Question 788312
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Think of a function as a little machine that looks like a box that sits on the table.  There is a slot on one end of the box, a hand crank on the side, and a little tray like in the front of a gumball machine on the other end.


You put a number into the input slot (that's the *[tex \LARGE x] you chose) and turn the crank.  Presuming the function was actually defined for the particular *[tex \LARGE x] chosen, you get some number that drops into the little tray.


Now, imagine that you have two of these function boxes, and each one represents the set of ordered pairs that constitute a straight line.  Then you have three possibilities:


1.  There is exactly one particular value of *[tex \LARGE x] that will produce the same output from both machines.


2.  Any value you pick for *[tex \LARGE x] will produce the same value out of both machines.


3.  There is no value of *[tex \LARGE x] for which both machines produce the same value.


The first case is where the two lines intersect in one point.  The *[tex \LARGE x] coordinate of that point is the particular value of *[tex \LARGE x] that causes the two machines to have the exact same output.  That output, by the way, is the *[tex \LARGE y]-coordinate of the point.


The second case is where the two lines are actually the same line, so every point on one line is on the other line as well.  And the answer to the question "Is the *[tex \LARGE y]-coordinate for the point on line 1 for a particular value of *[tex \LARGE x] the same as the *[tex \LARGE y]-coordinate for the point on line 2 with that same *[tex \LARGE x]-coordinate?" is yes (i.e. True) for any value of *[tex \LARGE x] you happen to choose.


The third case is that you must answer No (False) to the same question no matter what value of *[tex \LARGE x] you decide to use.  That is because they are parallel lines and never intersect.  Hence, 'no solution'.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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