<PRE><B>Question 349915</B>
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Let *[tex \Large x] be the measure of the longer board.  Then the shorter board is *[tex \Large \frac{x}{3}].  6 times the sum of the short board and -10 is then: *[tex \Large 6\left(\frac{x}{3}\ -\ 10)\ =\ 2x\ -\ 60].  The long board minus 11 is *[tex \Large x\ -\ 11]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ -\ 60\ =\ x\ -\ 11]


Solve for *[tex \Large x].


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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