# Find the equations of the tangent line(s) to $\large{\frac{(x-1)}{(x+1)}}$ that are parallel to the line $y=2(x-5)$

Find the equations of the tangent line(s) to $\large{\frac{x-1}{x+1}}$ that are parallel to the line $y=2(x-5)$.

I've found the slope at $2$, and set the derivative equal to it, but I cant think how to solve for $x$. It seams simple but i'm drawing a blank.

So far I'm at $\large{\frac{2}{(x+1)^2}} = 2$.

• Since $\ x \$ cannot equal -1 , you can multiply both sides of the equation by $\ ( x + 1)^2 \$ (so that you are not multiplying through by zero). Commented Jun 19, 2015 at 4:27
• so then x would = -2 because after multiplying it out to get 2 = 2x squared + 4x +2, subtracting the two makes 0 = 2x squared +4x, subtract 4x and divide by 2x which makes x= -2
– Alec
Commented Jun 19, 2015 at 4:43
• The equation becomes $$\ ( x + 1 )^2 \ = \ 1 \ \ \Rightarrow \ \ x + 1 \ = \ \pm 1 \ \ ,$$ which has two solutions. as also discussed below. So there are two possible tangent lines. Commented Jun 19, 2015 at 4:55

Following what you have(in fact, you already did the important part), $x=-2, 0$.
For $x=-2$, $\frac{x-1}{x+1}=3$, so the line is $y-3=2(x+2)$, so $y=2x+7$;
For $x=0$, $\frac{x-1}{x+1}=-1$, so the line is $y+1=2(x-0)$, so $y=2x-1$.