Question 204859
Well, I tried 7 cookies and that seems to work, and I know that this is not an algebraic answer, but I think that it can be worked out.
Start with 7.
To Jodie, half of 7 is 3.5 plus 0.5 = 4 so there are 3 cookies left.
To Beth, half of 3 is 1.5 pus 0.5 = 2, so now there is 1 cookie left.
To Michelle, half of 1 is 0.5 plus o.5 = 1, no more cookies left and no cookies needed to be broken.
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How did I get 7?  Read on!
Let x = the original number of whole cookies on the plate.
To Jodie, you gave: 
{{{(x/2 + 1/2)}}} what was left after that can be expressed by:
{{{x-(x/2+1/2)}}} let's call this quantity y, so...
{{{y = (x-(x/2+1/2))}}} 
To Beth, you gave:
{{{(y/2+1/2)}}} what was left after that can be expressed by:
{{{y-(y/2+1/2)}}} let's call this quantity z, so...
{{{z = (y-(y/2)+1/2)}}}
To Michelle, you gave:
{{{z/2+2/2}}} what was left after that can be xepressed by:
{{{z-(z/2+1/2)}}}...and, because there were no cookies left, we'll call this zero, so...
{{{z-(z/2+1/2) = 0}}} Now we can solve this for z: First simplify it.
{{{(2z-z-1)/2 = 0}}} Multiply both sides by 2.
{{{z-1 = 0}}} Add 1 to both sides.
{{{highlight(z = 1)}}} Now we know z, we can solve for y.
{{{1 = (y-(y/2+1/2))}}}
{{{(2y-y-1)/2 = 1}}}
{{{y-1 = 2}}}
{{{highlight(y = 3)}}} Finally, we can solve for x, the original number of cookies.
{{{3 = (x-(x/2+1/2))}}}
{{{(2x-x-1)/2 = 3}}}
{{{x-1 = 6}}}
{{{highlight(x = 7)}}}