AMC 10 · 2002 · #13

Grade 8 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
polynomial-factoringlinear-equations-one-var convert-to-algebra ↑ Prerequisites: polynomial-factoring
📏 Medium solution 💡 2 insights
Problem
There is one equation, 8xy - 12y + 2x - 3 = 0, that mixes two letters x and y. Find the value(s) of x that make the equation hold no matter what number y is.

Pick an answer.

(A)
$\frac23$
(B)
$\frac32 \text{ or } -\frac14$
(C)
$-\frac23 \text{ or } -\frac14$
(D)
$\frac32$
(E)
$-\frac32 \text{ or } -\frac14$

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Organize Information in More Ways

The equation looks tangled because x-terms and y-terms are shuffled together. Tool #15 (Organize Information in More Ways) re-sorts it by the letter that is allowed to change, y, collecting everything into 'a number times y' plus 'a leftover number'. Once it is sorted that way, Tool #13 (Convert to Algebra) reads the phrase 'true for all y' as a hard condition: the number multiplying y must be 0 and the leftover must be 0, which turns into two little equations for x. Tool #3 (Eliminate Possibilities) is the backup — because it is multiple choice, the true x can also be found by testing which choice keeps the equation balanced for two different y values.

1STEP 1

Sort the terms by y

Bundle the two y-terms: 8xy - 12y = (8x - 12)y, so the left side becomes (8x - 12)y + (2x - 3).

8xy - 12y + 2x - 3 = (8x - 12) y + (2x - 3)
2STEP 2

Read what 'true for all y' demands

A line in y stays glued to 0 for every y only if it is flat and already at 0: 8x - 12 = 0 and 2x - 3 = 0.

(8x - 12) y + (2x - 3) = 0 for all y → 8x - 12 = 0 and 2x - 3 = 0
3STEP 3

Solve and check both conditions agree

Both conditions agree: 8x = 12 and 2x = 3 each give x = 3/2, a single value — choice (D).

8x - 12 = 0 → x = 3/2; 2x - 3 = 0 → x = 3/2 → x = 3/2 (D)
Answer
3/2
Plug x = 3/2 back in: 8(3/2)y - 12y + 2(3/2) - 3 = 12y - 12y + 3 - 3 = 0, which is 0 no matter what y is — exactly what the problem wanted. The choices offering a second value like -1/4 (B, C, E) are traps: no single x other than 3/2 can kill both the y-term and the constant, so those extra values fail. Only (D), the lone value 3/2, works.
💡Key takeaway

To make an equation hold for every value of a changing letter, sort the terms by that letter and force whatever multiplies it — and the leftover — to both be zero.

  • Sort the terms by y
  • Read what 'true for all y' demands
  • Solve and check both conditions agree

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