Plot y=x^2 and y=x+6 on the same graph. The first is a U-shaped curve sitting on the origin (0,0), the second is a straight line crossing the x axis (y=0) at x=-6 and crossing the y axis (x=0) at y=6. Where the line cuts through the U curve represents the solution to the equation x^2-x-6=0 (x^2=x+6). The solution to this quadratic is x=-2, +3. The intersection points are therefore (-2, 4) and (3, 9).
The remaing part of your question involves straight lines only, linear equations. For (a) draw a straight line joining y=-4 on the y axis to x=2 on the x axis. Also draw a line joining y=3 1/3 to x=-1 2/3. These points on the axes are where x=0 and y=0 for the two functions. The lines don't stop at the axes, so just continue them after they cut the axes. What may have been confusing to you is that the first equation starts y=..., but the second equation has x and y together in an expression. However, you can move things around in an equation and, if you want, you can make the equation look like y=... or x=... or just combine x and y in an expression. It doesn't matter. What you'll find is that the lines are parallel (have the same slope) so that means they never cross and that means there's no solution.
In (b) the two lines have different slopes so we would expect them to intersect. The first equation means joining (0,3) to (9/2,0) and (0,11/2) to (11/3,0). Again, extend the lines to beyond where they cut the axes. Note that the second equation simplifies to 3x+2y=11. The solution to the equations solved simultaneously or by substitution give a single intersection point at (3,1).
