Question 851996

Need help solving this equation.     

{{{4/r=(5r-5)/(r^2-3r)}}}
<pre>
{{{4/r = (5r - 5)/(r^2 - 3r)}}}_______{{{r<> 0}}}   AND    {{{r <> 3}}}
{{{r(5r - 5) = 4(r^2 - 3r)}}} ------- Cross-multiplying
{{{5r^2 - 5r = 4r^2 - 12r}}}
{{{5r^2 - 4r^2 - 5r + 12r = 0}}}
{{{r^2 + 7r = 0}}}
{{{r(r + 7) = 0}}}
r = 0 (ignore)       OR      r + 7 = 0____{{{highlight_green(r = - 7)}}}