If $a, b$ are the roots of the equation $x^2-10cx-11d=0$ and $c,d$ are the roots of the equation $x^2-10ax-11b=0$ (where $a\ne b\ne c\ne d\ne 0$), then find the value of $a+b+c+d$.
I have the following data:
- $a+b=10c$
$c+d=10a$
$ab=-11d$
$cd=-11b$
$\frac{c-a}{b-d}=\frac1 {11}$
Only $c\neq a$ is needed, as shown below.
Let $p=10$ and $q=11$. Then we have
$$ x^2-pcx-qd=(x-a)(x-b) \\ x^2-pax-qb=(x-c)(x-d) \tag{1} $$
so that
$$ a+b=pc, ab=-qd, c+d=pa, cd=-q \tag{2} $$
We deduce that
$$ b=pc-a, d=pa-c \tag{3} $$
so that (2) gives $a(pc-a)=-q(pa-c)$ and $c(pa-c)=-q(pc-a)$, and hence
$$ a^2=(pq)a-(q)c+(p)ac,\\ c^2=(pq)c-(q)a+(p)ac \tag{4} $$
Substracting, we see that $c^2-a^2=(pq)(c-a)+q(c-a)$ and hence $a+c=pq+q=q(p+1)$. Finally we have
$$ a+b+c+d=a+c+(pc-a)+(pa-c)=p(a+c)=p(p+1)q $$
In your example, one obtains $a+b+c+d=10\times 11 \times 11=1210$.