System of equations

Revision as of 18:36, 14 March 2019 by Shootingstar8 (talk | contribs) (Solution)

A system of equations is a set of equations which share the same variables. An example of a system of equations is

$2a - 3b$ $= 4$
$3a - 2b$ $= 3$

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 $(x,y)$ for which

$x - 12y$ $= 2$
$3x + 6y$ $= 6$

Solution

We can eliminate $y$ by adding twice the second equation to the first:

$x - 12y= 2$
$+2($ $3x + 6y = 6)$
$\overline{7x + 0=14}$

Thus $x=2$. We can then plug in for $x$ in either of the equations:

$(2)-12y = 2 \Rightarrow y = 0$.

Thus, the solution to the system is $(2,0)$.

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 $(x,y)$ for which

$x - 12y$ $= 2$
$3x + 6y$ $= 6$

Solution

The first equation can be solved for $x$:

$x = 12y + 2.$

Plugging this into the second equation yields

$3(12y + 2) + 6y = 6 \Leftrightarrow 42 y = 0.$

Thus $y=0$. Plugging this into either of the equations and solving for $x$ yields $x=2$.

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 $(x,y)$ for which

$x - 12y$ $= 2$
$3x + 6y$ $= 6$

Solution

We graph the two lines as follows:

Twolineintersect.PNG

From the graph, we can see that the solution to the system is $(2,0)$.

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:

$a+$ $b+$ $c+$ $d$ $=4$
$a+$ $b+$ $c+$ $e$ $=8$
$a+$ $b+$ $d+$ $e$ $=12$
$a+$ $c+$ $d+$ $e$ $=16$
$b+$ $c+$ $d+$ $e$ $=20$

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 $4+8+12+16+20 = 60$. Thus

$4(a+b+c+d+e) = 60 \Leftrightarrow a + b + c + d + e = 15.$

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

See also

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