How do you multiply $((r+5)/(r^2-2r)) - 1= 1/(r^2-2r)$ ?

Question

5208 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-21T03:12:45

$((r+5)/(r^2-2r)) - 1= 1/(r^2-2r)$

$((r+5)/(r^2-2r)) - 1 xx ((r^2-2r)/(r^2-2r))= 1/(r^2-2r)$

$((r+5)/(r^2-2r)) - ((r^2-2r)/(r^2-2r))= 1/(r^2-2r)$

$((r+5 - r^2+2r)/cancel(r^2-2r))= 1/cancel(r^2-2r)$

$(r+5 - r^2+2r)= 1$
$- r^2+ 3r +4= 0$

$r^2 - 3r -4= 0$
Factorising by splitting the middle term:

$r^2 - 4r +r -4= 0$
$r(r - 4) +1(r -4)= 0$
$(r+1)(r - 4) = 0$

$color(green)(r = -1,4$ are the two solutions

How do you multiply ((r+5)/(r^2-2r)) - 1= 1/(r^2-2r)((r+5)/(r^2-2r)) - 1= 1/(r^2-2r)?