Question 628054
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Let's see if your approach makes sense.  You said *[tex \LARGE \frac{2}{3}x\ =\ 45].  If a part of the quantity *[tex \LARGE x] is equal to 45, then *[tex \LARGE x] must be bigger than 45.   If I asked you for two-thirds of the money in your wallet and you handed me $45, you must have started with more than the 45 bucks, right?  That means, according to your theory, that one of the pieces of board is longer than the piece it was cut from.  Absurd in the extreme.


Read the problem carefully.  I'm going to assign *[tex \LARGE x] to represent the measure of the longer piece, just because it is easier to explain the problem that way.  Generally you will want to assign your variable to represent the quantity asked for by the problem, but that is just a little inconvenient this time.


Let *[tex \LARGE x] represent the measure of the longer piece.  Then we are given that *[tex \LARGE \frac{2}{3}x] is the measure of the shorter piece.  Together (except for a negligible saw kerf) the two boards should add up to the original 45 feet.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ \frac{2}{3}x\ =\ 45]


Solve for *[tex \LARGE x] and then calculate *[tex \LARGE \frac{2}{3}x]


Step 1:  Multiply both sides by 3:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 2x\ =\ 105]


Step 2:  Collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ =\ 135]


Step 3:  Multiply by the reciprocal of the coefficient on the variable:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{5}\,\cdot\,5x\ =\ \frac{1}{5}\,\cdot\,135]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 27]


Which is the measure of the larger piece.  You can either calculate *[tex \LARGE \frac{2}{3}\,\cdot\,27] or *[tex \LARGE 45\ -\ 27] to find the measure of the shorter piece.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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