Difference between revisions of "2004 AMC 10A Problems/Problem 4"
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Alternatively, we can solve by casework (a method which should work for any similar problem involving [[absolute value]]s of [[real number]]s). If <math>x \leq 1</math> then <math>|x - 1| = 1-x</math> and <math>|x - 2| = 2 - x</math> so we must solve <math>1 - x = 2 - x</math> which has no solutions. Similarly, if <math>x \geq 2</math> then <math>|x - 1| = x - 1</math> and <math>|x - 2| = x - 2</math> so we must solve <math>x - 1 = x- 2</math>, which also has no solutions. Finally, if <math>1 \leq x \leq 2</math> then <math>|x - 1| = x - 1</math> and <math>|x - 2| = 2-x</math> so we must solve <math>x - 1 = 2 - x</math>, which has the unique solution <math>x = \frac32</math>. | Alternatively, we can solve by casework (a method which should work for any similar problem involving [[absolute value]]s of [[real number]]s). If <math>x \leq 1</math> then <math>|x - 1| = 1-x</math> and <math>|x - 2| = 2 - x</math> so we must solve <math>1 - x = 2 - x</math> which has no solutions. Similarly, if <math>x \geq 2</math> then <math>|x - 1| = x - 1</math> and <math>|x - 2| = x - 2</math> so we must solve <math>x - 1 = x- 2</math>, which also has no solutions. Finally, if <math>1 \leq x \leq 2</math> then <math>|x - 1| = x - 1</math> and <math>|x - 2| = 2-x</math> so we must solve <math>x - 1 = 2 - x</math>, which has the unique solution <math>x = \frac32</math>. | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 02:38, 11 September 2007
Problem
What is the value of if ?
Solution
is the distance between and ; is the distance between and .
Therefore, the given equation says is equidistant from and , so .
Alternatively, we can solve by casework (a method which should work for any similar problem involving absolute values of real numbers). If then and so we must solve which has no solutions. Similarly, if then and so we must solve , which also has no solutions. Finally, if then and so we must solve , which has the unique solution .
See also
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 |