AMC 10 · 2020 · #14
Grade 7 arithmeticProblem
Real numbers x and y satisfy x+y=4 and x⋅y=−2. What is the value of
x+y2x3+x2y3+y?
Pick an answer.
AMC 10 2020 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: Real numbers $x$ and $y$ satisfy $x + y = 4$ and $xy = -2$. Compute the value of $x + \dfrac{x^3}{y^2} + \dfrac{y^3}{x^2} + y$.
Givens: $x + y = 4$; $xy = -2$; Target expression: $E = x + \dfrac{x^3}{y^2} + \dfrac{y^3}{x^2} + y$; Answer choices: $360,\; 400,\; 420,\; 440,\; 480$
Unknowns: Value of $E$
Understand
Restated: Real numbers $x$ and $y$ satisfy $x + y = 4$ and $xy = -2$. Compute the value of $x + \dfrac{x^3}{y^2} + \dfrac{y^3}{x^2} + y$.
Givens: $x + y = 4$; $xy = -2$; Target expression: $E = x + \dfrac{x^3}{y^2} + \dfrac{y^3}{x^2} + y$; Answer choices: $360,\; 400,\; 420,\; 440,\; 480$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #13 Convert to Algebra, #3 Eliminate Possibilities
Tool #7 (Subproblems): the target $E$ has four terms that don't obviously combine. Group them as $\big(x + \dfrac{y^3}{x^2}\big) + \big(\dfrac{x^3}{y^2} + y\big) = \dfrac{x^3 + y^3}{x^2} + \dfrac{x^3 + y^3}{y^2}$, then factor out $x^3 + y^3$ to get $E = (x^3 + y^3)\big(\dfrac{1}{x^2} + \dfrac{1}{y^2}\big) = \dfrac{(x^3 + y^3)(x^2 + y^2)}{x^2 y^2}$. Tool #13 (Algebra): compute the symmetric sums $x^2 + y^2, x^3 + y^3, x^2 y^2$ from $x + y = 4, xy = -2$. Tool #3 (Eliminate): match the final number against the five choices.
Execute — Answer: D
7.EE.A.1 Step 1 - Group the four terms of $E$ by which power sits on top.
- Pair $x$ with $\dfrac{y^3}{x^2}$ (both have $x^2$ as denominator when written over $x^2$) and pair $y$ with $\dfrac{x^3}{y^2}$.
- $E = x + y + \dfrac{x^3}{y^2} + \dfrac{y^3}{x^2} = \dfrac{x^3 + y^3}{x^2} + \dfrac{x^3 + y^3}{y^2}$.
💡 Grade 7 expressions: rewrite $x = \dfrac{x^3}{x^2}$ so each term has a clean denominator.
7.EE.A.1 Step 2 Factor out the common $x^3 + y^3$ and combine the two fractions over $x^2 y^2$: $E = (x^3 + y^3) \left( \dfrac{1}{x^2} + \dfrac{1}{y^2} \right) = (x^3 + y^3) \cdot \dfrac{x^2 + y^2}{x^2 y^2}$.
💡 Grade 7 expressions: factor the common piece, then add fractions with a common denominator.
7.EE.A.2 Step 3 - Compute $x^2 + y^2$ from $(x + y)^2 = x^2 + 2xy + y^2$.
- So $x^2 + y^2 = (x+y)^2 - 2xy = 4^2 - 2(-2) = 16 + 4 = 20$.
💡 Grade 7 identity: expand $(x+y)^2$ and rearrange to free $x^2 + y^2$.
7.EE.A.2 Step 4 - Compute $x^3 + y^3$ from $(x+y)^3 = x^3 + 3x^2 y + 3 x y^2 + y^3 = (x^3 + y^3) + 3xy(x+y)$.
- So $x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 4^3 - 3(-2)(4) = 64 + 24 = 88$.
💡 Grade 7 identity: cube the sum and subtract the cross terms.
6.EE.A.1 Step 5 Compute $x^2 y^2 = (xy)^2 = (-2)^2 = 4$.
💡 Grade 6 exponents: squaring the product is the same as squaring each factor.
6.EE.A.2 Step 6 Substitute into the boxed formula: $E = \dfrac{(88)(20)}{4} = \dfrac{1760}{4} = 440$.
💡 Grade 6 expression evaluation: plug the three pieces into one formula.
4.NBT.A.2 Step 7 - Match $440$ to the choices: option (D).
