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#1**+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**+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

#2**+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