Question 791886
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Let's say you had the equation


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -5z\ -\ 6\ =\ 4]


But you were told that the variable *[tex \LARGE z] was restricted to non-negative real numbers.  How many solutions would this equation have under that constraint?  If z is positive, the -5z is negative, and adding -6 just makes it more negative.  If z is zero, then -5z equals zero, and the entire left side equals -6, still negative and nowhere near 4.  You can't get there from here, is the message.


Well, *[tex \LARGE |3x\ +\ 5|\ \geq\ 0\ \forall\ x\ \in\ \mathbb{R}], so there is no number that you can substitute for *[tex \Large x] that would make *[tex \LARGE -5|3x\ +\ 5|\ -\ 6\ =\ 4] a true statement.


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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