Difference between revisions of "System of equations"
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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> | 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)=( | + | 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)=(-5,-1,3,7,11).</math> |
== See also == | == See also == | ||
* [[Algebra]] | * [[Algebra]] |
Revision as of 17:36, 14 March 2019
A system of equations is a set of equations which share the same variables. An example of a system of equations is
Contents
[hide]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
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
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:
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 . 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