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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.

No Solution Yet Submitted by K Sengupta    
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Solution solution | Comment 1 of 3
(6a+6b+6c+6d+6e)/6 = (6+12+24+48+96)/6 = (a+b+c+d+e) = 31

We can now solve for d and e without having to solve for
a, b, c (though solving for such is simple).

(a+b+c+2d+e) = (a+b+c+d+e) + d = 31 + d
(a+b+c+2d+e) = 48
31 + d = 48
d = 17

(a+b+c+d+2e) = (a+b+c+d+e) + e = 31 + e
(a+b+c+d+2e) = 96  
31 + e = 96
e = 65

3(17)+2(65) = 181
  Posted by Dej Mar on 2014-12-16 11:05:52
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