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

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.

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

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

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

~Charles3829

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 5: / Line 5: @@
 <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>
 == Solution ==
-<math>2x + 7 = 3 \iff x = -2</math>, so we require <math>-2b-10 = -2 \iff -2b = 8 \iff b = \boxed{\mathrm{(B) } \ -4}</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>.
 ==Video Solution 1==