Both equations should be given with variables of y and x. They may however not be given in the form y=mx+c where m and c are constants, the equation could be given in the form of a circle (y-a)2+(x-b)2=r2 where (a,b) is the coordinates of the centre and r is the radius. They may also give equations that need to rearranged algebraically, such as x2-14=3y+2x. Whatever formulae style is given to you, it should be dealt with in the same way algebraically. The coordinates of where the functions meet will be where the x and y values satisfy both equations. These can be found by substituting in an expression for y in terms of one equation into another.
EG If 3y=6x2+18 and 8x=y-16 you would rearrange boths equations to
y=2x2+6 (by dividing by 3) and y=8x+16 (by adding 16 to both sides) NOTE I chose to make y the subject of the equation to avoid having to deal with the x2
Now these equations can be made equal to one another 2x2+6=8x+16. This can rearranged into the quadratic formula x2-4x-5=0 which gives x=5 and x=-1
Substitute these values back into y=8x+16 (you can do with either as these are points that both functions go through it is just easier to not use the x2 but would still create the same answer. This gives y=56 and y=8 respectively, so the coordinates where the two functions meet are (5, 56) and (-1, 8). These values can be checked by substituting back into y=2x2+6 and making sure they are consistent.