AMC 10 · 2013 · #14
Grade 8 algebraDefine a♣b=a2b−ab2. Which of the following describes the set of points (x,y) for which x♣y=y♣x?
Pick an answer.
AMC 10 2013 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: A made-up operation is defined by $a \clubsuit b = a^2b - ab^2$. Find every point $(x, y)$ where using $x$ first gives the same value as using $y$ first, that is, where $x \clubsuit y = y \clubsuit x$, and describe what that set of points looks like.
Givens: The rule $a \clubsuit b = a^2b - ab^2$; The condition $x \clubsuit y = y \clubsuit x$; Answer choices: (A) a finite set of points, (B) one line, (C) two parallel lines, (D) two intersecting lines, (E) three lines
Unknowns: The shape of the set of all points $(x, y)$ that satisfy the condition
Understand
Restated: A made-up operation is defined by $a \clubsuit b = a^2b - ab^2$. Find every point $(x, y)$ where using $x$ first gives the same value as using $y$ first, that is, where $x \clubsuit y = y \clubsuit x$, and describe what that set of points looks like.
Givens: The rule $a \clubsuit b = a^2b - ab^2$; The condition $x \clubsuit y = y \clubsuit x$; Answer choices: (A) a finite set of points, (B) one line, (C) two parallel lines, (D) two intersecting lines, (E) three lines
Plan
Primary tool: #13 Convert to Algebra
Secondary: #7 Identify Subproblems, #1 Draw a Diagram
The strange $\clubsuit$ symbol is just shorthand for an algebra expression, so Tool #13 (Convert to Algebra) turns $x \clubsuit y = y \clubsuit x$ into a plain equation you can simplify and factor. Once it is factored into a product that equals zero, Tool #7 (Identify Subproblems) splits it into separate small equations, one for each factor. Tool #1 (Draw a Diagram) then reads each of those equations as a line in the plane, so counting the lines answers the question.
Execute — Answer: E
6.EE.A.2 Step 1 Write both sides with the rule
- Replace each $\clubsuit$ with what the rule says.
- For $x \clubsuit y$ let $a=x$ and $b=y$, giving $x^2y - xy^2$.
- For $y \clubsuit x$ let $a=y$ and $b=x$, giving $y^2x - yx^2$, which is the same as $xy^2 - x^2y$.
- So the left side is $x^2y - xy^2$ and the right side is $xy^2 - x^2y$.
💡 A new symbol is only scary until you swap it for the plain expression it stands for.
7.EE.A.1 Step 2 Set equal and factor
- Set the two sides equal: $x^2y - xy^2 = xy^2 - x^2y$.
- Move everything to the left by adding $x^2y$ and subtracting $xy^2$ from both sides, which collects the like terms into $2x^2y - 2xy^2 = 0$.
- Both terms share the common factor $2xy$, so pull it out to get $2xy(x - y) = 0$.
💡 Getting one side to zero lets you factor, and a factored product is far easier to read than a spread-out equation.
6.EE.B.5 Step 3 Split into three cases
- A product equals zero only when one of its factors is zero.
- The constant $2$ is never zero, so the equation $2xy(x - y) = 0$ holds exactly when $x = 0$, or $y = 0$, or $x - y = 0$ (that is, $x = y$).
- These three separate equations together capture every solution point.
💡 If several numbers multiply to zero, at least one of them had to be zero to begin with.
8.F.A.3 Step 4 Read each case as a line
- Each of the three equations describes a straight line in the $xy$-plane.
- $x = 0$ is the vertical $y$-axis, $y = 0$ is the horizontal $x$-axis, and $y = x$ is the slanted line through the origin at $45^\circ$.
- The solution set is the union of these three lines, so the picture is three lines, which is choice (E).
💡 An equation like $x=0$ or $y=x$ is a rule every point on one straight line obeys.
6.EE.A.2 Replace each $\clubsuit$ with what the rule says. For $x \clubsuit y$ let $a=x$ 7.EE.A.1 Set the two sides equal: $x^2y - xy^2 = xy^2 - x^2y$. Move everything to the lef 6.EE.B.5 A product equals zero only when one of its factors is zero. The constant $2$ is 8.F.A.3 Each of the three equations describes a straight line in the $xy$-plane. $x = 0$ Review
Reasonableness: Test a point off all three lines, say $(2, 1)$: $x \clubsuit y = 4\cdot1 - 2\cdot1 = 2$ while $y \clubsuit x = 1\cdot2 - 1\cdot4 = -2$, so they differ, as expected since $(2,1)$ lies on none of $x=0$, $y=0$, $y=x$. Now test $(3, 3)$ on the line $y=x$: both sides give $9\cdot3 - 3\cdot9 = 0$, so they match. The three lines all pass through the origin, meaning they intersect rather than stay parallel, which rules out (C) and (D); having three of them rules out (A) and (B), leaving (E).
Alternative: Skip the coordinate picture and reason from symmetry. The equation $2xy(x-y)=0$ is a product of three factors $x$, $y$, and $x-y$; the whole product is zero exactly when at least one factor is zero. Each factor set to zero is a linear equation, and the three factors are genuinely different, so they give three distinct lines. Counting the distinct linear factors immediately gives three lines, confirming (E).
CCSS standards used (min grade 8)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting into the rule $a \clubsuit b = a^2b - ab^2$ to write $x \clubsuit y = x^2y - xy^2$ and $y \clubsuit x = xy^2 - x^2y$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Collecting like terms into $2x^2y - 2xy^2 = 0$ and factoring out $2xy$ to get $2xy(x - y) = 0$.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Recognizing that $2xy(x-y)=0$ holds exactly when $x=0$, $y=0$, or $x=y$.)8.F.A.3Interpret the equation y = mx + b as defining a linear function (Reading $x=0$, $y=0$, and $y=x$ as three straight lines in the coordinate plane.)
⭐ Turn the weird symbol into plain algebra, get one side to zero and factor; each factor that can be zero is its own line, and here there are three of them.
⭐ Turn the weird symbol into plain algebra, get one side to zero and factor; each factor that can be zero is its own line, and here there are three of them.
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