How do you solve the system using elimination $x-y=12$ and $2x+y=3?$

Hint: We will add or subtract the equations accordingly. Before that, we need to check if this helps us to eliminate either of the unknown variables. If not, we will multiply any of the equations with a scalar so that we can eliminate either of the equations using addition or subtraction. When we will get the value of one of the variables, we will apply that in any of the equations to find the second variable.

Complete step by step answer:
Let us consider the given system of equations, $x-y=12$ and $2x+y=3.$
We are asked to solve the system of equations using elimination.
By inspection, we can see that the elimination of the variable $y$ can be done using the addition of these equations.
When we add the equations, we can eliminate $y.$ Because, $-y+y=0.$
So, the addition will give us $x-y+2x+y=12+3.$
We know that $x+2x=3x.$
Now, we will see that $3x-0=15.$
That is, $3x=15.$
Let us transpose $3$ from the left-hand side to the right-hand side to get all the constant terms on one side and the variable on the other side.
So, we will get $x=\dfrac{15}{3}=5.$
Now, we have found the value of the variable $x=5$ by the elimination process.
Now, we will apply the value in any of the equations. We can find the value of $y$ by applying the value of $x$ in the equation $x-y=12.$
We will get $5-y=12.$
Let us transpose $5$ to get $-y=12-5=7.$
Let us multiply the whole equation with $-1$ to get $y=-7.$
Hence the solution is $x=5$ and $y=-7.$

Note:
If we apply the obtained value of $x$ in the second equation also, we will be able to find the value of $y$ as $2\times 5+y=10+y=3.$ When we transpose $10,$ we will get $y=3-10=7.$