Wacky explanation in a student solutions manual for manipulating an equation

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Discussion Overview

The discussion revolves around a specific manipulation in a distance/rate/time problem from Stewart's College Algebra 4th Edition. Participants are examining the reasoning behind the introduction of a factor of 2 in the common denominator procedure used to solve the equation 4/(r+8) + 2.5/(r) = 1. The focus is on understanding the steps taken in the solution process, particularly the justification for multiplying by 2r(r+8).

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant questions the origin of the factor of 2 in the common denominator procedure, expressing confusion over its necessity.
  • Another participant asserts that multiplying both sides of an equation by any non-zero number is valid, emphasizing that the author's goal is to arrive at a solution.
  • A different perspective suggests that the author likely aimed for integer coefficients, explaining that rewriting 2.5 as 5/2 leads to the need for the factor of 2 in the common denominator.
  • One participant proposes that if the equation is multiplied by r(r+8) without the factor of 2, it still leads to a valid equation, but the introduction of 2 simplifies the coefficients.

Areas of Agreement / Disagreement

Participants express varying views on the necessity and reasoning behind the factor of 2, with some agreeing that it simplifies the process while others question its introduction. The discussion remains unresolved regarding whether the factor is essential or merely a stylistic choice.

Contextual Notes

There are unresolved assumptions regarding the preferences for integer coefficients versus maintaining the original fractional form, which may affect the clarity of the solution process.

chr1s
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In the answer book to Stewart's College Algebra 4th Edition, question 47 in Review for Chapter 2, it takes me, in a distance/rate/time problem, from 4/(r+8) + 2.5/(r) = 1 (which I got), to this common denominator procedure: "Multiplying by 2r(r+8), we get..." WHERE DID THEY GET THE "2"? It continues on to a quadratic procedure, all of which follows logically, and the answer, r = [-3 + (sq rt of 329)]/4, which seems to be right when I plug it back in. Can't figure out that 2... Thanks for anybody's help.
 
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First of all, the following property does indeed hold for all real numbers $x$, $y$ and $z$: if $x=y$, then $xz=yz$. (Note that it is not the case that the converse is true for all $x$, $y$ and $z$.) Therefore, the author of a proof or a solution has the right to multiply a true equation by any number he or she wants. This is not an error. The author's responsibility is to arrive at the solution. The reader has the right to ask, "Why is this true?", but the question "Why did the author do this?" is secondary.

Now, $2.5/r$ can be represented as $$\frac{5}{2r}$$. The author probably wanted to arrive at an equation with integer coefficients after multiplication.
 
chr1s said:
In the answer book to Stewart's College Algebra 4th Edition, question 47 in Review for Chapter 2, it takes me, in a distance/rate/time problem, from 4/(r+8) + 2.5/(r) = 1 (which I got), to this common denominator procedure: "Multiplying by 2r(r+8), we get..." WHERE DID THEY GET THE "2"? It continues on to a quadratic procedure, all of which follows logically, and the answer, r = [-3 + (sq rt of 329)]/4, which seems to be right when I plug it back in. Can't figure out that 2... Thanks for anybody's help.
The "2" is just because they want integer coefficients. If you just multiply both sides by r(r+ 8) you get 4r+ 2.5(r+ 8)= r(r+ 8). Multiplying by 2 gives 8r+ 5(r+ 8)= 2r(r+ 8).

Another way of looking at it is that 2.5= \frac{5}{2} so that original form can be written as 4/(r+ 8)+ 5/2r+ 1. Now the "common denominator" is 2r(r+ 8).
 
Thanks everybody. Certainly makes sense now.
 

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