Explanation:
for such simple equations and only 2- dimensional problems, a request like this (which of the following points is a solution for the system ?) can be solved (often) the fastest by just putting the point coordinates in, calculate and see, if the equations hold or not.
the formal way to solve the system is usually done, when there are no solution options, and we need to find them ourselves.
to be sure you understand what I mean, I am demonstrating here both for the first 2 systems :
2x - y = 7
3x + y = 3
(2, -3)
1. the formal method. with experience I see immediately that when just adding the 2 equations, y is eliminated, and we can solve only for x.
this looks like
2x - y = 7
3x + y = 3
----------------
5x 0 = 10
so,
5x = 10
x = 2
and we get from e.g. the first equation
2×2 - y = 7
-y = 3
y = -3
2. the trial method. often accompanied with the mentioned experience factor that helps what option to try first.
but I am picking a wrong one first to show you what happens in that case.
let's pick (-2, -3) :
2×-2 - -3 = 7
-4 + 3 = 7
-1 = 7
no, the equation does not hold, -1 is definitely not equal to 7. so, that cannot be the answer.
mostly we find that already after trying the first equation, but sometimes only after the second equation (both need to hold for an actual solution).
so, when we pick (2, -3) we get
2×2 - -3 = 7
4 + 3 = 7
7 = 7 OK
3×2 + -3 = 3
6 - 3 = 3
3 = 3 OK
yeah, (2, -3) is the solution.
now for the second system.
2x + 3y = 8
3x - 2y = -14
(-2, 4)
1. formal method.
we can multiply or divide whole equations by a factor before adding them up.
since we have no clear option for elimination (as we had for the previous system), we need to bring them to factors that when added a whole variable disappears. this is similar to bringing fractions to the same denominator.
y still has a + and a - term in the 2 equations. so, I am going to eliminate y.
we multiply the first equation by 2, and the second by 3, and then we add then up.
this looks like
4x + 6y = 16
9x - 6y = -42
----------------------
13x 0 = -26
13x = -26
x = -2
using e.g. the first equation with that x :
2×-2 + 3y = 8
-4 + 3y = 8
3y = 12
y = 4
2. the trial method.
because the equating looked different in their structure compared to the first system, I picked the option to try that I considered the most different to the solution of the 1st system. and that was (-2, 4)
2×-2 + 3×4 = 8
-4 + 12 = 8
8 = 8 OK
3×-2 - 2×4 = -14
-6 - 8 = -14
-14 = -14 OK
yeah, (-2, 4) is the solution.
and now to the other 2 systems that actually do not have a defined solution (as explained) :
3x - 2y = 12
6x - 4y = -24
no solution (because when multiplying the first equation by 2 and then subtracting the second from the first we get 0 = 36, which is impossible, a contradiction, so, there is not a single solution possible).
2x - 3y = 6
-4x + 6y = -12
infinitely many solutions, as the second equation is just the first equation multiplied by -2. so, it is in its core the same equation as the first, and therefore we have only one equation but with 2 variables, and that gives infinitely many solutions.