text: Consider a simple branched pathway with all three steps irreversible. Such a pathway will admit two elementary modes which are indicated in thicked (or red) reaction lines.
  Because both  and  are irreversible, and elementary mode lying on both these reactions is not possible since it would mean one reactions going against its thermodynamic direction. Each mode in this system satisfies the three conditions described above. The first condition is steady state, that is for each mode , it has to be true that .
  Algebraically the two modes are given by:
  By substituting each of these vectors into , it is easy to show that condition one is satisfied. For condition two we must ensure that all reactions that are irreversible have positive entries in the corresponding elements of the elementary modes. Since all three reactions in the branch are irreversible and all entries in the elementary modes are positive, condition two is satisfied.
  Finally, to satisfy condition three, we must ask whether we can decompose the two elementary modes into other paths that can sustain a steady state while using the same non-zero entries in the elementary mode. In this example, it is impossible to decompose the elementary modes any further without disrupting the ability to sustain a steady state. Therefore, with all three conditions satisfied, we can conclude that the two vectors shown above are elementary modes.
  All possible flows through a network can be constructed from linear combinations of the elementary modes, that is:
  such that the entire space of flows through a network can be described.  must be greater than or equal to zero to ensure that irreversible steps aren't inadvertently made to go in the reverse direction. For example, the following is a possible steady-state flow in the branched pathway.
  If one of the outflow steps in the simple branched pathway is made reversible, an additional elementary mode becomes available, representing the flow between the two outflow branches. An additional mode emerges because, with only the first two modes, it is impossible to represent a flow between the two branches because the scaling factor, , cannot be negative (which would be required to reverse the flow).