your problem is:
y > 3x + 1
y < 2x + 4
this can only happen if 2x + 4 > 3x + 1
that's the equation you have to solve.
subtract 1 from both sides of that equation to get:
2x + 3 > 3x
subtract 2x from both sides of that equation to get:
3 > x
this is the same as:
x < 3
to test this out, set up a table of values where the value of x is less than 3 and equal to 3 and greater than 3 and solve for each equation.
in this table, make:
y1 = 3x + 1
y2 = 2x + 4
since 2x + 4 needs to be greater than 3x + 1, this means that y2 needs to be greater than y1.
your table will look like this:
x y1 y2 remarks
0 1 4 y2 > y1 (ok)
2 7 8 y2 > y1 (ok)
3 10 10 y2 not > y1 (no good)
4 13 12 y2 not > y1 (no good)
you can see from the table that the original statement is only true when y2 > y1 which only happens when x < 3.
you can graph the equations of y1 = 3x + 1 and y2 = 2x + 4 and the graph will also show you that you can only find a y between those 2 equations when the value of x is smaller than 3.
the graph is shown below:
the graph of y = 3x + 1 intersects the y-axis at y = 1.
the graph of y = 2x + 4 intersects the y-axis at y = 4.
a value of y < 2x + 4 and > 3x + 1 can only happen in the region where the line of 3x + 1 is lower than the line of 2x + 4.
that only happens when the value of x is smaller than 3.
you find that by picking any value of x < 3 and solving for each equation at that value of x.
example:
when x = -5, your equations of:
y > 3x + 1 and y < 2x + 4 become:
y > -14 and y < -6
you can see that, when x = -5, any value of y < -6 and > -14 can be found between the lines of the equations.
when x = 5, your equations of:
y > 3x + 1 and y < 2x + 4 become:
y > 16 and y < 14
you can see that this is an impossible situation, so, when x = 5, the system of equation is false.
the system of equations is true only if x < 3.
if you were able to, you would shade the region between the 2 lines for all values of x < 3.
that would be the region of compatibility with the system of equations.