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Using some algebraic manipulation the expression $\displaystyle\frac{P-W_1P_x}{W_2}$ is made into $\displaystyle \frac{1-W_1 \frac{P_x}{P}}{W_2}$. It says it is simply multiplied by $\displaystyle \frac{1}{P}$. I am not terribly fluent in algebra as I should be, but online resources are failing me in helping me see what rules were applied to make this happen. Help please?

2012ssohn
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Marcus
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2 Answers2

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Consider $\frac{P - W_{1}P_{x}}{W_{2}} = (P - W_{1}P_{x}) * \frac{1}{W_{2}}$. Now dividing by $P$ is equivalent to $\frac{(P - W_{1}P_{x})}{P} * \frac{1}{W_{2}} = (\frac{P}{P} - \frac{W_{1}P_{X}}{P}) * \frac{1}{W_{2}}$. The cancellations should be obvious from here.

ml0105
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It is the distributive property of multiplication and addition.

$$\frac{1}{P}\frac{P-W_1P_x}{W_2} = \frac{\frac{1}{P}\left(P-W_1P_x\right)}{W_2}= \frac{1-W_1\frac{P_x}{P}}{W_2}$$

Note that this is also the same as just writing

$$\frac{P-W_1P_x}{PW_2}$$

Erik M
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  • Cool, so that is certainly equivalent to the answer below, which made more sense to me initially but I guess it's just because I'm not used to thinking of multiplying the numerator alone by a fraction. Usually if I saw some fraction a/b times c/d I would think it would become ac/bd but I guess you can also do a/b(c)/d, if that is equivalent the solution you've just done? – Marcus Mar 20 '14 at 04:15
  • Yeah, any time a term is in front of a fraction it is implied that it is multiplied by the numerator only. That is $a\frac{b}{c} = \frac{ab}{c}$. The product that you wrote is indeed true since $\frac{a}{b}\frac{c}{d} = \frac{\left(\frac{a}{b}c\right)}{d} = \frac{\left(\frac{ac}{b}\right)}{d} = \frac{ac}{bd}$. Things can get a little ambiguous (visually) at times, so it is best to use parentheses to clarify if there could be some confusion. – Erik M Mar 20 '14 at 04:24
  • Another (potentially tricky) manipulation is $\frac{a}{b}=\frac{\frac{1}{b}}{\frac{1}{a}}$. – Erik M Mar 20 '14 at 04:46
  • Great, that's very helpful. Thanks for your time and thoughtful response, that will help me plenty to know in the future. – Marcus Mar 20 '14 at 04:49