what is going on here? (x/(x+3)) / ((x/(x+3) + x)

allegansveritatem

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Here is the problem:problem.PNG

Here is what I did with it:work.jpg

My solution: 1/(2(x+3))

Book solution:book solution.PNG

Where the **** do they get that denominator?
 
Here is the problem:View attachment 10385

Here is what I did with it:View attachment 10386

My solution: 1/(2(x+3))

Book solution:View attachment 10387

Where the **** do they get that denominator?

You did the sort of silly thing we all do from time to time (but a lot less with experience): you took x + x(x+3) and saw it as if it were (x + x)(x+3) = 2x(x + 3).

Don't forget the order of operations! The addition is not done first.

It is really x + x^2 + 3x = x^2 + 4x. Can you finish?
 
Let u = x + 3

(x/u) / (x/u + x)

continue (will save you quite a bit of writing).
Substitute back in at end...
 
You did the sort of silly thing we all do from time to time (but a lot less with experience): you took x + x(x+3) and saw it as if it were (x + x)(x+3) = 2x(x + 3).

Don't forget the order of operations! The addition is not done first.

It is really x + x^2 + 3x = x^2 + 4x. Can you finish?

Thanks. I went over this problem again last night and again today before I looked at your post I realized finally what you point out here, namely that I had incorporated the first X into the system as a factor when it was actually an addend. But it took a lot of fooling around to discover it.
 
Last edited:
Let u = x + 3

(x/u) / (x/u + x)

continue (will save you quite a bit of writing).
Substitute back in at end...

I did something like this while fooling again with this problem--even before looking at your post. Here is how I applied the principle you are suggesting:

example.jpg

The beauty of this is: You already know what the right answer is going to be, which will guide you in working it out the long way. I am going to try the way you suggest, however, because it has a certain elegant leanness to it.Thanks for pointing it out.
 
Here is the problem:View attachment 10385 solution:View attachment 10387
Where the **** do they get that denominator?
Why don't you use simple elementary algebra? \(\displaystyle \dfrac{{\frac{x}{{x + 3}}}}{{\frac{x}{{x + 3}} + x}}\)=\(\displaystyle \dfrac{{(x + 3)\left( {\frac{x}{{x + 3}}} \right)}}{{(x + 3)\left( {\frac{x}{{x + 3}} + x} \right)}} = \dfrac{x}{{x + ({x^2} + 3x)}}\)
\(\displaystyle \dfrac{x}{{x + ({x^2} + 3x)}} = \dfrac{1}{{x + 4}}\)
 
Why don't you use simple elementary algebra? \(\displaystyle \dfrac{{\frac{x}{{x + 3}}}}{{\frac{x}{{x + 3}} + x}}\)=\(\displaystyle \dfrac{{(x + 3)\left( {\frac{x}{{x + 3}}} \right)}}{{(x + 3)\left( {\frac{x}{{x + 3}} + x} \right)}} = \dfrac{x}{{x + ({x^2} + 3x)}}\)
\(\displaystyle \dfrac{x}{{x + ({x^2} + 3x)}} = \dfrac{1}{{x + 4}}\)
Thank you verry much for this lesson and tutorial.
 
tha
Why don't you use simple elementary algebra? \(\displaystyle \dfrac{{\frac{x}{{x + 3}}}}{{\frac{x}{{x + 3}} + x}}\)=\(\displaystyle \dfrac{{(x + 3)\left( {\frac{x}{{x + 3}}} \right)}}{{(x + 3)\left( {\frac{x}{{x + 3}} + x} \right)}} = \dfrac{x}{{x + ({x^2} + 3x)}}\)
\(\displaystyle \dfrac{x}{{x + ({x^2} + 3x)}} = \dfrac{1}{{x + 4}}\)

Well, sometimes it helps to make a kind of model of a problem--I like to knock down all elements ,coefficients, variables and constants, to one digit numbers so I can compute what the result is going to be without having to even work it out. That way, when I do work it out, I know I'm on the right track if I get the answer I have already determined is correct. I also use this method a lot when working with exponents because I tend to forget certain rules, ie, the rules governing negative exponents, so I build a simple model and test the result. Thus 10^-2 = .01 and so does (1)/(10^2). I find it pleasing to prove this stuff this way to myself.
 
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