- The others would correspond to dropping the $x^2 + y^2$ factor ($88$ alone, too small) or mishandling signs in $x^3 + y^3$ ($x^3 + y^3 = 40$ would give $200$, etc.).
💡 Grade 4 comparison: only one option equals $440$.
7.EE.A.1 Group the four terms of $E$ by which power sits on top. Pair $x$ with $\dfrac{y^ 7.EE.A.1 Factor out the common $x^3 + y^3$ and combine the two fractions over $x^2 y^2$: 7.EE.A.2 Compute $x^2 + y^2$ from $(x + y)^2 = x^2 + 2xy + y^2$. So $x^2 + y^2 = (x+y)^2 7.EE.A.2 Compute $x^3 + y^3$ from $(x+y)^3 = x^3 + 3x^2 y + 3 x y^2 + y^3 = (x^3 + y^3) + 6.EE.A.1 Compute $x^2 y^2 = (xy)^2 = (-2)^2 = 4$. 6.EE.A.2 Substitute into the boxed formula: $E = \dfrac{(88)(20)}{4} = \dfrac{1760}{4} = 4.NBT.A.2 Match $440$ to the choices: option (D). The others would correspond to dropping Review
Reasonableness: The two roots of $t^2 - 4t - 2 = 0$ are $x = 2 + \sqrt{6}, y = 2 - \sqrt{6}$ (since $x + y = 4, xy = -2$). Then $x \approx 4.449, y \approx -0.449$, so $\dfrac{x^3}{y^2} \approx \dfrac{88.0}{0.202} \approx 436$ and $\dfrac{y^3}{x^2} \approx \dfrac{-0.091}{19.79} \approx -0.005$. Adding $x + y = 4$ gives $E \approx 440$. ✓ Matches.
Alternative: Tool #6 (Guess and Check) using Newton's power sums. Let $P_n = x^n + y^n$. Then $P_1 = 4, P_2 = 20, P_n = 4 P_{n-1} + 2 P_{n-2}$ (from $x, y$ being roots of $t^2 = 4t + 2$). This gives $P_3 = 4 \cdot 20 + 2 \cdot 4 = 88$, $P_4 = 4 \cdot 88 + 2 \cdot 20 = 392$, $P_5 = 4 \cdot 392 + 2 \cdot 88 = 1744$. After noting $E = (x + y) + \dfrac{x^5 + y^5}{x^2 y^2} = 4 + \dfrac{1744}{4} = 4 + 436 = 440$. Same answer (D).
CCSS standards used (min grade 7)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Matching the computed value $440$ to the answer choices.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Computing $(xy)^2 = (-2)^2 = 4$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating $E = \dfrac{(x^3 + y^3)(x^2 + y^2)}{x^2 y^2}$ after the three symmetric sums are known.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Grouping $E$ as $\dfrac{x^3 + y^3}{x^2} + \dfrac{x^3 + y^3}{y^2}$ and combining over a common denominator.)7.EE.A.2Rewrite an expression in different forms to shed light on the problem (Using $x^2 + y^2 = (x+y)^2 - 2xy$ and $x^3 + y^3 = (x+y)^3 - 3xy(x+y)$ to convert to symmetric sums.)
⭐ This AMC 10 problem only needs Grade 7 algebra you already know! Rewrite $x = \dfrac{x^3}{x^2}$ and $y = \dfrac{y^3}{y^2}$ so the four terms factor as $E = \dfrac{(x^3 + y^3)(x^2 + y^2)}{x^2 y^2}$. The identities $x^2 + y^2 = (x+y)^2 - 2xy = 20$ and $x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 88$ and $x^2 y^2 = 4$ are all you need. Plug in: $E = \dfrac{88 \cdot 20}{4} = \mathbf{440}$, answer (D).
⭐ This AMC 10 problem only needs Grade 7 algebra you already know! Rewrite $x = \dfrac{x^3}{x^2}$ and $y = \dfrac{y^3}{y^2}$ so the four terms factor as $E = \dfrac{(x^3 + y^3)(x^2 + y^2)}{x^2 y^2}$. The identities $x^2 + y^2 = (x+y)^2 - 2xy = 20$ and $x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 88$ and $x^2 y^2 = 4$ are all you need. Plug in: $E = \dfrac{88 \cdot 20}{4} = \mathbf{440}$, answer (D).
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