AMC 10 · 2002 · #12
Grade 8 algebraFor which of the following values of k does the equation x−2x−1=x−6x−k have no solution for x?
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: The equation $\frac{x-1}{x-2}=\frac{x-k}{x-6}$ has a number $k$ sitting inside it. For most choices of $k$ you can solve for $x$, but for exactly one of the listed values the equation cannot be satisfied by any $x$. Find that value of $k$.
Givens: The equation is $\frac{x-1}{x-2}=\frac{x-k}{x-6}$; $x=2$ and $x=6$ are forbidden, since they make a denominator $0$; Answer choices for $k$: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: The value of $k$ for which no $x$ makes the equation true
Understand
Restated: The equation $\frac{x-1}{x-2}=\frac{x-k}{x-6}$ has a number $k$ sitting inside it. For most choices of $k$ you can solve for $x$, but for exactly one of the listed values the equation cannot be satisfied by any $x$. Find that value of $k$.
Givens: The equation is $\frac{x-1}{x-2}=\frac{x-k}{x-6}$; $x=2$ and $x=6$ are forbidden, since they make a denominator $0$; Answer choices for $k$: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #3 Eliminate Possibilities
The phrase 'has no solution' is a statement about an algebra equation, so Tool #13 (Convert to Algebra) is the lever: clear the fractions by cross-multiplying, then simplify until $x$ stands alone. What comes out is a single linear equation in $x$ of the form (coefficient)$\cdot x=$ (number), with both the coefficient and the number depending on $k$. A linear equation fails to have a solution only when the coefficient of $x$ collapses to $0$ while the other side is not $0$. Tool #3 (Eliminate Possibilities) then confirms that the other four values of $k$ really do give a usable $x$, so exactly one choice survives.
Execute — Answer: E
6.EE.B.6 Step 1 Cross-multiply to clear the fractions
- As long as $x\neq2$ and $x\neq6$, two equal fractions can be cross-multiplied.
- Multiply both sides by $(x-2)(x-6)$: the equation $\frac{x-1}{x-2}=\frac{x-k}{x-6}$ becomes $(x-1)(x-6)=(x-k)(x-2)$.
- This trades the fractions for a plain polynomial equation.
💡 Cross-multiplying is just multiplying both sides by both denominators at once, which is legal whenever neither denominator is zero.
6.EE.A.3 Step 2 Expand both sides
- Multiply out each side.
- The left side is $(x-1)(x-6)=x^2-7x+6$.
- The right side is $(x-k)(x-2)=x^2-(k+2)x+2k$.
- Both sides start with $x^2$, so those terms will cancel in the next step.
💡 Expanding turns the two products into standard $x^2+\dots$ form so matching terms can be lined up.
8.EE.C.7 Step 3 Cancel the square term and collect x
- Subtract $x^2$ from both sides; it disappears entirely, leaving $-7x+6=-(k+2)x+2k$.
- Now gather the $x$-terms on the left and the constants on the right: add $(k+2)x$ to both sides and subtract $6$, giving $(k+2)x-7x=2k-6$, that is $(k-5)x=2k-6$.
💡 Once the $x^2$ pieces match and cancel, what is left is a straight-line equation whose whole behavior sits in the coefficient $(k-5)$.
8.EE.C.7 Step 4 Find when no x can work
- The equation $(k-5)x=2k-6$ can be solved by dividing by $k-5$ — but only when $k-5\neq0$.
- If $k=5$, the left side becomes $0\cdot x=0$ for every $x$, while the right side is $2(5)-6=4$.
- That reads $0=4$, which is false no matter what $x$ is, so there is no solution.
- For any other $k$ the coefficient $k-5$ is nonzero, so $x=\frac{2k-6}{k-5}$ is a genuine value.
💡 A linear equation has no answer exactly when the $x$ term vanishes but a nonzero number is left demanding to equal zero.
8.EE.C.7 Step 5 Confirm the other four choices do have solutions
- For $k=1,2,3,4$ the formula $x=\frac{2k-6}{k-5}$ gives $x=1,\tfrac{2}{3},0,-2$ in turn.
- None of these hits the forbidden $x=2$ or $x=6$ (in fact $x=\frac{2k-6}{k-5}=2$ would force $-6=-10$, impossible, and $x=6$ would need $k=6$, which is not offered), so each of those four values of $k$ really does give a valid solution.
- Only $k=5$ leaves the equation unsolvable, which is choice (E).
💡 Testing the leftover choices shows they all produce a legal $x$, so the one that breaks is the answer.
6.EE.B.6 As long as $x\neq2$ and $x\neq6$, two equal fractions can be cross-multiplied. M 6.EE.A.3 Multiply out each side. The left side is $(x-1)(x-6)=x^2-7x+6$. The right side i 8.EE.C.7 Subtract $x^2$ from both sides; it disappears entirely, leaving $-7x+6=-(k+2)x+2 8.EE.C.7 The equation $(k-5)x=2k-6$ can be solved by dividing by $k-5$ — but only when $k 8.EE.C.7 For $k=1,2,3,4$ the formula $x=\frac{2k-6}{k-5}$ gives $x=1,\tfrac{2}{3},0,-2$ i Review
Reasonableness: Plug $k=5$ back into the original equation: $\frac{x-1}{x-2}=\frac{x-5}{x-6}$. Cross-multiplying gives $(x-1)(x-6)=(x-5)(x-2)$, i.e. $x^2-7x+6=x^2-7x+10$, which simplifies to $6=10$ — a contradiction with no $x$ at all. That matches the claim exactly. And the other four choices each produced a clean value of $x$ that avoids $2$ and $6$, so among the five options only $k=5$ has no solution.
Alternative: Instead of solving generally, notice the structure: the equation loses all its $x^2$ and $x$ information at once only when the two linear expressions on top differ by the same shift as the bottoms. Both denominators differ by $4$ ($x-2$ vs $x-6$), so the equation degenerates when the numerators also differ by $4$, i.e. $x-1$ vs $x-5$, giving $k=5$. You can also just test each choice by cross-multiplying; only $k=5$ produces a false numeric statement like $6=10$.
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Treating $k$ as a fixed unknown parameter and cross-multiplying the equation into a polynomial statement in $x$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $(x-1)(x-6)$ and $(x-k)(x-2)$ into $x^2-7x+6$ and $x^2-(k+2)x+2k$.)8.EE.C.7Solve linear equations in one variable (Reducing to $(k-5)x=2k-6$ and recognizing that $k=5$ makes it $0\cdot x=4$, a linear equation with no solution.)
⭐ Clear the fractions and simplify; the equation has no solution exactly when the $x$-term disappears but a nonzero number is left behind, which happens at $k=5$.
⭐ Clear the fractions and simplify; the equation has no solution exactly when the $x$-term disappears but a nonzero number is left behind, which happens at $k=5$.
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