How to solve system of 4 equations with four unknowns? I have got system of 4 equations as shown below and I am considering if there is any other method than brute force to solve them.
B + C + D = S1
A + C + D = S2
A + B + D = S3
A + B + C = S4

Values of S1-S4 are given:
S1 = 70
S2 = 75
S3 = 80
S4 = 75

I tried to to solve that, but always I have infinite equation what means to solve A I need B, to solve B I need C, to solve C I need D and to solve D I need.. A. I know the answer, but need the solution how to solve it. Answer:
A = 30
B = 25
C = 20
D = 25

 A: Use Gaussian elimination/method:
Your equation becomes:
$$\left[\begin{array}{cccc|c}
      1 & 1 & 1 & 0 & 75         \\[0.55ex]
      1 & 1 & 0 & 1 & 80        \\[0.55ex]
      1 & 0 & 1 & 1 & 75\\[0.55ex]
      0 & 1 & 1 & 1 & 70
\end{array}\right]$$
You can multiply rows and add/subtract them from another row, which will not affect the solution. So you can subtract the first row from the second and the thirdrow and the matrix will become:
$$\left[\begin{array}{cccc|c}
      1 & 1 & 1 & 0 & 75         \\[0.55ex]
      0 & 0 & -1 & 1 & 5        \\[0.55ex]
      0 & -1 & 0 & 1 & 0\\[0.55ex]
      0 & 1 & 1 & 1 & 70
\end{array}\right]$$
Now add the third row to the fourth:
$$\left[\begin{array}{cccc|c}
      1 & 1 & 1 & 0 & 75         \\[0.55ex]
      0 & 0 & -1 & 1 & 5        \\[0.55ex]
      0 & -1 & 0 & 1 & 0\\[0.55ex]
      0 & 0 & 1 & 2 & 70
\end{array}\right]$$
And at last add the second row to the fourth row:
$$\left[\begin{array}{cccc|c}
      1 & 1 & 1 & 0 & 75         \\[0.55ex]
      0 & 0 & -1 & 1 & 5        \\[0.55ex]
      0 & -1 & 0 & 1 & 0\\[0.55ex]
      0 & 0 & 0 & 3 & 75
\end{array}\right]$$
Now from the last row we get that $3D = 75 \implies D=25$
From the third row we get that $D-B = 0 \implies D=B \implies B=25$
From the second row we get that $D-C = 5 \implies C=25-5 \implies C=20$
And from the first row we get that: $A+B+C = 75 \implies A=30$
A: If you subtract the last from the third, you get $D-C=-1$ or $D=C-1$.  Now you can substitute this into three equations and eliminate $D$.  Similarly subtracting the first two allows you to eliminate $B$  You will then have two equations in two unknowns, and again can solve one equation for one unknown in terms of the other.  This is a standard technique for simultaneous equations, which works very well with this set because of the structure.
A: If you add all systems to each other you will get:
3(A+B+C+D)=300 so A+B+C+D=100
Then use this equation to find out A, B, C and D
For example: B+C+D=70 and A+B+C+D=100 
subtract the systems to each other you'll find A=30.
A: B=D
So use the first equation b+c+d=b+c+b=70
and the second a+c+d=75
and the third a+b+d=80
so b-c=5
(b+c+b+5) divided by 3 equals b and d
(70+5) divided by 3 equals 25(b and d)
25-5=20(c)
(a+c+b)-(b+c+b)=5(a-b)
25+5=30(a)
a=30
b=25
c=20
d=25
