Difference between revisions of "1986 AIME Problems/Problem 4"
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
x_2-x_1&=6\\ | x_2-x_1&=6\\ | ||
| − | x_3- | + | x_3-x_1&=18\\ |
x_4-x_1&=42\\ | x_4-x_1&=42\\ | ||
x_5-x_1&=90 | x_5-x_1&=90 | ||
| Line 21: | Line 21: | ||
Thus | Thus | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | 2x_1+x_2+x_3+x_4+x_5&=6 | + | 2x_1+x_2+x_3+x_4+x_5&=6\\ |
| − | 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, | ||
| + | <cmath>3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}</cmath> | ||
| + | |||
| + | ~ Nafer | ||
== See also == | == See also == | ||
Latest revision as of 15:25, 17 November 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,
![\[3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}\]](http://latex.artofproblemsolving.com/6/b/8/6b8d2d669668b1e3ef66e7b2c3b17f484ed79172.png)
~ Nafer
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. 
