# math

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Dave & Jim separated the peanuts they had collected into two piles. One pile was for doubles (2 peanuts inside) and the other for triples (3 peanuts inside). When they finished shelling peanuts, Dave & Jim had 35 shells and 78 peanuts. How many shells were double?

Looks like a pretty fun problem to me. What have you tried so far?
Ok, well let's try to formulate this into a system of 2 equations in 2 unknowns.
Let x denote the number of peanuts with 2 inside and y the number with 3 inside.
Then we know that x+y=the number of peanuts = 35
We also know that 2*(peanuts with 2 inside) + 3*(peanuts with 3 inside) = total peanuts = 78 Thus our system is
x + y = 35
2x + 3y = 78
Now add -2 times the first equation to the second to get
-2x + 2x + -2y + 3y = -2*35 + 78 or
y = 8
You should be able to finish and verify the solution for the number of double shells.
Did you follow all the steps I did?

Let X = # double shells.
and Y = # triple shells.
========================
You have two equations and two unknowns.
X + Y = 35 (total shells)
2X + 3Y = 78 (total peanuts)

Solve for X and Y. I hope this helps. Post your work if you get stuck and require more assistance.

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