First you want to solve for y in both equations
subtract x from both sides of the equation and then divide by 3
3y+x -x > 2-x ⇒ 3y > 2-x
3y/3 > (2-x)/3 ⇒ y > 2/3 - x/3 ⇒ y > (-1/3)x + 2/3
x + 2y < 3
substract x from both sides of the equations and then divide by 2
x+2y - x < 3 -x ⇒ 2y < 3-x
2y/2 < (3-x)/2 ⇒ y < 3/2 - x/2 ⇒ y < (-1/2) x + 3/2
These equations give you the boundary lines that defined the region that y is bounded by.
"Solving" systems of linear inequalities means "graphing each individual inequality, and then finding the overlaps of the various solutions". So I would graph each inequality, and then find the overlapping portions of the solution regions.
So the first equation which is in y intercept format tells you that the y intercept is 2/3 and the slope of the line is (-1/3). The region than that is bounded by the line would be all the values of (x,y) above this line as y greater than the equation
From the second eqation, the y intercept is 3/2 and the slope is (-1/2) but since y is less than this equation it would be all the values of (x,y) below this line.
If you can visualize the graph, the portion of the graph that is valid for these two equations would be the triangle region that goes out from the intersection of the lines to x ⇒ -∞
(-1/2) x + 3/2 < y > (-1/3)x + 2/3
Its hard to show this without being able to graph