How do you solve #4(r+1)=6-2(1-2r) #?

1 Answer
Aug 31, 2017

Answer:

The given equation is true for all values of #r#

Explanation:

#4(r+1)#
#color(white)("XXX")=4r+4#

#6-2(1-2r)#
#color(white)("XXX")=6-(2-4r)#

#color(white)("XXX")=6-2+4r#

#color(white)("XXX")=4+4r#

So if
#color(white)("XXX")4(r+1)=6-2(1-2r)#
then
#color(white)("XXX")4r+4=4+4r#

...but this is true for any value of #r#