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Help 12,13

NotSoSmart  May 16, 2018
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4+0 Answers

 #1
avatar+7056 
+2

12.

     \( \frac{4-x}{x^2+5x-6}\div\frac{x^2-11x+28}{x^2+7x+6}\)

                                                    Factor the numerators and denominators.

=   \(\frac{-(x-4)}{(x-1)(x+6)}\div\frac{(x-7)(x-4)}{(x+1)(x+6)}\\\)

 

All the values of  x  that cause a zero in a denominator are:   1,  -6,  7,  4,  -1

 

So the restrictions are:   x ≠ 1 ,  x ≠ -6 ,  x ≠ 7 ,  x ≠ 4 ,  x ≠ -1

 

Now let's flip the second fraction and change the sign to multiplication.

 

\(\frac{-(x-4)}{(x-1)(x+6)}\cdot\frac{(x+1)(x+6)}{(x-7)(x-4)}\)

                                                    Multiply the fractions.

=   \(\frac{-(x-4)(x+1)(x+6)}{(x-1)(x+6)(x-7)(x-4)}\)

                                                    Cancel the common factors.

=   \(\frac{-{\color{purple}(x-4)}(x+1){\color{NavyBlue}(x+6)}}{(x-1){\color{NavyBlue}(x+6)}(x-7){\color{purple}(x-4)}}\)

 

=   \(\frac{-(x+1)}{(x-1)(x-7)}\)

 

(Hmm...  is   x ≠ -1  not actually a restriction?? Is the expression defined when  x = -1 ?)

 
hectictar  May 17, 2018
 #3
avatar+1961 
+2

Just to clear up any confusion, hecticar, \(x\neq -1\). Any time a division-by-zero error arises, the entire expression is immediately deemed as undefined--no matter where. 

 

\(x=-1;\\ \frac{-(x-4)}{(x-1)(x+6)}\div\frac{(x-7)(x-4)}{(x+1)(x+6)}\\\) Let's just do a quick substitution.
\(\frac{5}{-10}\div\frac{40}{0}\)  
   

 

We have reached a problem. Do not go any further! Here is another way to think about, I guess.

 

Any result in the format \(\frac{x}{0}\) is given the name "undefined" because it literally does not have a meaning; it is pure nonsense; it means nothing at all. By taking the reciprocal, you are attempting to attach meaning to nonsense, which you should never do.

 
TheXSquaredFactor  May 21, 2018
 #4
avatar+7056 
+1

Thanks for the response! I didn't think anybody was going to see my question this late.

 

I also thought that  x ≠ -1  should be a restriction, but since it is not listed in the restrictions in the answer option I wanted to make sure...  smiley

 
hectictar  May 22, 2018
 #2
avatar+7056 
+1

13.

\(\hphantom{=\,}\ \dfrac{\frac{x}{x+4}}{\frac{1}{x}+\frac{1}{x+4}}\\~\\ =\,\dfrac{\frac{x}{x+4}}{\frac{1}{x}+\frac{1}{x+4}}\cdot\dfrac{x+4}{x+4}\\~\\ =\,\dfrac{x}{\frac{x+4}{x}+1}\\~\\ =\,\dfrac{x}{\frac{x+4}{x}+1}\cdot\dfrac{x}{x}\\~\\ =\,\dfrac{x^2}{x+4+x}\\~\\ =\,\dfrac{x^2}{2x+4}\)

 
hectictar  May 17, 2018

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