All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Equation Exercise (Posted on 2014-12-16) Difficulty: 2 of 5
a,b,c,d and e satisfy this system of equations:

2a+b+c+d+e = 6
a+2b+c+d+e = 12
a+b+2c+d+e = 24
a+b+c+2d+e = 48
a+b+c+d+2e = 96

Find 3d+2e without solving the above system.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solving as little as possible | Comment 2 of 3 |
Summing the 5 equations gives
6(a+b+c+d+e)=186
multiplying by 5/6 gives
5a+5b+5c+5d+5e=155   (1)

Add three times the fourth to twice the fifth gives
5a+5b+5c+8d+7e=336 (2)

Subtracting (1) from (2) gives
3d+2e=336-155=181



  Posted by Jer on 2014-12-16 13:48:49
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information