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February 1, 2015

February 1, 2015

Posted by **Sandra** on Wednesday, October 17, 2012 at 12:02am.

- math -
**Anonymous**, Wednesday, October 17, 2012 at 12:35amLet x = 15$ paint

- math -
**Reiny**, Wednesday, October 17, 2012 at 9:17amNumber of $15 cans --- x

number of $20 cans --- y

15x + 20y = 320

divide by 5

**3x + 4y = 64**

we are looking at the equation of a straight line, where the solutions can only be integers, (can't buy a partial can of paint)

x-intercept = 64/3 , x < 21

y-intercept = 64/4 , y < 16

3x = 64-4y

x = (64-4y)/3

64-4y must be a multiple of 3

with y as an integer, 64-4y can only be

60 56 52 48 44 40 36 32 28 24 20 16 12 8 4

of those the only multiples of 3 are:

60 48 36 24 and 12

so, assuming they buy at least one of each can, possible solutions for (x,y) are

(20,1)

(16,4)

(12,7) ---- your answer

(8,10)

(4,13)

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