Difference between revisions of "2005 AMC 12A Problems/Problem 2"
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<math>2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = -5\ \mathrm{(B)}</math> | <math>2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = -5\ \mathrm{(B)}</math> | ||
| − | == See | + | == See also == |
{{AMC12 box|year=2005|num-b=1|num-a=3|ab=A}} | {{AMC12 box|year=2005|num-b=1|num-a=3|ab=A}} | ||
{{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]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:35, 2 February 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 
and 
have the same solution. What is the value of 
?
Solution
See also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
