AMC 10 · 2002 · #20
Grade 8 arithmeticLet a, b, and c be real numbers such that a−7b+8c=4 and 8a+4b−c=7. Then a2−b2+c2 is
Pick an answer.
AMC 10 2002 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: Two equations tie together three numbers $a$, $b$, and $c$: $a - 7b + 8c = 4$ and $8a + 4b - c = 7$. Find the value of $a^2 - b^2 + c^2$.
Givens: The equation $a - 7b + 8c = 4$; The equation $8a + 4b - c = 7$; $a$, $b$, $c$ are real numbers; Answer choices: (A) $0$, (B) $1$, (C) $4$, (D) $7$, (E) $8$
Unknowns: The value of $a^2 - b^2 + c^2$
Understand
Restated: Two equations tie together three numbers $a$, $b$, and $c$: $a - 7b + 8c = 4$ and $8a + 4b - c = 7$. Find the value of $a^2 - b^2 + c^2$.
Givens: The equation $a - 7b + 8c = 4$; The equation $8a + 4b - c = 7$; $a$, $b$, $c$ are real numbers; Answer choices: (A) $0$, (B) $1$, (C) $4$, (D) $7$, (E) $8$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #15 Organize Information in More Ways, #6 Guess and Check
Two equations with three letters cannot lock down $a$, $b$, and $c$ separately, but they are enough to lock down two letters once the third is chosen. Tool #4 (Introduce a Variable) turns that freedom into a lever: keep $a$ as a free stand-in and solve the two linear equations for $b$ and $c$ in terms of $a$. Then Tool #15 (Organize Information in More Ways) rewrites the target $a^2 - b^2 + c^2$ smartly — grouping $c^2 - b^2$ as a difference of squares $(c-b)(c+b)$ — so the messy fractions collapse and the leftover $a$ cancels. Tool #6 (Guess and Check) is the safety net: because the value is promised to be constant, plugging in one convenient choice of a letter must give the same number, so it doubles as a fast confirmation.
Execute — Answer: B
8.EE.C.8 Step 1 Solve for b and c using a
- Treat $a$ as a known stand-in and solve the two equations for $b$ and $c$.
- From the second equation, $8a + 4b - c = 7$, isolate $c$: $c = 4b - 7 + 8a$.
- Substitute this into the first equation, $a - 7b + 8c = 4$: $a - 7b + 8(4b - 7 + 8a) = 4$, which is $a - 7b + 32b - 56 + 64a = 4$, so $25b + 65a - 56 = 4$, giving $25b = 60 - 65a$ and $b = \frac{12 - 13a}{5}$.
- Then $c = 4b - 7 + 8a = \frac{48 - 52a}{5} - 7 + 8a = \frac{48 - 52a - 35 + 40a}{5} = \frac{13 - 12a}{5}$.
💡 Two equations can pin down two of the letters once you let the third one float free.
6.EE.A.3 Step 2 Group c squared minus b squared
- The target is $a^2 - b^2 + c^2 = a^2 + (c^2 - b^2)$.
- Instead of squaring the ugly fractions for $b$ and $c$ separately, use the difference-of-squares shortcut $c^2 - b^2 = (c - b)(c + b)$, because $c - b$ and $c + b$ come out clean.
- Subtracting, $c - b = \frac{(13 - 12a) - (12 - 13a)}{5} = \frac{1 + a}{5}$.
- Adding, $c + b = \frac{(13 - 12a) + (12 - 13a)}{5} = \frac{25 - 25a}{5} = 5(1 - a)$.
💡 Difference of squares trades two hard squarings for one easy subtraction times one easy addition.
6.EE.A.3 Step 3 Multiply and cancel the a squared
- Multiply the two clean pieces: $c^2 - b^2 = (c - b)(c + b) = \frac{1 + a}{5} \cdot 5(1 - a) = (1 + a)(1 - a) = 1 - a^2$.
- Now put it back into the target: $a^2 - b^2 + c^2 = a^2 + (c^2 - b^2) = a^2 + (1 - a^2) = 1$.
- The $a^2$ terms erase each other, so the free letter leaves no trace and the value is exactly $1$, which is choice (B).
💡 When the only leftover of the free letter cancels itself, what remains must be the same for every valid $a$, $b$, $c$.
8.EE.C.8 Treat $a$ as a known stand-in and solve the two equations for $b$ and $c$. From 6.EE.A.3 The target is $a^2 - b^2 + c^2 = a^2 + (c^2 - b^2)$. Instead of squaring the ugl 6.EE.A.3 Multiply the two clean pieces: $c^2 - b^2 = (c - b)(c + b) = \frac{1 + a}{5} \cd Review
Reasonableness: The answer should not depend on which valid triple you pick, so test one. Let $a = 0$: the equations become $-7b + 8c = 4$ and $4b - c = 7$; solving gives $b = \frac{12}{5}$ and $c = \frac{13}{5}$, and $a^2 - b^2 + c^2 = 0 - \frac{144}{25} + \frac{169}{25} = \frac{25}{25} = 1$. Try another: let $c = 0$, giving $a - 7b = 4$ and $8a + 4b = 7$, which solve to $a = \frac{13}{12}$, $b = -\frac{5}{12}$, and $a^2 - b^2 + c^2 = \frac{169}{144} - \frac{25}{144} + 0 = \frac{144}{144} = 1$. Both land on $1$, matching choice (B).
Alternative: Since the value is guaranteed constant, use Tool #6 (Guess and Check) directly: pick the most convenient choice, $a = 0$, solve the resulting two-equation system for $b = \frac{12}{5}$ and $c = \frac{13}{5}$, and compute $a^2 - b^2 + c^2 = \frac{169 - 144}{25} = 1$ in one shot. This skips the general algebra entirely — a legitimate contest shortcut precisely because the problem structure forces a single answer for every valid triple.
CCSS standards used (min grade 8)
8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving the system $a - 7b + 8c = 4$ and $8a + 4b - c = 7$ for $b$ and $c$ in terms of the free letter $a$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Rewriting $c^2 - b^2$ as $(c - b)(c + b)$, multiplying the clean factors, and cancelling the $a^2$ terms to reach $1$.)
⭐ When two equations share more letters than they can pin down, let one letter float free, solve for the rest, and watch the leftover cancel — the answer that survives is the one that was fixed all along.
⭐ When two equations share more letters than they can pin down, let one letter float free, solve for the rest, and watch the leftover cancel — the answer that survives is the one that was fixed all along.
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