Really hard math

A three-part problem:

1) Let x represent the difficulty of pulling two 3 1/2-year-olds in a Radio Flyer wagon.  Using the following equation, solve for x.

[A (weight of child A) + B (weight of child B)]*[the likelihood that, at any given time, either A or B will hamper your progress by grabbing the wheels and you will keep wondering what is wrong with the stupid wagon until you finally turn around at just the right time and catch them at it]² = x.

2) Let y represent the difficulty of walking to the local pond with two 3 1/2-year-olds who are not confined to a mother-controlled conveyance.  Solve for y.

[(number of times A will run ahead because THERE IS ANOTHER ANTHILL UP THERE, YAY!)+(number of times B will lag behind because Mom, the curb is a balance beam!  Look at me, I’m balancing!)]*[how loudly A & B will fight over who gets a piggyback ride home when they get tired of walking]³ + (the difficulty of providing said piggyback ride while keeping up with the other variable) = y.

3) Which is greater, x or y?


About Grape

I've got the world's best kids and husband. Great house, steady job. I'm living the American dream. The trick is to appreciate it. I'm working on that part.
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1 Response to Really hard math

  1. Most definitely y.

    Or possibly X.

    Or maybe the coefficient past participle of the anthill times the Krebs cycle time elapse for z.

    Please write a book! Please?


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