# Algebraic Structure to solve a linear system of 2 variables?

#### ksmith630

##### New member

If i were given two linear equations and needed to solve for x and y using the methods of "substitution" and "elimination", what algebraic structure should i use? I can solve the systems just fine, but i'm trying to "explain" each step using properties of groups, rings, fields etc... and it's difficult without deciding on the structure.

1. I must pick from (group, ring, field, integral domain) to solve a system.
2. Once i have chosen the algebraic structure, i'll cite theorems/properties for EACH step of the solutions.

Take this system, which yields one solution (3, 2).

x + y = 5
x - y = 1

Here's the work anyone would do to solve the system in the two methods:

***************************************************************
For the "elimination" method:

x+y=5
x-y=1

1(1x+1y)=(5)1 Multiply the top equation (both sides) by 1
-1(1x11y)=(1)(-1) Multiply the bottom equation (both sides) by -1

note 1+(-1)=0

(1x-1x)+(1y+1y)=5-1
(101)x+(1+1)y=5-1
1+(-1)x+(1+1)y=5-1
0+(1+1)y=5-1
2y=4

divide both sides by 2...
2y/2=4/2
y=2

plug into eqn x+y=5
1x+1(2)=5
1x=5-2 subtract 2 from both sides
1x=3
(1/1)(1)x=(3)(1/1) mult both sides by (1/1)
x=3

So solution is (3,2)

********************************************

By the "substitution" method:

x+y=5
x-y=1

solve for y in eq

1y=5-1x Subtract 1x from both sides
y=(5-1x) Divide both sides by 1
y=5-1x

substitute

1x+(-1)(5-1x)=1
1x-1(5)-1(-1)x=1 Distribute -1 to 5-1x
1x-5+1x=1

1x-5+1x+5=1+5 Add 5 to both sides
1x+1x=6
2x=6

(1/2)(2/1)x=(6/1)(1/2) Mult both sides by (1/2)
x=3

Substitute into eq

1(3)-1y=1
3-1y=1
-1y=1-3 Subtract both sides by 3
-1y=-2
(1/(-1))(-1)y=(-2/1)(1/(-1)) Mult both sides by 1/(-1)
y=-2/(-1)
y=2

So solution is yet again (3,2)

**********************************************

If i had to solve an equation with 1 variable and 1 solution x, like x + 2 = 5 i would solve it using properties in a "group" <Z,+> where + is the normal binary op of addition in Z.

But since (as my work shows above for each method of solving a system with 2 eqns) i now have two binary operations (and rational numbers) so i cannot solve the system in a "group" of integers.
So when solving a system like x + y = 5 and x - y = 1 for x and y, would i then be in a field (of rationals) since i have two binary operations and fractions?

Once i know what structure i'm working in, i can then cite specific properties and theorems from that structure at each step of my work. Note i need to do this twice; once for the "elimination" method and once for the "substitution" method.

The start of my solution will start with a line that looks like:

To solve this system of linear equations, i will be in the (group, ring, field, integral domain) represented by <Z,R,Q,+,x,?,?> where + and x are the ordinary binary ops of the (integers, real numbers, rational numbers).

Can anyone help me fill in the blanks so it flows better? i'm new to this material so i'm not fluent just yet...

Then once the structure is decided, i will cite theorem from that structure that look like:

For any a and b in a (group, ring, field, integral domain) <Z,R,Q,+,x,?,?> if we know a=b, then for any c we know that a*c=b*c...

Can anyone tell me if i'm on the right track? I'm thinking field of rational numbers, but i'm not sure if that's the easiest... Thanks in advance!

#### Ackbach

##### Indicium Physicus
Staff member
This may not be official enough, but with systems of equations, you're typically talking about inverting operators on vector spaces over fields. So I would pick the field as the fundamental algebraic structure.

#### ksmith630

##### New member
Ahh perfect. I'll choose the field then- but how do i write the notation? I know if i was working in a group i'd write:

"To solve, i will be in the group <Z,+> where + is the ordinary binary operation in Z."

So for this system in the field with rationals as well as integers, would i write:

"To solve, i will be in the field <Q,+,x> where + and x are the ordinary binary operations in Q." ?

#### Deveno

##### Well-known member
MHB Math Scholar
it turns out that in this case, only the ring structure of Z is needed. in fact, since Z admits a rather obvious description as a Z-module, in truth, only the abelian group structure of Z is needed.

in general, however, most systems of Z-linear equations require the field Q to solve (this will be the case if the solutions are not in Z itself). the reason we can get away with solving these in Z, is that Z is an integral domain, so we can apply the cancellation laws in lieu of dividing.

one caveat: it is not explicitly stated what system 5 and 1 reside in. the integers are probably what was intended.

#### Swlabr

##### New member
A statement like $2x+2y+2z=0$ in $\langle\mathbb{Z}, +\rangle$ is equivalent to $x^2y^2z^2=1$ when we write the group multiplicatively. Equations like this have been much studied, and quite frankly the solutions to these problems, in the 'easiest' case (of free groups), go way over my head! (The solutions use, I believe, Sela's $\mathbb{R}$-trees, whatever they are!)

However, it has been shown recently that if your group is something called a right-angled Artin group (which includes both free groups and free abelian groups) then if $x^py^2q^r=1$ then there exists some $g\in G$ such that $x, y, z\in\langle{g}\rangle$. Which is pretty neat! (The case of free groups and $p=q=r=2$ has been known since 1959.)

#### ksmith630

##### New member
So i'm only working in the ring <Z,x,+> of integers?