Difference between revisions of "2004 AMC 10A Problems/Problem 4"
m (→Solution) |
|||
| Line 7: | Line 7: | ||
<math>|x-1|</math> is the distance between <math>x</math> and <math>1</math>; <math>|x-2|</math> is the distance between <math>x</math> and <math>2</math>. | <math>|x-1|</math> is the distance between <math>x</math> and <math>1</math>; <math>|x-2|</math> is the distance between <math>x</math> and <math>2</math>. | ||
| − | Therefore, the given equation says <math>x</math> is | + | Therefore, the given equation says <math>x</math> is equidistant from <math>1</math> and <math>2</math>, so <math>x=\frac{1+2}2=\frac{3}{2}\Rightarrow\boxed{\mathrm{(D)}\ \frac{3}{2}}</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>. | 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>. | ||
Revision as of 21:40, 20 July 2014
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
