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Wacky explanation in a student solutions manual for manipulating an equation

chr1s

New member
Jan 28, 2019
2
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.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,485
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 \(\displaystyle \frac{5}{2r}\). The author probably wanted to arrive at an equation with integer coefficients after multiplication.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
344
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 [tex]2.5= \frac{5}{2}[/tex] so that original form can be written as [tex]4/(r+ 8)+ 5/2r+ 1[/tex]. Now the "common denominator" is 2r(r+ 8).
 

chr1s

New member
Jan 28, 2019
2
Thanks everybody. Certainly makes sense now.