View source for System of equations

A '''sytstem of equations''' is a set of [[equation]]s which share the same [[variable]]s.  An example of a system of equations is

{| class="wikitable" style="margin: 1em auto 1em auto"
| <math>2a - 3b</math> || <math>= 4</math>
|-
| <math>3a - 2b</math> || <math>= 3</math>
|}

== 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 ===
[[Gauss]]ian 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 <math>(x,y)</math> for which

{| class="wikitable" style="margin: 1em auto 1em auto"
| <math>x - 12y</math> || <math>= 2</math>
|-
| <math>3x + 6y</math> || <math>= 6</math>
|}

==== Solution ====
We can eliminate <math>y</math> by adding twice the second equation to the first:

{| class="wikitable" style="margin: 1em auto 1em auto"
| || <math>x - 12y= 2</math>
|-
| <math>+2(</math> || <math>3x + 6y = 6)</math>
|-
| || <math>\overline{7x + 0=14}</math>
|}

Thus <math>x=2</math>.  We can then plug in for <math>x</math> in either of the equations:  <center><math> (2)-12y = 2 \Rightarrow y = 0</math>.</center>

Thus, the solution to the system is <math>(2,0)</math>.

=== 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 <math>(x,y)</math> for which

{| class="wikitable" style="margin: 1em auto 1em auto"
| <math>x - 12y</math> || <math>= 2</math>
|-
| <math>3x + 6y</math> || <math>= 6</math>
|}

==== Solution ====
The first equation can be solved for <math>x</math>:  <center><math> x = 12y + 2. </math></center>

Plugging this into the second equation yields <center><math>3(12y + 2) + 6y = 6 \Leftrightarrow 42 y = 0. </math></center>

Thus <math>y=0</math>.  Plugging this into either of the equations and solving for <math>x</math> yields <math>x=2</math>.

=== 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 <math>(x,y)</math> for which

{| class="wikitable" style="margin: 1em auto 1em auto"
| <math>x - 12y</math> || <math>= 2</math>
|-
| <math>3x + 6y</math> || <math>= 6</math>
|}

==== Solution ====
We graph the two lines as follows: <center>[[Image:Twolineintersect.PNG]]</center>

From the graph, we can see that the solution to the system is <math>(2,0)</math>.

=== Advanced Methods ===
[[Matrix | 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 ==
Convenient systems usually seem very tough to solve at first.  Often times a clever insight will make things much easier.

=== Problem ===
Solve the following system:

{| class="wikitable" style="margin: 1em auto 1em auto"
| <math>a+</math> || <math>b+</math> || <math>c+</math> || <math>d</math> || || <math>=4</math>
|-
| <math>a+</math> || <math>b+</math> || <math>c+</math> || || <math>e</math> || <math>=8</math>
|-
| <math>a+</math> || <math>b+</math> || || <math>d+</math> || <math>e</math> || <math>=12</math>
|- 
| <math>a+</math> || || <math>c+</math> || <math>d+</math> || <math>e</math> || <math>=16</math>
|-
| || <math>b+</math> || <math>c+</math> || <math>d+</math> || <math>e</math> || <math>=20</math>
|}

=== Solution ===
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 <math>4+8+12+16+20 = 60</math>.  Thus <center><math> 4(a+b+c+d+e) = 60 \Leftrightarrow a + b + c + d + e = 15. </math></center>

So then subtracting the first equation from this leaves <math>e</math> on the LHS and <math>15-4=11</math> on the RHS.  Subtracting this equation from the second equation leaves <math>d</math> on the LHS and <math>15-8=7</math> on the RHS.  And thus we continue on in this way to find that <math>(a,b,c,d,e)=(11,7,3,-1,-5).</math>

== See also ==
* [[Algebra]]