Predicting the evolution of a linear model depending on a parameter I have sets of data depending on a experimental parameter (wt%, in the legend) that can be fitted linearly.
I would like to get the line corresponding to another value of that parameter, for example 10.

I thought that maybe I could make a linear fit for both the slope and the intercept as
$$slope = a*wt + b$$
$$intercept = c*wt + d$$
but maybe there is an actual statistical method to do so.
EDIT:
I understood the sense of @Tim answer: making just one regression. But I think there are a few mistakes though, so probably the correct answer is:
$$\begin{align}
y &= intercept + slope \times x + \varepsilon \\
&= a \times wt + b + c \times wt \times x + d \times wt +\varepsilon \\
&= \beta_0 + \beta_1 \times wt + \beta_2 \times x +  \beta_3 \times wt \times x + \varepsilon 
\end{align}$$
with this additional regressor I get a great fit (R-sq = 0.995).


What you are describing is just a regular regression model with an interaction term
$$\begin{align}
y &= intercept + slope \times x + \varepsilon \\
&= (b+d) + a \times wt + c \times wt \times x + \varepsilon \\
&= \beta_0 + \beta_1 \times wt + \beta_2 \times wt \times x + \varepsilon 
\end{align}$$
You can't have $b$ and $d$ as two intercepts, they wouldn't be identifiable because are redundant.
As you noticed in the edit, you may add the $\beta_2 x$ term to the model. It might improve performance or interpretability.     The term would tell you how much influence $x$ has on $y$ regardless of the experimental condition $wt$. There may be scenarios though where the additional parameter may not be necessary or desirable.
