Question 156967
{{{3/y=y/(5y-12)}}} Start with the given equation




{{{3(5y-12)=y(y)}}} Cross multiply



{{{15y-36=y(y)}}} Distribute



{{{15y-36=y^2}}} Multiply



{{{15y-36-y^2=0}}} Subtract {{{y^2}}} from both sides



{{{-y^2+15y-36=0}}} Rearrange the terms.



Notice we have a quadratic equation in the form of {{{ay^2+by+c}}} where {{{a=-1}}}, {{{b=15}}}, and {{{c=-36}}}



Let's use the quadratic formula to solve for y



{{{y = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{y = (-(15) +- sqrt( (15)^2-4(-1)(-36) ))/(2(-1))}}} Plug in  {{{a=-1}}}, {{{b=15}}}, and {{{c=-36}}}



{{{y = (-15 +- sqrt( 225-4(-1)(-36) ))/(2(-1))}}} Square {{{15}}} to get {{{225}}}. 



{{{y = (-15 +- sqrt( 225-144 ))/(2(-1))}}} Multiply {{{4(-1)(-36)}}} to get {{{144}}}



{{{y = (-15 +- sqrt( 81 ))/(2(-1))}}} Subtract {{{144}}} from {{{225}}} to get {{{81}}}



{{{y = (-15 +- sqrt( 81 ))/(-2)}}} Multiply {{{2}}} and {{{-1}}} to get {{{-2}}}. 



{{{y = (-15 +- 9)/(-2)}}} Take the square root of {{{81}}} to get {{{9}}}. 



{{{y = (-15 + 9)/(-2)}}} or {{{y = (-15 - 9)/(-2)}}} Break up the expression. 



{{{y = (-6)/(-2)}}} or {{{y =  (-24)/(-2)}}} Combine like terms. 



{{{y = 3}}} or {{{y = 12}}} Simplify. 



So the answers are {{{y = 3}}} or {{{y = 12}}}