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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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The equations <math>2x + 7 = 3</math> and <math>bx - 10 = - 2</math> have the same solution. What is the value of <math>b</math>?
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The equations <math>2x + 7 = 3</math> and <math>bx - 10 = -2</math> have the same solution <math>x</math>. What is the value of <math>b</math>?
  
 
<math>
 
<math>
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(\mathrm {A}) \ -8 \qquad (\mathrm {B}) \ -4 \qquad (\mathrm {C})\ 2 \qquad (\mathrm {D}) \ 4 \qquad (\mathrm {E})\ 8
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\textbf{(A) } -8 \qquad \textbf{(B) } -4 \qquad \textbf{(C) } -2 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8
 
</math>
 
</math>
  
 
== Solution ==
 
== Solution ==
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<math>2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = -4\ \mathrm{(B)}</math>
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<math>2x + 7 = 3 \iff x = -2</math>, so we require <math>-2b-10 = -2 \iff -2b = 8 \iff b = \boxed{\textbf{(B) } -4}</math>.
  
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==Video Solution 1==
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https://youtu.be/GmOEQzJVAn4
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==Video Solution 2==
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https://youtu.be/c_Zxp8iwCD4
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~Charles3829
  
 
== See also ==
 
== See also ==

Latest revision as of 17:01, 1 July 2025

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 $x$. What is the value of $b$?

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

Solution

$2x + 7 = 3 \iff x = -2$, so we require $-2b-10 = -2 \iff -2b = 8 \iff b = \boxed{\textbf{(B) } -4}$.

Video Solution 1

https://youtu.be/GmOEQzJVAn4

Video Solution 2

https://youtu.be/c_Zxp8iwCD4

~Charles3829

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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