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

Revision as of 12:38, 24 October 2015

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 $a^2-b^2+c^2=\boxed{(\text{B})1}$ .

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 7: / Line 7: @@
 == Solution ==
-<math>a+8c=7b+4</math> and <math>8a-c=7-4b</math>
+Rearranging, we get <math>a+8c=7b+4</math> and <math>8a-c=7-4b</math>
 Squaring both, <math>a^2+16ac+64c^2=49b^2+56b+16</math> and <math>64a^2-16ac+c^2=16b^2-56b+49</math> are obtained.