# 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.

• @dxiv Good catch, missed that one. I have deleted my comments. Jul 17, 2022 at 4:12

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 $$q+r+s+t=0$$ .

Hint: $$\,$$ the equations are of the form: $$\frac{1}{x+a} + \frac{1}{x+b} + \frac{1}{x+c}+ \frac{1}{x+d} = \frac{1}{x} \quad\quad\text{for}\;\; x = \frac{1}{q}, \frac{1}{r}, \frac{1}{s}, \frac{1}{t} \tag{1}$$

Let $$P(x) = (x+a)(x+b)(x+c)(x+d)$$ then $$(1)$$ can be written as:

$$P(x) - x P'(x) = 0 \quad\quad\text{for}\;\; x = \frac{1}{q}, \frac{1}{r}, \frac{1}{s}, \frac{1}{t} \tag{2}$$

The LHS of $$(2)$$ is a polynomial of degree $$4$$ which is zero for the distinct values $$x = \frac{1}{q}, \frac{1}{r}, \frac{1}{s}, \frac{1}{t}$$, so those are its $$4$$ roots, and the rest follows from Vieta's relations.