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

Revision as of 10:34, 7 December 2021

The following problem is from both the 2005 AMC 12A #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 $b$ ?

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

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

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

@@ Line 9: / Line 9: @@
 == 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 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = -4\ \boxed{\textbf{(B)}-4}</math>
 ==Video Solution==