I have four equations in four unknowns, namely, $w_1,w_2,x_1,x_2$.
$$w_1+w_2=2 \tag{1}$$ $$w_1x_1+w_2x_2=0 \tag{2}$$ $$w_1x_1^2+w_2x_2^2=2/3 \tag{3}$$ $$w_1x_1^3+w_2x_2^3=0 \tag{4}$$
How can I solve the system of linear equations?
from Equation $(1)$, $$w_2=2-w_1\ldots(5)$$
from Equation $(2)$, $$w_1x_1+(2-w_1)x_2=0$$ $$\Rightarrow w_1x_1+2x_2-w_1x_2=0$$
On
.I want to rearrange these equations. Call $w_1 = a,w_2=b,x_1=c,x_2=d$. We will get the system of equations: $$ b=2-a, bd=-ac, bd^3=-ac^3, bd^2 = \frac{2}{3} - ac^2 $$ To solve these, first eliminate $b$ from all the equations: $$ (2-a)d=-ac,(2-a)d^3=-ac^3, (2-a)d^2 = \frac{2}{3} - ac^2 $$ Next, note that $\frac{2-a}{-a} = \frac{d}{c} = \frac{d^3}{c^3}$, so $d = \pm c$. But since $a = \frac{2d}{d-c}$ from the first equation, $d$ cannot be equal to $c$, hence $d=-c$ and $a = 1$.
Finally, $b=1$ also. Now, the important equation is: $$ \frac{2}{3} = d^2 + c^2 = 2d^2 \implies d,c = \pm \frac{1}{\sqrt{3}} $$
Hence the solutions are $a=b=1, c = -d = \pm \frac{1}{\sqrt{3}}$.
From your second eqution you get:
$$w_1x_1=-w_2x_2=:c \tag{2}$$
Pluging this into the equation (3) and (4) you get:
\begin{align*} (x_1-x_2)\cdot c&= 2/3 \tag{3}\\ (x_1+x_2)(x_1-x_2)\cdot c=(x_1^2-x_2^2)\cdot c&= 0 \tag{4} \end{align*}
From this we deduce $(x_1+x_2)=0$ and hence $x_1= -x_2$. Insert this again into (2) we get $w_1=w_2$ and from (1) we get $w_1=1=w_2$ and from (3) $x_1= \pm \sqrt{1/3}=-x_2$.