Solving Systems

# Systems

A system of equations is a list of equations with more than one variable that work together to show specific data. Systems use equations to portray data, and what the equations have in common is what the equations are ultimately showing. Typically, one equation in a system cannot be solved. To solve for all the variables in a system, you have to connect them to find a common point. One example of a system is this.

Say you have a rectangular field. The field is 20 meters around, and the length of the field is 3 meters longer than the width. What are the dimensions?
With this, we can set up to equations. Our perimeter, 20=2l+2w, and our length-width relation, l=w+3. This is a system of equations, and with them, you can solve for both variables. Now, we will discuss the different types of systems.

## Linear

A linear system is a system of equations with all variables with a degree of one or lower. There are no terms with a degree greater than one. Linear systems have one intersection point, and thus one solution (unless lines are parallel). Here are a few examples of linear systems:

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

y=2x-3
x+y=2

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

Each of these systems can be solved for x and y. Each equation represents a line on a coordinate plane, and where the lines intersect is the solution. When pertaining to systems of more than two dimensions, it will refer to where the planes or spaces intersect (a little advanced…).

## Nonlinear

A nonlinear system pertains to systems with terms of more than one. In these systems, you might find variables like x2, y3, or xy. The graphs for these are not lines, but they are curves rather. In these systems, you might encounter more than one solution. Here are a few examples of nonlinear systems.

10=2l+2w
40=lw

x2-y=17
3x+2y2=8

(2x+y)(-x+2y)=0
(3x2+6xy-4y2)2=143

These equations can get really complex. You can get no solution quite often in nonlinear systems. These systems can be a pain, and they are hard to solve. Again, this are equations, a lot like functions. When you end up with more than one variable for each equation, then things start becoming very abstract and unpredictable.

## Three or more Dimensions

These types of systems have more than two variables and more than two equations. Because of this, these equations cannot be graphed on a two dimensional plane. Rather, they exist in three or more dimensions. To obtain a solution, you must have at least the same number of equations as there are variables. Not only are these equations complex, they are very time consuming. Here are some examples of such systems:

2x+3y-4z=31
x-y+2z=12
-2x+6y-2z=35

a+b=c-d
2a+b=3c+2d
7+a-b=c+d
-a+2c=b-2d+2

(a-b)(2c+d)=1
a2-2abc+3d=-4
(2a+2)(-2b-3)(4c-5)(7d+2)=0
2a2+ab2c2-3d3=25

These types of systems are very abstract and theoretical. Solving them can be ridiculous and tiresome as well, and these types of systems will not be thoroughly discussed in this article.

# Solving a System

The solution of a system is the values of the variables that are true for all the equations in the system. The solution is always written as a coordinate. To solve a system, you must work simultaneously with each equation. Here are some popular methods on solving systems of equations.

## Substitution

This is probably one of the most useful ways of solving a system. The reason is that no matter what, it will always work, until you begin reaching high degree polynomials. The way it works is that you can solve for certain variables which can be plugged into the other equation. That gets rid of a variable. For example, if you have two equations with x and y, you can solve one of the equations for y (getting y by itself), and then plug that expression into the y part in the other equation. That eliminates the y leaving only x's, which can then be solved. Let's say we had a system.

(1)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \begin{align} 2x+3y = 13 \\ x-y = -1 \end{align} \end{align}

First, identify what variable you want to solve for. Notice how simple the bottom equation is. It would be very easy to solve for a variable there. Solve for x by adding y to both sides. This will give you x=y-1. That equation is solved for x. Now we know what x equals, so we can put that value into x's spot in the top equation. We will put y-1 into x. We then get 2(y-1)+3y=13. Now, all we have is y, and we can solve for it. Begin by distributing your two to y and -1. Combine your y terms to get 5y-2=13. Now solve. When we finish, we get y=3. We are not finished yet! Now we have to get x. Substitute 3 into y in any equation. We will use the bottom because it is simpler. You end up with x-3=-1. When we solve, we get x=2. So, putting our answer in coordinate form, our solution to the system is (2,3).

The addition method is my favorite method. Although it becomes cumbersome in many equations, it can be real simple with just two. What it does is it easily gets rid of a variable while keeping your equations small. Let's say you have two equations.

(2)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \begin{align} x+2y = 7 \\ 3x-2y = -3 \end{align} \end{align}

Notice that each equation has a 2y term, one is positive and one is negative. What you can do is add the equations vertically, adding the coefficients and constants with their respective term. So, you can add the x's, the y's, and the constant numbers. So, for this system, we can add x+3x, 2y+(-2)y, and 7+(-3). This gives you one equation, 4x+0y=4. This becomes 4x=4. Notice, the y term is now gone. The reason this happened is because the y terms for each equation were opposites, and they added each other out. This does not work for if each y is positive and one is negative, but one must be positive while the other is the negative of equivalent absolute value. Be careful. Now with 4x=4, we find that x=1, and we can then plug that value into one of the equations and get y=3. So our answer is (1,3).

Now, we know that this method works when the coefficients are set up correctly. However, the coefficients are usually never set up correctly. So, does this mean we can practically never use the addition method? Of course not! We can actually manipulate the equations in a system to where they can be set up for this method. Let's say we had a system:

(3)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \begin{align} 4x-2y = 5 \\ 3x-5y = -7 \end{align} \end{align}

How can we manipulate this system to make it set up for the addition method? What we need to do is get one set of variable coefficients, say y in this instance, to be opposites of equivalent absolute value. So we have -2 and -5. Find a number that each can be easily divided into. For this case, that number is 10. So, we want one value to be -10 and the other to be 10. What number can we multiply by -2 to get -10? When we multiply -2 by 5, we get -10. Similarly, if we multiply -5 by -2, we can get 10. Now, we have our multiplicands 5 for the top and -2 for the bottom. Multiply every term in each equation by its respective multiplicand. Now, our system becomes

(4)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \begin{align} 20x-10y = 25 \\ -6x+10y = 14 \end{align} \end{align}

Notice, now we are ready to add our equations! When we add our equations, we get 14x=39. So x is 39/14. When we put this rather ugly value into x, we can solve for y which becomes 43/14. Our answer is (39/14,43/14).

Don't try doing this with systems of more than two equations.

## Matrix Inversion

Matrices are very useful when it comes to systems of equations. First, you must understand that a system of equations is actually the product of two matrices. Think of this: what do you know about matrix multiplication? Each row is multiplied by each column. So, that means that each column in the product was multiplied by a row in the second matrix. In other words, every column in the product has a bit from the second matrix row. So, knowing this, we can factor our system into a matrix. Here is the relation.

(5)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \left( \begin{align} 2x+3y = 18 \\ 4x-y = 22 \end{align} \right) = \left( \left[ \begin{array}{cc} 2 & 3 \\ 4 & -1 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} 18 \\ 22 \end{array} \right] \right) \end{align}

As you can see, it can get pretty intense. How does this thing work? Remember how to multiply matrices? First, multiply row one in matrix one by column one in matrix two. This gives 2x+3y. Familiar? Then do the bottom row by the column, and you get 4x-y. So hopefully you see how this can actually work.

So, how do we use this to solve a system? First, you should know that the opposite of multiplication, when pertaining to matrices, is inversion. So, to solve a system of equations using matrix inversion, you simply take the inverse matrix on both sides. On the left side, you get an identity matrix, which is technically "1".