Vrftsjtryk

I have to solve this apparently simple system of linear equations:

$$ a t^5+b t^4+c t^3+d t^2+e t=0 \
a (t/2)^5+b (t/2)^4+c (t/2)^3+d (t/2)^2+e (t/2)=0 \
a/6 t^6+b/5 t^5+c/4 t^4+d/3 t^3+e t^2/2=0 \
a/42 t^7+b/30 t^6+c/20 t^5+d/12 t^4+e/6 t^3=v \
a/336 t^8+b/210 t^7+c/120 t^6+d/60 t^5+e/24 t^4=y $$

in the unknowns $a,b,c,d,e$ where I assume $t=0.8$, $y=0.1$ and $v=0.45$.
Even if the coefficient matrix is only 5x5, the system is ill-conditioned and I cannot easily find a solution in MATLAB with the methods of matrix inversion $A^{-1} b$, linsolve and lsqr.

Do you have any suggestion to solve this problem?
Thanks,

Regards,

E.

To make the problem simpler and assuming $t neq 0$, let us reawrite the equations as
$$a t^4+b t^3+c t^2+d t+e=0tag 1$$
$$frac{a t^4}{32}+frac{b t^3}{16}+frac{c t^2}{8}+frac{d t}{4}+frac{e}{2}=0tag 2$$
$$frac{a t^4}{6}+frac{b t^3}{5}+frac{c t^2}{4}+frac{d t}{3}+frac{e}{2}=0 tag3$$
$$frac{a t^4}{42}+frac{b t^3}{30}+frac{c t^2}{20}+frac{d t}{12}+frac{e}{6}=frac v {t^3} tag4$$
$$frac{a t^4}{336}+frac{b t^3}{210}+frac{c t^2}{120}+frac{d t}{60}+frac{e}{24}=frac y {t^4} tag5$$

Now, solve equations $(2)$, $(3)$, $(4)$, $(5)$ for $b,c,d,e$. This gives
This gives as solutions
$$b=-frac{5 left(a t^8+1120 t v-2240 yright)}{2 t^7}qquad c=frac{15 left(a t^8+2016 t v-3920 yright)}{7 t^6}qquad d=-frac{5 left(a t^8+2604 t v-4704 yright)}{7 t^5}qquad e=frac{a t^8+2800 t v-3920 y}{14 t^4}$$

Plug these values in $(1)$ to get
$$a t^4+b t^3+c t^2+d t+e=-frac{140 (t v-2 y)}{t^3}$$ which can be $0$ only if $tv=2y$ what ever could be the value assigned to $a$ !