AMC 10 · 2002 · #13
Grade 8 arithmeticFind the value(s) of x such that 8xy−12y+2x−3=0 is true for all values of y.
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The equation $8xy - 12y + 2x - 3 = 0$; It must be true for every possible value of $y$; Answer choices: (A) $\frac23$, (B) $\frac32$ or $-\frac14$, (C) $-\frac23$ or $-\frac14$, (D) $\frac32$, (E) $-\frac32$ or $-\frac14$
Unknowns: The value(s) of $x$ for which the equation is true for all $y$
Understand
Restated: 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.
Givens: The equation $8xy - 12y + 2x - 3 = 0$; It must be true for every possible value of $y$; Answer choices: (A) $\frac23$, (B) $\frac32$ or $-\frac14$, (C) $-\frac23$ or $-\frac14$, (D) $\frac32$, (E) $-\frac32$ or $-\frac14$
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #13 Convert to Algebra, #3 Eliminate Possibilities
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.
Execute — Answer: D
6.EE.A.3 Step 1 Sort the terms by y
- Group the two terms that carry a $y$ and the two that do not.
- From $8xy - 12y + 2x - 3$, the $y$-terms are $8xy$ and $-12y$, and the rest are $2x$ and $-3$.
- Factor $y$ out of the first pair: $8xy - 12y = (8x - 12)y$.
- So the whole left side becomes $(8x - 12)y + (2x - 3)$.
- Nothing has changed except the bookkeeping — it is the same expression, now written as 'a number times $y$' plus 'a plain number'.
💡 Collecting every $y$ into one bundle turns a messy mix into a simple 'slope-times-$y$ plus offset' shape.
8.EE.C.7 Step 2 Read what 'true for all y' demands
- The line $(8x - 12)y + (2x - 3) = 0$ must hold for every $y$.
- Think of the left side as depending on $y$: it is a straight-line rule in $y$ with slope $8x - 12$ and starting value $2x - 3$.
- A straight line can only stay glued at $0$ for every $y$ if it is flat AND already sitting on $0$.
- That forces two separate conditions at once: the slope must vanish, $8x - 12 = 0$, and the starting value must vanish, $2x - 3 = 0$.
💡 If a term still contains $y$, changing $y$ would tip the balance — so the only way to be safe for every $y$ is to erase that term entirely.
8.EE.C.7 Step 3 Solve and check both conditions agree
- Solve each condition.
- From $8x - 12 = 0$: $8x = 12$, so $x = \frac{12}{8} = \frac32$.
- From $2x - 3 = 0$: $2x = 3$, so $x = \frac32$.
- Both requirements point to the exact same value, so $x = \frac32$ really does clear the whole equation for every $y$.
- There is a single answer, $x = \frac32$, which is choice (D).
💡 When two independent conditions land on the same number, that number is locked in as the answer.
6.EE.A.3 Group the two terms that carry a $y$ and the two that do not. From $8xy - 12y + 8.EE.C.7 The line $(8x - 12)y + (2x - 3) = 0$ must hold for every $y$. Think of the left 8.EE.C.7 Solve each condition. From $8x - 12 = 0$: $8x = 12$, so $x = \frac{12}{8} = \fra Review
Reasonableness: Plug $x = \frac32$ back in: $8(\frac32)y - 12y + 2(\frac32) - 3 = 12y - 12y + 3 - 3 = 0$, which is $0$ no matter what $y$ is — exactly what the problem wanted. The choices offering a second value like $-\frac14$ (B, C, E) are traps: no single $x$ other than $\frac32$ can kill both the $y$-term and the constant, so those extra values fail. Only (D), the lone value $\frac32$, works.
Alternative: Factor first. Since $8xy - 12y = 4y(2x - 3)$ and $2x - 3 = 1\cdot(2x-3)$, the whole left side factors as $(2x - 3)(4y + 1)$. For $(2x - 3)(4y + 1)$ to equal $0$ for every $y$, the factor $4y + 1$ can't be trusted to be $0$ (it changes with $y$), so the other factor must be zero: $2x - 3 = 0$, giving $x = \frac32$. Or, purely as a test, pick two values like $y = 0$ and $y = 1$: $y=0$ needs $2x - 3 = 0$ and $y=1$ needs $8x - 12 + 2x - 3 = 10x - 15 = 0$; both give $x = \frac32$.
CCSS standards used (min grade 8)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Regrouping $8xy - 12y + 2x - 3$ and factoring $y$ out of the first pair to write it as $(8x-12)y + (2x-3)$.)8.EE.C.7Solve linear equations in one variable, including cases with one solution, infinitely many, or none (Reading 'true for all $y$' as the two conditions $8x-12=0$ and $2x-3=0$, then solving each to get $x=\frac32$.)
⭐ 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.
⭐ 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.
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