2002 AMC 10B Problem 12

P11-15~3 minPrint

Linear Equations & Word Problems Use the answer choices (mod checks, size, form) to eliminate or select Set up the equation/formula and compute; no special trick needed

For which of the following values of $k$ does the equation $\frac{x-1}{x-2} = \frac{x-k}{x-6}$ have no solution for $x$ ?

Show solution without answering

Key insight

Cross-multiply. The x squared terms cancel, leaving (k-5)x = 2k-6. At k = 5 the left side is always 0 but the right side is 4.

Solution

1. The fractions only make sense when $x \ne 2$ and $x \ne 6$ (otherwise we divide by zero). Keep that in mind.
2. Clear the fractions by cross-multiplying:
$(x - 1)(x - 6) = (x - k)(x - 2).$
3. Multiply out both sides:
$x^2 - 7x + 6 = x^2 - 2x - kx + 2k.$
4. The $x^2$ on each side cancels. Move the $x$ terms to one side and the numbers to the other:
$kx + 2x - 7x = 2k - 6, \qquad\text{so}\qquad (k - 5)x = 2k - 6.$
5. Look at $k = 5$ . The left side becomes $0 \cdot x = 0$ and the right side becomes $10 - 6 = 4$ . The equation says $0 = 4$ , which is never true, so there is no solution.
6. For the other choices, $k - 5$ is not zero, so we can divide:

$k$	$1$	$2$	$3$	$4$
$x = \dfrac{2k - 6}{k - 5}$	$1$	$\tfrac23$	$0$	$-2$

None of these is $2$ or $6$ , so each one really solves the original equation.

The answer is $\boxed{\textbf{(E)}\ 5}$ .

Why this works

After cross-multiplying, the equation became "some number times $x$ equals some number." That kind of equation has no solution only when the number in front of $x$ is $0$ and the other side is not $0$ . It is also a good habit to check that an answer does not make a denominator zero, since those values were never allowed.

Alternative approach

Split each fraction. Since $x - 1 = (x - 2) + 1$ , we get $\frac{x-1}{x-2} = 1 + \frac{1}{x-2}$ . For $k = 5$ , the same trick gives $\frac{x-5}{x-6} = 1 + \frac{1}{x-6}$ . For these to be equal we would need $\frac1{x-2} = \frac1{x-6}$ , that is $x - 2 = x - 6$ , which is impossible.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The equation is defined for $x \ne 2, 6$ . On that domain, cross-multiply:
$(x-1)(x-6) = (x-k)(x-2).$
Expanding, $x^2 - 7x + 6 = x^2 - (k+2)x + 2k$ . The $x^2$ terms cancel, leaving a linear equation:
$(k+2)x - 7x = 2k - 6 \quad\Longrightarrow\quad (k-5)\,x = 2k - 6.$

If $k = 5$ the left side is $0$ while the right side is $4$ , so there is no $x$ at all.

For the other choices the coefficient $k - 5$ is nonzero and $x = \frac{2k-6}{k-5}$ is a candidate: $k = 1, 2, 3, 4$ give $x = 1, \frac23, 0, -2$ . None of these is $2$ or $6$ , so each is a genuine solution and those equations are solvable.

The answer is $\boxed{\textbf{(E)}\ 5}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 7: Proportion (§7.4)