2013 AMC 10B Problem 14

First factor the operation: $a \clubsuit b = ab(a - b)$. Then
$$x \clubsuit y = xy(x - y), \qquad y \clubsuit x = yx(y - x) = -xy(x - y).$$
Setting them equal gives $xy(x - y) = -xy(x - y)$, i.e.
$$2xy(x - y) = 0.$$
A product is zero when some factor is zero, so the solution set is the union of $x = 0$ (the $y$-axis), $y = 0$ (the $x$-axis), and $x = y$. These are three distinct lines, all passing through the origin.

The answer is $\boxed{\textbf{(E)}\ \text{three lines}}$.

Custom operations are just polynomials in disguise; factoring exposes their structure. An equation of the form (product of linear factors) $= 0$ describes the union of the lines given by each factor. Do not divide by a variable expression: each factor you would cancel is an entire line of solutions.

Expand: $x^2y - xy^2 = y^2x - yx^2$ gives $2x^2y - 2xy^2 = 0$, the same as $2xy(x-y) = 0$. Sanity check with points: $(1, 0)$, $(0, 5)$ and $(3, 3)$ all satisfy the equation, and they lie on three different lines through the origin.

• Cancelling xy from both sides and keeping only the line x = y, throwing away the axes.
• Seeing $x^2 y - x y^2$ and concluding the graph is a curve; the equation is homogeneous of degree $3$ and factors completely into lines.