Question 604033
<font face="Times New Roman" size="+2">


If your solution were incorrect, then when you substitute into either of your equations, one or both of the statements would be false.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\frac{20}{3}\right)\ +\ 2\left(\frac{22}{3}\right)\ =\ \left(-\frac{20}{3}\right)\ +\ \left(\frac{44}{3}\right)\ =\ \frac{24}{3}]


Which is 8 for as long as I've been doing arithmetic.  I'll leave it to you to check the other side.


I know, your diabolical teacher/instructor/professor gave you a problem that came out with a solution that is a pair of fractions that have an ugly odd number in the denominator and are irreducible.  Just doesn't seem fair, does it?  Welcome to the real world, except things are even uglier out here.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>