Question 24549
{{{2/(y+3)+ 3/(y-4)= 5/(y+6)}}}


OK...need to get rid of the denominators..the fractions. To do this, we multiply every term by (y+3)(y-4)(y+6) and cancel out the ones we can, to leave:


{{{(2(y+3)(y-4)(y+6))/(y+3)+ (3(y+3)(y-4)(y+6))/(y-4)= (5(y+3)(y-4)(y+6))/(y+6)}}}
{{{2(y-4)(y+6)+ 3(y+3)(y+6) = 5(y+3)(y-4)}}}
{{{2(y^2+2y-24)+ 3(y^2+9y+18) = 5(y^2-y-12)}}}
{{{2y^2+4y-48+ 3y^2+27y+54 = 5y^2-5y-60}}}
{{{5y^2+31y+6 = 5y^2-5y-60}}}
31y+6 = -5y-60
36y + 6 = -60
36y = -66
y = -66/36
y = -11/6


there you go!


jon.