Question 25543
Since y=3-2x, and  y=2-3x,  you can substitute the y = of the first equation into the y = of the second equation, and you have:

3-2x= 2-3x


Add +3x to each side:
3-2x + 3x = 2 -3x + 3x
3+x = 2


Subtract 3 from each side:
x=-1


Substitute back into the first equation:
y = 3-2x
y = 3-2(-1)
y = 5


Check in the second equation:
y = 2-3x
5 = 2-3(-1)
5=5 
It checks!


The solution is the point (-1,5).


R^2 at SCC