Editing System of linear equations (section)

===Describing the solution===
When the solution set is finite, it is reduced to a single element. In this case, the unique solution is described by a sequence of equations whose left-hand sides are the names of the unknowns and right-hand sides are the corresponding values, for example <math>(x=3, \;y=-2,\; z=6)</math>. When an order on the unknowns has been fixed, for example the [[alphabetical order]] the solution may be described as a [[vector space|vector]] of values, like <math>(3, \,-2,\, 6)</math> for the previous example.

To describe a set with an infinite number of solutions, typically some of the variables are designated as '''free''' (or '''independent''', or as '''parameters'''), meaning that they are allowed to take any value, while the remaining variables are '''dependent''' on the values of the free variables.

For example, consider the following system:
:<math>\begin{alignat}{7}
 x &&\; + \;&& 3y &&\; - \;&& 2z &&\; = \;&& 5 & \\
3x &&\; + \;&& 5y &&\; + \;&& 6z &&\; = \;&& 7 &
\end{alignat}</math>
The solution set to this system can be described by the following equations:
:<math>x=-7z-1\;\;\;\;\text{and}\;\;\;\;y=3z+2\text{.}</math>
Here ''z'' is the free variable, while ''x'' and ''y'' are dependent on ''z''. Any point in the solution set can be obtained by first choosing a value for ''z'', and then computing the corresponding values for ''x'' and ''y''.

Each free variable gives the solution space one [[degree of freedom]], the number of which is equal to the [[dimension]] of the solution set. For example, the solution set for the above equation is a line, since a point in the solution set can be chosen by specifying the value of the parameter ''z''. An infinite solution of higher order may describe a plane, or higher-dimensional set.

Different choices for the free variables may lead to different descriptions of the same solution set. For example, the solution to the above equations can alternatively be described as follows:
:<math>y=-\frac{3}{7}x + \frac{11}{7}\;\;\;\;\text{and}\;\;\;\;z=-\frac{1}{7}x-\frac{1}{7}\text{.}</math>
Here ''x'' is the free variable, and ''y'' and ''z'' are dependent.