Question 149800
I can't tell if you mean {{{1/(4(8y+4))}}} or {{{(8y+4)/4}}}

Parens are cheap - no need to be stingy with them.

I'll assume you mean
{{{(1/4)(8y+4)-17=(1/2)(4y-8)}}}
{{{(2y+1) - 17 = (2y - 4)}}}
{{{-16 = -4}}} 
This is not true. So my original assumption on the problem must have been wrong.

Now let's assume you meant

{{{(1/(4(8y+4)))-17=(1/(2(4y-8)))}}}

{{{(1/(16(2y+1)))-17=(1/(8(y-2)))}}}  Multiply by 16

{{{(1/(2y+1))-272=(2/(y-2))}}}

{{{(1/(2y+1))-272 -(2/(y-2))= 0}}} Get a common denominator
{{{( (y-2) -272(y-2)(2y+1) - 2(2y+1) ) / ((y-2)(2y+1))= 0}}} Don't forget the denominator. We'll need to check the answers we get to make sure they don't result in a 0 in the denominator.

{{{( (y-2) -272(y-2)(2y+1) - 2(2y+1) )= 0}}}
{{{ y - 2 - 272(2y^2 -3y - 2) - 4y - 2 = 0 }}}
{{{-544y^2 + 816y + 544 -3y -4 = 0}}}
{{{544y^2 -813y -540 = 0}}}
*[invoke quadratic "x", 544, -813, -540]

The solutions are 'close to the zeroes' in the original denominator, but not equal