Question 55198
{{{ (4y+2)(1/6)y }}}
First thing here is to realize there are actually three fractions here.
Remeber anything over 1 equals itself.  So re-write the 4y + 2 and the y as fractions.
{{{ ((4y+2)/1)(1/6)(y/1) }}}
since we are multiplying, there is no need for common denominators,  just place all numerators and denominators into a single fraction
{{{ ((4y+2)(1)(y))/((1)(6)(1)) }}}
With this fraction I would pull a 2 out of the 4y + 2 so that we can reduce.
{{{ (2(2y+1)(1)(y))/((1)(6)(1)) }}}
reduce
{{{ ((2y+1)(1)(y))/((1)(3)(1)) }}}
multiply
{{{ (2y^2+y)/3 }}}