How do you solve #(r+4)/3=r/5# and find any extraneous solutions?

1 Answer
Dec 24, 2016

Answer:

#r=-10#

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"#