Toon, your statement is wrong.
x isn't equals 3.
Here is the solution :
12-2/3x = 1e
12-2/3x = e
12-2/3x = 12-2/3x
(12-2/3
x)/
x = 12-2/3
12-2/3 = 12-2/3
(36-2)/3 = (36-2)/3
34/3 = 34/3
So x = ]-∞;+∞[ excluding 0
Same thing goes for e.
That is a system of 1 equation with 2 unknowns. Every equation containing 2 unknowns has it's unknowns varying between -∞ and +∞
That's why we said it's unsolvable, since we can never know the exact number, because the equation will
always be valid with both x and e varying between -∞ and +∞, but of course, in this case, x has to be =/= than 0 cause the denominator cannot be 0 else the fraction would become an ∞ as well.
To get the exact number, the system has to be of 2 equations with 2 unknowns.
Last edited by EJM; Apr 7, 2012 at 01:00 AM.