Don't understand x''=x^2-y-xy,\ \ \ y''=y-x as a system of first order

1. Let, ${\displaystyle x_1=x,x_2=x^{\prime }}$ then the ${\displaystyle x{}^{″}=x^2-y-xy}$ equation converted to ${\displaystyle x_2^{\prime }=x_1^2-y-x_{1}y}$ where ${\displaystyle x_1^{\prime }=x_2}$.
Let,
$y_1=y,y_2=y^{\prime }$ then the equation $y^{″}=y-x$ converted to $y_2^{\prime }=y_1-x$  where ${\displaystyle y_2=y_1^{\prime }}$.
2. ${\displaystyle x{}^{‴}=tx}$ is already a non-autonomous system. We need to convert it to first order equation.
Let, ${\displaystyle x_1=x,x_2=x^{\prime },x_3=x{}^{″}}$, the above system transformed to ${\displaystyle x_3^{\prime }=t{x}_1}$ which is a non-autonomous system of first order differential equation.
3. ${\displaystyle x{}^{⁗}=tx}$ is not a autonomous system of linear equation. First transfer it to autonomous.
Put, ${\displaystyle x=tv}$.
The equation reduces to ${\displaystyle \frac{3d^{2}v}{{dt}^2}+\frac{x}{v{d}^3}\frac{v}{{dt}^3}=\frac{x^2}{v}wheret=\frac{x}{v}}$.
Now let, ${\displaystyle v=v_1,v^{\prime }=v_2,v{}^{″}=v_3}$ then the above equation reduces to ${\displaystyle 3v_3+\frac{x}{v_1},v_3^{\prime }=\frac{x^2}{v_1}}$ which is an autonomous system of first order differential equation.