<s>[INST] How to find the sum of constants given the following systems of equations? Given that $q,r,s$, and $t$ are different constant values in the following systems of equations containing $a,b,c$, and $d$. Find the sum $q+r+s+t$.
$\frac{1}{qa+1} + \frac{1}{qb+1} + \frac{1}{qc+1}+ \frac{1}{qd+1} = 1$
$\frac{1}{ra+1} + \frac{1}{rb+1} + \frac{1}{rc+1}+ \frac{1}{rd+1} = 1$
$\frac{1}{sa+1} + \frac{1}{sb+1} + \frac{1}{sc+1}+ \frac{1}{sd+1} = 1$
$\frac{1}{ta+1} + \frac{1}{tb+1} + \frac{1}{tc+1}+ \frac{1}{td+1} = 1$
I tried using a simpler version of the problem but I can't find the pattern.
I am thinking that there might be a pattern that I can get from finding the sum $q+r$ from this simpler version of the problem.
$\frac{1}{qa+1} + \frac{1}{qb+1} = 1$
$\frac{1}{ra+1} + \frac{1}{rb+1} = 1$
I also tried adding the four equations that leads me to factor out the sum $q+r+s+t$ but it becomes more complicated. But I noticed that the sum of the each set of denominators per equation is equal to the product of all those denominators. However, I am stuck at looking for the pattern where I can use that fact.
Thanks in advance for your comments and suggestions on how to solve this particular algebra problem.
 [/INST] Begin by noting that $q,r,s,t$ are the four roots of the equation $$\frac{1}{ax+1}+ \frac{1}{bx+1} + \frac{1}{cx+1} + \frac{1}{dx+1} =1$$
Simplifying, we get $$(bx+1)(cx+1)(dx+1)+(ax+1)(bx+1)(cx+1)+(cx+1)(dx+1)(ax+1)+(dx+1)(ax+1)(bx+1)=(ax+1)(bx+1)(cx+1)(dx+1).$$ Now, we know that for a general cubic equation $ux^4+vx^3+wx^2+m$, the sum of the roots is $-\dfrac vu$. So we do not need to simplify the whole equation, but just find the coefficients of $x^3$ and $x^4$.
Coefficient of $x^4=abcd$Coefficient of $x^3=0$
So $q+r+s+t=0$  .
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