When you start with three equations and three unknowns (variables), you may think you have enough information to solve for all the variables. However, when solving a system of linear equations using the elimination method, you may find that the system is not sufficiently determined to find one unique answer, and instead an infinite number of solutions is possible. This occurs when the information in one of the equations in the system is redundant to information contained in the other equations.
A 2x2 Example
3x+2y=5 6x+4y=10 This system of equations is clearly redundant. You can create one equation from the other by just multiplying through by a constant. In other words, they convey the same information. Despite there being two equations for the two unknowns, x and y, the solution of this system can’t be narrowed down to one value for x and one value for y. (x,y)=(1,1) and (5/3,0) both solve it, as do many more solutions. This is the sort of “problem,” this insufficiency of information, that leads to an infinite number of solutions in larger systems of equations as well.
A 3x3 Example
x+y+z=10 x-y+z=0 x_+_z=5 [Underscores are used merely to maintain spacing.] By the elimination method, eliminate x from the second row by subtracting the second row from the first, giving x+y+z=10 _2y=10 x_+z=5 Eliminate x from the third row by subtracting the third row from the first. x+y+z=10 _2y=10 y=5 Clearly the last two equations are equivalent. y equals 5, and the first equation can be simplified by eliminating y. x+5+z=10 y__=5 or x+z=5 y=5 Note that the elimination method won’t produce a nice triangular shape here, as it does when there is one unique solution. Instead, the last equation (if not more) will itself be absorbed into the other equations. The system is now of three unknowns and only two equations. The system is called “underdetermined,” because there are not enough equations to determine the value of all the variables. An infinite number of solutions are possible.
How to Write the Infinite Solution
The infinite solution for the above system can be written in terms of one variable. One way of writing it is (x,y,z)=(x,5,5-x). Since x can take on an infinite number of values, the solution can take on an infinite number of values.
About the Author
Paul Dohrman's academic background is in physics and economics. He has professional experience as an educator, mortgage consultant, and casualty actuary. His interests include development economics, technology-based charities, and angel investing.