How do you solve #(r-4)/(5r)=1/(5r)+1# and check for extraneous solutions?

1 Answer
Jul 9, 2017

Answer:

Solution : #r= -5/4#

Explanation:

# (r-4)/(5r) = 1/(5r) +1 #. multiplying by #5r# on both sides , we get,

# r-4 = 1 +5r or 5r-r = -4-1 or 4r = -5 or r= -5/4#

Solution : #r= -5/4#

Check: L.H.S # = (r-4)/(5r) =( -5/4 - 4)/ (5* (-5/4)) =( -21/ cancel4)/ (-25/ cancel4) = 21/25#

R.H.S # = 1/(5r) +1 = 1/(5* (-5/4)) +1 = -4/25+1= 21/25#

#:. R.H.S = L.H.S# {checked) [Ans]