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"
