Wacky explanation in a student solutions manual for manipulating an equation

[chr1s]

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]

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.

 

[HOI]

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).

 

[chr1s]

Thanks everybody. Certainly makes sense now.