2006-07-01 04:48:24 · 11 answers · asked by shermala143 2 in Science & Mathematics ➔ Mathematics
Yes. One possibility is that two or more of the equations are really just the same equation in a different form. In this case, the system may still be solvable.
If you have more equations than variables and all the equations are independent, then there will not be a solution to the system.
A *consistent* system, can only have as many *independent* equations as there are variables. If that's what you meant, then the answer to your question is no.
2006-07-01 04:51:00 · answer #1 · answered by mathsmart 4 · 1⤊ 0⤋
Yes. There are the following possibilities:
- The system may be consistent, i.e. there is a set of values of the variables that satisfies all the equations. In this case, some of the equations are redundant and can be discarded without loss.
- The system may be inconsistent: there does not exist a set of values of the variables that can satisfy all the equations. In this case, one may be able to eliminate some equations and thus obtain a consistent set, but there is typically no way to determine, a priori, which equations to eliminate. Consider:
x+y=3
x+2y=5
x+3y=6
Any two of these equations yield solutions for x and y, but there is no solution that can satisfy all three of them.
-- Robert A. Saunders, Lake Stevens, WA.
2006-07-01 05:01:09 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Yes, they may contain more number of equations. But certainly some of them are mutually dependent. The number of independent equations to find the values of variables is exactly the number of variables. If there are less number of independent equations, we find values of only some variables and relationship between the other variables. This is based on the equations.
2006-07-01 05:30:26 · answer #3 · answered by K N Swamy 3 · 0⤊ 0⤋
Yes, it is still a system of equations.
If they are all linear equations, then it is possible that there will not be any one solution that satisfies all the equations at the same time.
Imagine that each equation describes a plane in space. If 3 planes come together, there could be a single point intersection. If you add a 4th plane, it might not pass through that point.
This means the system is non-solvable. But, it is still a system.
2006-07-01 04:52:58 · answer #4 · answered by tbolling2 4 · 0⤊ 0⤋
Yes, but this is called an " inconsistant system" and there are no unique solutions. If there are m equations and n variables,m>n, you may obtain n solutions for the variables, and the remaining m-n solutions will be functions in terms of the first n solutions/variables. I know this sounds confusing but I hope this explains it.
2006-07-01 05:34:31 · answer #5 · answered by don1n8 4 · 0⤊ 0⤋
Technically, an equation doesn't have to contain any variables (a numerical equation). Therefore, yes.
2006-07-01 04:54:44 · answer #6 · answered by Merovign 2 · 0⤊ 0⤋
Yes, but it will either be an inconsistent system (one without a solution), or some of the equations will be redundant (i.e., combinations of other equations in the system).
Example of the first:
x + y = 3
x + 2y = 5
2x + 3y = 11
Example of the second:
x + y = 3
x + 2y = 5
3x + 4y = 11
Hope that helps!
2006-07-01 04:56:55 · answer #7 · answered by Jay H 5 · 0⤊ 0⤋
Yes a sytem of equations can have ore quations than variables it is just that some may not be solveable but they are still a system of equations.
2006-07-01 05:02:02 · answer #8 · answered by JohnDoe 1 · 0⤊ 0⤋
absolutely
2006-07-01 04:53:38 · answer #9 · answered by bryan4pres 2 · 0⤊ 0⤋
yes.
2006-07-01 04:51:20 · answer #10 · answered by Halo 5 · 0⤊ 0⤋