How do you solve #(r+4)/3=r/5# and find any extraneous solutions?
1 Answer
Dec 24, 2016
Explanation:
To ' eliminate' the fractions on both sides of the equation, multiply by the lowest common multiple (LCM ) of 3 and 5 which is 15
#cancel(15)^5xx(r+4)/cancel(3)^1=cancel(15)^3xxr/cancel(5)^1#
#rArr5(r+4)=3rlarr" no fractions"#
#rArr5r+20=3r# subtract 3r from both sides.
#5r-3r+20=cancel(3r)cancel(-3r)#
#rArr2r+20=0# subtract 20 from both sides.
#2rcancel(+20)cancel(-20)=0-20#
#rArr2r=-20# To solve for r, divide both sides by 2
#(cancel(2) r)/cancel(2)=(-20)/2#
#rArrr=-10" is the only solution"#
