Difference between revisions of "2002 AMC 10B Problems/Problem 20"

Revision as of 12:28, 30 November 2019

Let a, b, and c be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$ . Then $a^2-b^2+c^2$ is

$\mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8$

Rearranging, we get $a+8c=7b+4$ and $8a-c=7-4b$

Squaring both, $a^2+16ac+64c^2=49b^2+56b+16$ and $64a^2-16ac+c^2=16b^2-56b+49$ are obtained.

Adding the two equations and dividing by $65$ gives $a^2+c^2=b^2+1$ , so $a^2-b^2+c^2=\boxed{(\text{B})1}$ .

The easiest way is to assume a value for $a$ and then solving the system of equations. For $a = 1$ , we get the equations

$-7b + 8c = 3$ and

$4b - c = -1$

Multiplying the second equation by $8$ , we have

$32b - 8c = -8$

Adding up the two equations yields

$25b = -5$ , so $b = -\frac{1}{5}$

We obtain $c = \frac{1}{5}$ after plugging in the value for $b$ .

Therefore, $a^2-b^2+c^2 = 1-\frac{1}{25}+\frac{1}{25}=\boxed{1}$ which corresponds to $\text{(B)}$

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

@@ Line 28: / Line 28: @@
 <math>25b = -5</math>, so <math>b = -\frac{1}{5}</math>
+We obtain <math>c = \frac{1}{5}</math> after plugging in the value for <math>b</math>.
+Therefore, <math>a^2-b^2+c^2 = 1-\frac{1}{25}+\frac{1}{25}=\boxed{1}</math> which corresponds to <math>\text{(B)}</math>
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