System of equations
A system of equations is a set of equations which share the same variables. Below is an example of a system of equations.
![\[\left\{\begin{array}{l}a^2 + b^2 = 25\\a+b=7\end{array}\right.\]](http://latex.artofproblemsolving.com/2/e/f/2ef87799b3c4b6c2c9fbba2e86aae36c95bf74e8.png)
Solving Linear Systems
A system of linear equations is where all of the variables are to the power 1. There are three elementary ways to solve a system of linear equations.
Gaussian Elimination
Gaussian elimination involves eliminating variables from the system by adding constant multiples of two or more of the equations together. Let's look at an example:
Problem
Find the ordered pair 
for which
![\[\left\{\begin{array}{l}x-12y=2\\3x+6y=6\end{array}\right.\]](http://latex.artofproblemsolving.com/b/8/9/b89cc5edaba92c28acdd060844345c54e35cf521.png)
Solution
We can eliminate 
by adding twice the second equation to the first:
| |
 |
|
|
Thus 
. We can then plug in for 
in either of the equations:
Thus, the solution to the system is 
.
Substitution
The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We'll show how to solve the same problem from the elimination section using substitution.
Problem
Find the ordered pair for which
Solution
The first equation can be solved for :
Plugging this into the second equation yields
Thus 
. Plugging this into either of the equations and solving for yields .
Graphing
The third method for solving a system of linear equations is to graph them in the plane and observe where they intersect. We'll go back to our same example to illustrate this.
Problem
Find the ordered pair for which
Solution
We graph the two lines as follows:
From the graph, we can see that the solution to the system is .
Advanced Methods
Matrices can also be used to solve systems of linear equations. In fact, they provide a way to make much broader statements about systems of linear equations.
There is a whole field of mathematics devoted to the study of linear equations called linear algebra.
Convenient Systems
Some systems can be solved by taking advantage of specific forms. Such systems can often seem tough to solve at first, however.
Symmetry
Consider the below system.
![\[\left\{\begin{array}{l}a+b+c+d=4\\a+b+c+e=8\\a+b+d+e=12\\a+c+d+e=16\\b+c+d+e=20\end{array}\right.\]](http://latex.artofproblemsolving.com/d/9/2/d9289246bf80332072af6b562fc22398e1fe1c02.png)
The key here is to take advantage of the symmetry. If we add up all 5 equations we will have a total of 4 of each variable on the LHS. On the RHS we will have 
. Thus
So then subtracting the first equation from this leaves 
on the LHS and 
on the RHS. Subtracting this equation from the second equation leaves 
on the LHS and 
on the RHS. And thus we continue on in this way to find that 
Clever Substitution
Consider the below system.
![\[\left\{\begin{array}{l}x+y+3xy = 5\\ 2x+2y+4xy=8 \end{array}\right.\]](http://latex.artofproblemsolving.com/4/f/b/4fb9cde2584c012424b80fa23a9d9fe243b1f696.png)
We can let 
and 
to get the two-variable linear system below.
![\[\left\{\begin{array}{l}a+3b = 5\\ 2a+4b=8 \end{array}\right.\]](http://latex.artofproblemsolving.com/8/6/5/8655c1ab7085e6456481abc1ca8c169bbdb9d740.png)
Solving the system results in 
and 
. Substituting that back results in 
and 
. We can do another substitution by letting 
and substituting to get 
. Rearranging results in 
, so 
. Finally, by substituting back in, we get 
. Plugging 
back satisfies the system.
