Question

Simplify the complex fraction.

Answer 1

\(\Large \frac{x-5}{x}-\frac{-12}{x-1}=\frac{(x-5)(x-1)+12x}{x(x-1)}\)

\(\Large \frac{x+1}{x}+\frac{x+1}{x-1}=\frac{(x-1)(x+1)}{x(x+1)}\)

\(\Large \Rightarrow \frac{\frac{x-5}{x}-\frac{-12}{x-1}}{\frac{x+1}{x}+\frac{x+1}{x-1}}=\frac{\frac{(x-5)(x-1)+12x}{x(x-1)}}{\frac{(x-1)(x+1)}{x(x+1)}}=\frac{(x-5)(x-1)+12x}{(x-1)(x+1)} \)

Answer 2

So the LCD is x(x-1) and and you put (x-1) on the numerator.

Answer 3

yeah sorry, I'm wrong

Answer 4

BTW, \(\Large \frac{x+1}{x}+\frac{x+1}{x-1}\) must be \(\Large \frac{(x-1)(x+1)+x(x+1)}{x(x-1)}\)

Answer 5

ok

Answer 6

\[\frac{ (x-5)(x-1)-(-12x) }{ (x+1)(x-1)+x(x+1) }\]
How do I simplify the numerator and denominator into trinomials?

Answer 7

\(\large (x-5)(x-1)+12x=x^2+6x+5\\\large =x^2+x+5x+5=x(x+1)+5(x+1)\\\large =(x+1)(x+5)\)

\(\large (x+1)(x-1)+x(x+1)=x^2-1+x^2+x=2x^2+x-1\\\large =2x^2+2x-x-1=2x(x+1)-(x+1)=(x+1)(2x-1)\)

\(\Large \Rightarrow \frac{ (x-5)(x-1)-(-12x) }{ (x+1)(x-1)+x(x+1) }=\frac{(x+1)(x+5)}{(x+1)(2x-1)}=\frac{x+5}{2x-1}\)

Answer 8

I didn't realize at first but if I am correct
\[(x^2+x)+(5x+5)\]
\[x(x+1)5(x+1)\]
is positive because when you foil there is two negatives?
The way I did it was like this
\[x^2-x-5x+5+12x=x^2+6x+5\]
\[=(x+5)(x+1)\]