Can substitution be used to solve a single equation that represent a partial order with two variables?

Question

I am still trying to wrap my head around partial orders and my algebra is not the best... My question is pertains to the following example:

ex.) Let a relation R be defined on the set of real numbers as follows: $ xRy ⇔ 2x +y = 3 $ Prove that this relation is antisymmetric.

Solution: aRb is $(2a+b=3)$, since bRa is $2b+a=3$
$(2a+b)−(2b+a) = 3−3 $
$2a + b − 2b − a = 0$
$a−b=0 $
$a = b $

My question is, why subtract the two equations? Do we subtract the two equations because a partial order is a way to quantify how objects in a set are different from each other?And is it possible to solve the same equation by substitution?

I solved: $2x + y = 3 $
$x= 3-y/2 $

and plugged x back into the equation and solved for y. $2(3-y/2)+y=3 $
$3-y+y=3$
$3+y=3+y $
$y=y$

I tested a few values to find that the equation 2x+y=3 is true for any value y that is added to 2(3-y/2).

Answer 1

In order to show antisymmetry we have to show that given $xRy$ and $yRx$ the equality $x=y$ follows. Here we have the situation that since $xRy$ is defined as $2x+y=3$ the system of equations \begin{align*} 2x+y&=3\tag{1}\\ x+2y&=3\tag{2} \end{align*} is valid.

We can do any admissible equivalence transformations with (1) and (2) to find out if $x=y$ is true. This means that we can subtract (1) from (2) to come to a conclusion as well as applying the substitution method. So, both approaches are fine.

Hint: Your calculation needs to be revised somewhat. In fact from $2x+y=3$ it follows $x=\frac{3-y}{2}$. Substitution of $x=\frac{3-y}{2}$ in $x+2y=3$ results in $y=1$ and we finally get $\color{blue}{x=y}=1$.