Difference between revisions of "1986 AIME Problems/Problem 4"
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2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\ | 2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\ | ||
6x_1+156&=6\\ | 6x_1+156&=6\\ | ||
| − | + | x_1=-25 | |
\end{align*}</cmath> | \end{align*}</cmath> | ||
| + | Using the previous equations, | ||
| + | <math></math>3x_4+2x_5=3(x_1+42)=2(x_1+90)=\boxed{ | ||
== See also == | == See also == | ||
Revision as of 14:16, 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 yields
Thus
Using the previous equations,
$$ (Error compiling LaTeX. Unknown error_msg)3x_4+2x_5=3(x_1+42)=2(x_1+90)=\boxed{
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. 
