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# Help 12,13 ​

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Help 12,13

NotSoSmart  May 16, 2018
#1
+7266
+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
+2202
+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
+7266
+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...

hectictar  May 22, 2018
#2
+7266
+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