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.

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

2 Answers 2


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.


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