+0  
 
0
139
4
avatar+3636 

Help 12,13

NotSoSmart  May 16, 2018
 #1
avatar+7324 
+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+2255 
+3

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+7324 
+2

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+7324 
+2

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

29 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.