# 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.