solution==
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CHECK OUT Video Solution: https://youtu.be/GmOEQzJVAn4
  
 
== See also ==
 
== See also ==
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{{AMC10 box|year=2005|ab=A|num-b=2|num-a=4}}
 
{{AMC10 box|year=2005|ab=A|num-b=2|num-a=4}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:00, 30 October 2020

The following problem is from both the 2005 AMC 12A #2 and 2005 AMC 10A #3, so both problems redirect to this page.

Problem

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

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

Solution

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

Video Solution

CHECK OUT Video Solution: https://youtu.be/GmOEQzJVAn4

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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