Solving of System of Two Equation with Two Variables. Method of Introducing New Variables

There are two ways to use method if introducing new variables:

  1. introduce one new variable only for one equation of the system;
  2. introduce two new variables for both equations of the system.

Example 1. Solve system `{(x/y+y/x=13/6),(x+y=5):}`.

Let `x/y=z` then `y/x=1/z` and first equation can be rewritten as `z+1/z=13/6`. Equivalent to this equation is equation `6z^2+6=13z` or `6z^2-13z+6=0`. Roots of this equation are `z=2/3,z=3/2`.

So, either `x/y=3/2`, i.e. `y=(3x)/2`; or `x/y=3/2`, i.e. `y=(2x)/3`.

Therefore, first equation of the system is equivalent to the set of two equation.

So, we now need to solve set of systems of equations: `{(y=(3x)/2),(x+y=5):}`, `{(y=(2x)/3),(x+y=5):}`.

These systems can be easily solved using substitution method. Solution of the first system is x=2, y=3. Solution of the second system is x=3, y=2.

Therefore, (2;3) and (3;2) are solutions of initial system.

Example 2. Solve system of equations `{(x^2+y^2+x+y=32),(xy+2(x+y)=26):}`.

Let `x+y=u,xy=v`. Then `x^2+y^2=(x+y)^2-2xy=u^2-2v`.

Now, system can be rewritten as `{(u^2-2v+u=32),(v+2u=26):}`.

From second equation `v=26-2u`. Plugging this value into first equation gives `u^2-2(26-2u)+u=32`. This equation can be rewritten as `u^2+5u-84=0`. It has two solutions: `u=-12,u=7`.

Now, use expression `v=26-2u`.

If u=-12 then `v=26-2*(-12)=50`.

If u=7 then `v=26-2*7=12`.

Now let's return to old variables: we obtain set of systems `{(x+y=-12),(xy=50):},{(x+y=7),(xy=12):}`.

These systems can be easily solved with substitution method (for example, expressing from first equation y in terms of x). First system doesn't have solutions. Second system has two solutions: (3;4) and (4;3). They are also solutions of the initial system.