How do you evaluate #\frac { 4r - 4} {r^2+ 4r - 12} + \frac { 4} { r + 6} = \frac { 2} { r - 2}#?

1 Answer
Jade
Mar 14, 2018

#r=4#
Refer to the explanation.

Explanation:

#(4r-4)/(r^2+4r-12) + 4/((r+6)) =2/((r-2)) #
#(4r-4)/(r^2+6r-2r-12) + 4/((r+6)) =2/((r-2)) #
#(4r-4)/(r(r+6)-2(r+6)) + 4/((r+6)) =2/((r-2)) #
#(4r-4)/((r-2)(r+6)) + 4/((r+6)) =2/((r-2)) #
#(4r-4+4(r-2))/((r-2)(r+6)) =2/((r-2)) #
#(4r-4+4(r-2))/(cancel((r-2))(r-6))=2/(cancel((r-2))#
#4r-4+4r-8=2(r+6)#
#8r-12=2r+12#
#6r=24#
#r=4#

Hope this helps :)

Related questions
Impact of this question
760 views around the world
You can reuse this answer
Creative Commons License