SOLUTION: .7a + .25b + .3c = a .2a + .6b + .05c = b .1a + .15b + .65c = c WHERE a + b + c = 1

Question 22332: .7a + .25b + .3c = a
.2a + .6b + .05c = b
.1a + .15b + .65c = c
WHERE a + b + c = 1
Answer by AnlytcPhil(1806) (Show Source): You can put this solution on YOUR website!

.7a + .25b + .3c = a
.2a + .6b + .05c = b
.1a + .15b + .65c = c
WHERE a + b + c = 1

1st equation:     .7a + .25b +  .3c = a
2nd equation:     .2a +  .6b + .05c = b
3rd equation      .1a + .15b + .65c = c
4th equation        a +    b    + c = 1

Subtract a, b, and c from both sides of the first three 
equations, respectively, and you have this system:

-.3a + .25b +  .3c = 0
 .2a -  .4b + .05c = 0
 .1a + .15b - .35c = 0
   a +    b +    c = 0

The augmented matrix of this system is

[-.3   .25    .3 | 0]
[ .2   -.4   .05 | 0]
[ .1   .15  -.35 | 0]
[  1     1     1 | 1]

which is row equivalent to its row reduced echelon form:

[1 0 0 53/111]
[0 1 0  10/37]
[0 0 1 28/111]
[0 0 0   0   ]

a = 53/111
b =  10/37
c = 28/111

Edwin
AnlytcPhil@aol.com

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