allegansveritatem
Full Member
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- Jan 10, 2018
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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...
Why don't you use simple elementary algebra? \(\displaystyle \dfrac{{\frac{x}{{x + 3}}}}{{\frac{x}{{x + 3}} + x}}\)=\(\displaystyle \dfrac{{(x + 3)\left( {\frac{x}{{x + 3}}} \right)}}{{(x + 3)\left( {\frac{x}{{x + 3}} + x} \right)}} = \dfrac{x}{{x + ({x^2} + 3x)}}\)Here is the problem:View attachment 10385 solution:View attachment 10387
Where the **** do they get that denominator?
Thank you verry much for this lesson and tutorial.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 are you here?Thank you verry much for this lesson and tutorial.
Of course the better question is Why are you here?WHY are you here?
Why don't you use simple elementary algebra? \(\displaystyle \dfrac{{\frac{x}{{x + 3}}}}{{\frac{x}{{x + 3}} + x}}\)=\(\displaystyle \dfrac{{(x + 3)\left( {\frac{x}{{x + 3}}} \right)}}{{(x + 3)\left( {\frac{x}{{x + 3}} + x} \right)}} = \dfrac{x}{{x + ({x^2} + 3x)}}\)
\(\displaystyle \dfrac{x}{{x + ({x^2} + 3x)}} = \dfrac{1}{{x + 4}}\)