[Math]\frac{10}{r-3}+\frac{4}{3-r}=6[/tex]

[MATh]\frac{10(3-r)}{(r-3)(3-r)}+\frac{4(r-3)}{(r-3)(3-r)}=\frac{6(r-3)(3-r)}{(r-3)(3-r)}[/tex]

Is it correct to this point? The latest fraction is 0. Then 30 -10r +4r -12 = 0, r= 3

My book says r=4 so at some point I must have made a mistake.