If paint is sold for $ 15 a can and $20 a can and they sell 320 dollars worth of paint, how many cans of paint do they sell? I know it takes 7 cans of paint at 20 equaling 140 and 12 cans at 15 equaling 180 but I', stumped at how to write the equation
Let x = 15$ paint
Number 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)
To write the equation for this problem, let's denote the number of cans sold at $15 each as 'x' and the number of cans sold at $20 each as 'y'.
According to the given information, the total amount earned from selling the paint is $320. So, we can write the equation:
15x + 20y = 320
This equation represents the total earnings from selling x cans at $15 each and y cans at $20 each, which should equal $320.
Now, to find the values of x and y, we need an additional equation or information. Unfortunately, the given information about 7 cans and 12 cans doesn't directly provide a second equation. We will need more information to solve for x and y.