Question

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

Answer 1

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]