AMC 10 · 2005 · #2

Grade 7 arithmetic
linear-equations-one-varsystems-of-equations identify-subproblemswork-backwards ↑ Prerequisites: linear-equations-one-var
📏 Short solution 💡 1 insight
Problem
Two equations are said to have the same solution. Find the value of the unknown coefficient that makes this happen.

Pick an answer.

(A)
-8
(B)
-4
(C)
-2
(D)
4
(E)
8
How to solve
Strategy Identify Subproblems

The phrase "same solution" splits the task into two clean subproblems, so Tool #7 (Identify Subproblems) leads: first find the single x that solves the self-contained first equation, then feed that number into the second equation. Tool #11 (Work Backwards) handles the second stage — the second equation gives the end result -2, and from it we reverse-engineer the coefficient b. Tool #3 (Eliminate Possibilities) guards the sign traps: dropping a minus sign turns the answer into (D) 4, and copying x by mistake gives (C) -2.

1STEP 1

Solve the first equation for x

The first equation has only one unknown, giving -2.

2x+7=3 → 2x=-4 → x=-2
2STEP 2

Use "same solution" to link the equations

Same solution means that value must fit the second equation too.

b(-2)-10=-2 → -2b-10=-2
3STEP 3

Solve for b

Solving with attention to the sign gives -4, choice (B).

-2b-10=-2 → -2b=8 → b=-4 → (B)
Answer
-4
Check by putting both numbers back in: with x=-2 and b=-4, the second equation reads (-4)(-2)-10=8-10=-2, which matches the required -2 exactly. The first equation also checks: 2(-2)+7=-4+7=3. Both hold, so b=-4 is right. The trap answers make sense as near-misses: (D) 4 is what you get from a sign slip when dividing by -2, and (C) -2 is just the value of x copied over by mistake.
💡Key takeaway

When two equations share a solution, solve the one that has a single unknown first, then plug that number into the other to unlock what is left.

  • Solve the first equation for x
  • Use "same solution" to link the equations
  • Solve for b