If you have a system fo linear equations, the solution is where the equations intersect Using row reduction you get a system of linear equations which still satisfies the intersection Why are free variables used? The values in which the free variable can be are limited in a range as for if the solution is a line and not a plane So what is the point in writing the solutions as a set of vector additions using free variables? If z in this case is a free variable, writing the solution in a vector equation Where , denotes a new row Solving for the pivot variables x and y and writing into decomposed vector form [x,y] = [4,0] - [0,3] + z[1,2] Z is said to be any real number zeR But looking at the graph, when z = 4, y=11/8 is not on the line of intersection So why do they say that z is a free variable when it isnt?

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If you have a system fo linear equations, the solution is where the equations intersect  Using row reduction you get a system of linear equations which still satisfies the intersection  Why are free variables used?  The values in which the free variable can be are limited in a range as for if the solution is a line and not a plane  So what is the point in writing the solutions as a set of vector additions using free variables?  If z in this case is a free variable, writing the solution in a vector equation  Where , denotes a new row  Solving for the pivot variables x and y and writing into decomposed vector form  [x,y] = [4,0] - [0,3] + z[1,2]  Z is said to be any real number zeR But looking at the graph, when z = 4, y=11/8 is not on the line of intersection So why do they say that z is a free variable when it isnt?

Derrick Woods · Accepted Answer

The idea of free variables is that every free variable can have any value whatsoever, so they are really FREE. Once the values of the free variables have been chosen, there is no more freedom at all. The values of the remaining variables are completely determined by those of the free variables. So the free variables are totally free and the non-free variables are totally imprisoned.

If you have a system fo linear equations, the solution is where the equations intersect Using ro

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