Take a look at the following system
$$ \begin{align} \dot{x}_1 &= x_2\\ \dot{x}_2 &= -x_1 + x^3_1 - x_2 \end{align} $$ which has three equilibrium points (0,0),(1,0), and (-1,0). In the book I'm reading, the author asks to define a new system that has equilibrium point as the origin for (1,0) and (-1,0). The procedure in the book is as follows: let $y=x-x_e$ where $x_e$ is an equilibrium point and let's perform the transformation for $x_e = (1,0)$, hence: $$ \begin{align} y_1 &= x_1 - (1) = x_1 - 1 \implies \dot{y}_1 = \dot{x}_1 \\ y_2 &= x_2 - (0) = x_2 \implies \dot{y}_2 = \dot{x}_2 \end{align} $$ The new system is now $$ \begin{align} \dot{y}_1 &= y_2 \\ \dot{y}_2 &= -(y_1+1) + (y_1+1)^3 - y_2 \end{align} $$ Now we need to show that the new system $\dot{y}=g(y)$ has an equilibrium point at the origin (i.e. $\dot{y}=g(y) \implies 0 = g(y_e)$), we get
$$ \begin{align} 0 &= y_2 \\ 0 &= -(y_1+1) + (y_1+1)^3 + y_2 \end{align} $$ $y_2=0, -(y_1+1) + (y_1+1)^3 = 0 \implies y_1=0,-2$. The new system has two equilibrium points and one of them are not at the origin. How to justify this issue. The actual question in the book states:
It is ok. When you transform $x\to y$ your old equilibrium $(1,0)$ becomes $(0,0)$ for the $y$-system. You only have to analyze that point for the new system. You cannot avoid by translation having several equilibria.