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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.

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:

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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

add eqns together

(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)

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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)

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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...

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!