Assume #x>y#.
Then, #xy = 30# and #x-y=20#
Isolating #x# in both equations yield:
#x=30/y# and #x=20+y#
By the transitive property, #30/y=20+y#.
Multiplying both sides by #y#, #30=y(20+y)#, which expands to #30=y^2+20y#
Thus, #y^2+20y-30=0#
By using the quadratic formula, we find that #y=(-20+-sqrt(520))/2#
Simplifying #sqrt(520)# gives us #2sqrt(130)# and so #y# may equal #(-20+2sqrt(130))/2# or #(-20-2sqrt(130))/2#.
Simplifying, y may equal #-10+-sqrt(130)#, which turns out to be approximately #1.40175425099137979136049055675# or #-21.401754250991379791360490255668#
If we substitute each of these values in the equation #x-y=20#, then we find that x may be (approximately) #21.401754250997379791360490255668# or #-1.40175425099379791360490255668#.
