AMC 10 · 2002 · #12
Grade 8 algebraPick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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) · 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.
Cross-multiply to clear the fractions
Since x ≠ 2 and x ≠ 6, cross-multiplying is legal, and the fractions vanish into (x-1)(x-6)=(x-k)(x-2).
Cross-multiplying is just multiplying both sides by both denominators at once, which is legal whenever neither denominator is zero.
6.EE.B.6Convert To AlgebraExpand both sides
Multiply out both sides: x²-7x+6=x²-(k+2)x+2k. Each side opens with x², so that term is doomed to cancel.
Expanding turns the two products into standard x²+… form so matching terms can be lined up.
6.EE.A.3Convert To AlgebraCancel the square term and collect x
Cancel x², then push the x-terms left and the constants right; everything collapses to (k-5)x=2k-6.
Once the x² pieces match and cancel, what is left is a straight-line equation whose whole behavior sits in the coefficient (k-5).
8.EE.C.7Convert To AlgebraFind when no x can work
Dividing by k-5 needs k-5 ≠ 0. At k=5 the line reads 0 · x=4 — false for every x, so no solution exists.
A linear equation has no answer exactly when the x term vanishes but a nonzero number is left demanding to equal zero.
A straight-line equation has no answer exactly when its variable term vanishes but a nonzero number is left.
▸ Why?
The whole behaviour sits in the coefficient, so a zero coefficient removes every solution at once.
▸ Why?
With the variable gone, what remains is a plain comparison of two fixed numbers that simply disagree.
Confirm the other four choices do have solutions
For k=1,2,3,4 the formula x=(2k-6)/(k-5) gives 1, 2/3, 0, -2 — all legal, so only k=5 breaks, choice (E).
Testing the leftover choices shows they all produce a legal x, so the one that breaks is the answer.
8.EE.C.7Eliminate PossibilitiesClear 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.
- Cross-multiply to clear the fractions
- Expand both sides
- Cancel the square term and collect x
- Find when no x can work
- Confirm the other four choices do have solutions