I want to show that the following system of equations does not have a solution, but I do not know how to do this
$$w_1+w_2=\frac{1}{2}$$ $$w_1s_1+w_2s_2=\frac{1}{6}$$ $$w_1t_1+w_2t_2=\frac{1}{6}$$ $$w_1s_1t_1+w_2s_2t_2=\frac{1}{24}$$ $$w_1s_1^2+w_2s_2^2=\frac{1}{12}$$ $$w_1t_1^2+w_2t_2^2=\frac{1}{12}$$ that all $t1,t2,w1,w2,s1,s2$ are unknown.
please help me, Thanks.
First note that we cannot have $w_1 = 0$ or $w_2 = 0$.
Subtracting the second and third give us
$$w_1(s_1 - t_1) = -w_2(s_2 - t_2)$$
Now if $s_1 = t_1$ then $s_2 = t_2$, and we can verify there is no solution, by comparing the fourth and fifth.
Subtracting the fifth and sixth gives us
$$w_1(s_1^2 - t_1^2) = -w_2(s_2^2 - t_2^2)$$
and thus from the above, we must have that
$$s_1 + t_1 = s_2 + t_2 = x \;(\text{say})$$
Adding second and third gives us
$$(w_1 + w_2) x = \frac{1}{3}$$
and so $$x = \frac{2}{3}$$ using the first equation.
Adding the fifth, sixth and two times the fourth gives us
$$w_1 x^2 + w_2 x^2 = \frac{1}{4}$$
and so $$x = \frac{\pm 1}{\sqrt{2}}$$
a contradiction.