Question

How do you solve `\frac { 1} { 3} - \frac { 2r ^ { 2} + 2r - 24} { 3r ^ { 2} - 6r } = \frac { 1} { r ^ { 2} - 2r }`?

Answer 1

`r = -7 and r =3`

Explanation:

With algebraic fractions, the first step is to see if any expressions can be factorised.

`1/3-color(red)( (2r^2 +2r-24))/(color(blue)(3r^2 -6r)) = 1/color(green)((r^2-2r))`

`1/3-color(red)(2 (r^2 +r-12))/(color(blue)(3r(r -2))) = 1/color(green)(r(r-2))`

Get rid of the denominators by multiplying by the LCM of the denominators, which will be: `color(purple)(3r(r-2))`

`(color(purple)(3r(r-2))xx1)/3-color(purple)(3r(r-2))xx(2 (r^2 +r-12))/(3r(r -2)) = (color(purple)(3r(r-2))xx1)/(r(r-2))`

After cancelling like factors we get:

`r(r-2) -2(r^2+r-12) = 3`

`r^2-2r -2r^2-2r+24 = 3`

`0 = r^2+4r-21" "larr` now factorise

`0 = (r+7)(r-3)`

Either of the factors could be equal to 0.

If `r +7 =0`, then `r = -7`

If `r-3 =0`, then `r =3`