Question

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

Answer 1

`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"`