Difference between revisions of "1986 AIME Problems/Problem 4"
(→Solution) |
(→Solution 2) |
||
| Line 11: | Line 11: | ||
== Solution 2 == | == Solution 2 == | ||
| + | |||
| + | Subtracting the first equation from every one of the other equations yield | ||
| + | <cmath>x_2-x_1=6</cmath> | ||
| + | <cmath>x_3-x_1=18</cmath> | ||
| + | <cmath>x_4-x_1=42</cmath> | ||
| + | <cmath>x-5-x_1=90</cmath> | ||
== See also == | == See also == | ||
Revision as of 14:08, 22 August 2019
Contents
Problem
Determine 
if 
, 
, 
, 
, and 
satisfy the system of equations below.
Solution
Adding all five equations gives us 
so 
. Subtracting this from the fourth given equation gives 
and subtracting it from the fifth given equation gives 
, so our answer is 
.
Solution 2
Subtracting the first equation from every one of the other equations yield
![\[x_2-x_1=6\]](http://latex.artofproblemsolving.com/f/e/7/fe7a0fe78b118f0d28fd846394621ccee2c93387.png)
![\[x_3-x_1=18\]](http://latex.artofproblemsolving.com/d/0/3/d03ee58b4d39ea290e321bfc56f58a3d4d997a71.png)
![\[x_4-x_1=42\]](http://latex.artofproblemsolving.com/5/e/8/5e8178d31488e7e44ae2763c1a9e0d0766fd8b95.png)
![\[x-5-x_1=90\]](http://latex.artofproblemsolving.com/e/d/6/ed68e282e9cd401e0e98a4b704d479c774b03412.png)
See also
| 1986 AIME (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 | ||
| All AIME Problems and Solutions | ||
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
