Question 709430
Because that system has fractional coefficients,
my first step would be to "eliminate denominators" in each equation,
by multiplying both sides of the equal sign times an appropriate number.
I would multiply times 2 for the first equation
and times 3 for the second equation
to transform the whole system into one that is easier for me.
{{{system((3/2)*x-2y=4,x+(1/3)*y=2)}}} --> {{{system(3x-4y=8,3x+y=6)}}}
 
That system has one and only one solution.
Why? Because the ratios of coefficients of x and y are different.
Let me explain by example
 
Obviously {{{system(3x+y=6,3x+y=6)}}} has many solutions
because both equations are the same,
and {{{3x+y=6)}}} represents one line
with an infinite number of (x,y) points that are solutions to that equation
and solutions to the system.
That system could appear "in disguise" and it would not be so obvious, as in
{{{system((3/2)x+(1/2)y=3,x+(1/3)y=2)}}} or {{{system(6x+2y=12,3x+y=6)}}}
All those equations are equivalent, because one can be obtained from another one
by multiplying both sides of the equal sign times an appropriate number.
 
On the other hand, the system {{{system(3x+y=6,3x+y=5)}}}
obviously has no solutions.
It could also be disguised to make it not so obvious.
 
Once your system was transformed into {{{system(3x-4y=8,3x+y=6)}}}
It was clear that it did not fit into the no-solution or infinite-solutions situations described above,
because both equations had 3 as the coefficient for {{{x}}} but had different coefficients for {{{y}}}.
 
The infinite number (x,y) data pairs that are solutions to {{{3x+y=6}}}
represent the points in the straight line that is the graph of {{{3x+y=6}}}
The infinite number (x,y) data pairs that are solutions to {{{3x-4y=8}}}
represent the points in the straight line that is the graph of {{{3x-4y=8}}}
The two lines intersect at just one common point,
and the coordinates of that point constitute the solution to the system {{{system(3x-4y=8,3x+y=6)}}} .
In this case, that point is (32/15, -2/5) and the solution to the system is
{{{highlight(x=32/15)}}}, {{{highlight(y=-2/5)}}}