2005 AMC 10A Problem 3

Only the first equation can be solved right away, because it has just one unknown. Start there.

1. Find $x$. From $2x + 7 = 3$, subtract $7$ from both sides: $2x = -4$. Divide by $2$: $x = -2$.
2. Use the same $x$ in the second equation. The problem says both equations have the same solution, so $x = -2$ must also make $bx - 10 = -2$ true:
$$b \cdot (-2) - 10 = -2.$$
3. Solve for $b$. Add $10$ to both sides: $-2b = 8$. Divide by $-2$: $b = -4$.

Check: with $b = -4$ and $x = -2$, we get $(-4)(-2) - 10 = 8 - 10 = -2$. Correct.

The answer is $\boxed{\textbf{(B)}\ -4}$.

When two equations share an answer, solve the easier one first and carry that answer into the other one. Once $x$ is a plain number, the second equation has only one unknown left, $b$, and becomes an ordinary one-step equation.

Try the choices. With $x = -2$, the second equation becomes $-2b - 10 = -2$. Testing $b = -4$ gives $8 - 10 = -2$, which works; the other choices do not.

The first equation involves only $x$: $2x + 7 = 3$ gives $2x = -4$, so $x = -2$.

That same $x$ satisfies the second equation. Substituting:
$$b(-2) - 10 = -2 \quad\Longrightarrow\quad -2b = 8 \quad\Longrightarrow\quad b = -4.$$

The answer is $\boxed{\textbf{(B)}\ -4}$.