2002 AMC 10B Problem 12

Question

For which of the following values of

k

does the equation

\dfrac{x - 1}{x - 2} = \dfrac{x - k}{x - 6}

have no solution for

x?

1

2

3

4

5

Cross multiply and expand both sides

The equation becomes linear; it has no solution when the coefficient of

x

vanishes

Cross multiplying gives

(x - 1)(x - 6) = (x - 2)(x - k),

which expands to

\begin{aligned} x^2 - 7x + 6 &= x^2 - (2 + k)x \\ &\quad {}+ 2k. \end{aligned}

Cancelling

x^2

leaves

(k - 5)x = 2k - 6.

For

k

equal to

1,

2,

3,

or

4,

this gives a valid value of

x

that is neither excluded denominator value

2

nor

6.

For

k=5,

the equation instead becomes

0\cdot x=4,

which has no solution.

Thus, the correct answer is E.

Problem 12 in other years