Question 162854
{{{(3y)/((3y-6))}}} + {{{8/((y^2-4))}}} = {{{(2y)/((2y+4))}}}
:
First, recognize that you can factor out 3 and 2 in the denominators
Note that y^2-4 is the "difference of squares"
{{{(3y)/(3(y-2))}}} + {{{8/((y-2)(y+2))}}} = {{{(2y)/(2(y+2))}}}
:
Note that the 3's and 2's will cancel leaving:
{{{y/((y-2))}}} + {{{8/((y-2)(y+2))}}} = {{{y/((y+2))}}}
:
Multiply equation by (x-2)(x+2), results:
y(y+2) + 8 = y(y-2)
:
y^2 + 2y + 8 = y^2 - 2y
:
arrange the variables on the left, and numbers on the right
y^2 - y^2 + 2y + 2y = -8
:
4y = -8
y = {{{(-8)/4}}}
y = -2
This should be the solution, but when we substitute -2 for y in the original
equation, we end up with 2 denominators of 0, which we can't have
therefore we have to say, " No solution"