Art of Problem Solving

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

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== Problem ==
== Problem ==
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>?
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>
(\mathrm {A}) \ -8 \qquad (\mathrm {B}) \ -4 \qquad (\mathrm {C})\ 2 \qquad (\mathrm {D}) \ 4 \qquad (\mathrm {E})\ 8
\textbf{(A) } -8 \qquad \textbf{(B) } -4 \qquad \textbf{(C) } -2 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8
</math>
</math>


== Solution ==
== Solution ==
<math>2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = -4\ \mathrm{(B)}</math>
<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>.


CHECK OUT Video Solution: https://youtu.be/GmOEQzJVAn4
==Video Solution 1==
https://youtu.be/GmOEQzJVAn4
 
==Video Solution 2==
https://youtu.be/c_Zxp8iwCD4
 
~Charles3829


== See also ==
== See also ==

Latest revision as of 16:01, 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 (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

These problems are copyrighted © by the Mathematical Association of America.