Can anybody help me with this algebra question?

Question

I am having a bit of trouble with this question; I know it is simple but I cannot get to the answer and I keep screwing up.

Here's the problem:

Find the values of the variables a, b, c, d, and e if they meet the following conditions:

a(b+c+d+e)=128
b(a+c+d+e)=155
c(a+b+d+e)=203
d(a+b+c+e)=243
e(a+b+c+d)=275
I think this has to be done by subtracting the total values from related conditions such as in the 1st 2 math sentences;
(ba+bc+bd+be)-(ab+ac+ad+ae)= 155-128= 27
However this keeps messing up and so far I could only think of this.
I need work too, not just the answer...
This is sort of urgent...

Answer 1

a(b+c+d+e)=128
b(a+c+d+e)=155
c(a+b+d+e)=203
d(a+b+c+e)=243
e(a+b+c+d)=275

c=1 or 7 or 29
a+b+c+d+e=c+203/c=204 or 36
a+b+c+d+e=b+155/b
b=1 or 5 or 31
a+b+c+d+e=b+155/b=156 or 36

so a+b+c+d+e=36 and c=7 or 29 and b=5 or 31

36=a+b+c+d+e=a+128/a
so a=4

36=a+b+c+d+e=d+243/d
so d=9 or 27

36=a+b+c+d+e=e+275/e
so e=11 or 25

a=4
b=5 or 31
d=9 or 27
c=7 or 29
e=11 or 25

so the only solution in integers is
a=4
b=5
c=7
d=9
e=11

Answer 2

(b+c+d+e)=128/a
(a+c+d+e)=155/b
(a+b+d+e)=203/c
(a+b+c+e)=243/d
(a+b+c+d)=275/e

Working with the 1st equation:
let a = 4 then b+c+d+e = 32

Working with the 2nd equation:
4 + 32 - b = 155/b
36 -b = 155/b
b^2 - 36b + 155 = 0
(b-5)(b-31) = 0
so b = 5 or 31
but b+c+d+e = 32 so b has to be 5

Working with the 3rd equation:
4 + 5 + (32 - b -c) = 203/c
4 + 5 + (27 -c) = 203/c
36 -c = 203/c
36c - c^2 = 203
c^2 - 36c + 203 = 0
(c-7)(c-29) = 9
By same reasoning as b, c is 7

Working with the 4th and 5th equations:
d(4 + 5 + 7 + e) = 243
e(4 + 5 + 7 + d) = 275

d(16 + e) = 243
e(16 + d) = 275

e = (243/d) - 16
((243/d) - 16)(16+d) = 275
(3888/d) - 256 + 243 - 16d = 275
3888 - 13d - 16d^2 = 275d
16d^2 + 288d - 3888 = 0
d^2 + 18d - 243 = 0
(d + 27)(d - 9) = 0
d = -27 or d = 9

e = (243/d) -16
b + c + d + e = 32
5 + 7 + d + e = 32
d + e = 20
d = 9
e = 11

So,
a=4
b=5
c=7
d=9
e=11

You can verify all of those. I verified them, and they all work.

The key to this problem was fixing a as 4. The reason I did this was because there were 4 numbers being summed in the first equation (b,c,d,e) and 128 is divisible by 4.

Answer 3

This is a system of equations, 5 unknowns with 5 equations, so it should be possible to solve that way. So far I have not been able to figure it out either. You could also put this into the form of a matrix. It would have to have 6 columns, 1 for each of the unknowns and 1 for the numbers.

Answer 4

a million distribute the 5 on the splendid facet. 3y-7=(10y-15)+4 2 minus 4 on the two sides 3y-7-4=(10y-15)+4-4 3y-11=10y-15 3 manage so which you have ys on one facet and numbers the different. 3y-11+15=10y-15+15 3y+4=10y 3y-3y+4=10y-3y 4=7y 4 divide via 7 to get y. 4/7=7y/7 4/7=y y=(approx)0.5714

Answer 5

This is not a trivial exercise. I am working on it too....there must be some trick as you indicate.