No, you don't need two equations. You can model that in one equation with z being a dummy variable. From the visual, both groups seem to follow a linear relationship with similar slope but varying intercepts. Or else you might conclude, that both are linear but differ slightly in slope and massively in intercept.
So a bad first approach to model them in one equation as one linear system would look like
$y \leftarrow \beta_0 + \beta_1\times x$
(with "$\leftarrow$" representing some kind of equation with errors left out for simplicity)
Now, far better would be to allow for a different intercept depending on $z$, which is in the following equation
$y \leftarrow \beta_0 + \beta_1\times z +\beta_2\times x$
If $z$ is zero, the intercept will be $\beta_0$, if $z$ is one, it will be $\beta_0+\beta_1$, a different value. If you also want the slope to be different, it's
$y \leftarrow \beta_0 + \beta_1\times z +\beta_2\times x+\beta_3\times x\times z$
You can easily see, that the slope is either $\beta_2$ or $\beta_2+\beta_3$.
So it depends on your assessment of the situation (background knowledge and plot), but through the power of dummy variables, both groups fit into one equation. With all of this being in one equation, you can then fit one modell to all the data and draw conclusions from which terms are significant or not significant.
R you would modell that as
lm(y ~ x + z) # different intercept
lm(y ~ x + z + x*z) #different intercept and different slope