Deducing the value of a variable resistor in configurable voltage divider
Setup system of equations

\$ \begin{cases}  
 { V  }_{ in  }   \dfrac{  { R  }_{ c  }    }{  { R  }_{ c  }  + { R  }_{ s  } ||  { R  }_{ e  }    }   =   { V  }_{ a  }      
\\ \\
{ V  }_{ in  }   \dfrac{  { R  }_{ s  } || { R  }_{ c  }  }{  { R  }_{ s  } || { R  }_{ c  }  + { R  }_{ e  }    }   =   { V  }_{ b  }    
\end{cases} 
\$

Solve for Rs

\$  { R  }_{ s  }   =   { R  }_{ c  }   \dfrac{  \left(  { V  }_{ in  }  - { V  }_{ a  }    \right)    }{  { V  }_{ a  }  - { V  }_{ b  }    } \$

Also can be solved for Re

\$  { R  }_{ e  }   =   { R  }_{ c  }   \dfrac{  \left(  { V  }_{ in  }  - { V  }_{ a  }    \right)    }{  { V  }_{ b  }    } \$

It can also be solved for any two of the three circuits.

\$ ({ V  }_{ a  },   { V  }_{ b  })  : { R  }_{ s  }   =   \dfrac{  { R  }_{ c  }   \left(  { V  }_{ in  }  - { V  }_{ a  }    \right)    }{  { V  }_{ a  }  - { V  }_{ b  }    },   { R  }_{ e  }   =   \dfrac{  { R  }_{ c  }   \left(  { V  }_{ in  }  - { V  }_{ a  }    \right)    }{  { V  }_{ b  }    }   \$

\$ ({ V  }_{ a  },   { V  }_{ c  }) :  { R  }_{ s  }   =   \dfrac{  { R  }_{ c  }   \left(  { V  }_{ in  }  - { V  }_{ a  }    \right)    }{  { V  }_{ c  }    }   ,{ R  }_{ e  }   =   \dfrac{  { R  }_{ c  }   \left(  { V  }_{ in  }  - { V  }_{ a  }    \right)    }{  { V  }_{ a  }  - { V  }_{ c  }    }   \$

\$ ({ V  }_{ b  },   { V  }_{ c  }) :  { R  }_{ s  }   =   \dfrac{ - { R  }_{ c  }   \left(  { V  }_{ b  }  + { V  }_{ c  }  - { V  }_{ in  }    \right)    }{  { V  }_{ c  }    } ,  { R  }_{ e  }   =   \dfrac{ - { R  }_{ c  }   \left(  { V  }_{ b  }  + { V  }_{ c  }  - { V  }_{ in  }    \right)    }{  { V  }_{ b  }    }  \$