Question 1184906: 2x+3y+2z=16
Find the value of x,y and z.
Found 2 solutions by Theo, greenestamps:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the sum of the pieces must be equal to 16.
the solution needs to be unique is what i think you are looking for.
we'll work with y first and then worry about x and z.
the equation is 2x + 3y + 2z = 16
3y can either be 3 or 6 or 9 or 12 or 15.
3y = 3 or 9 are no good because 16 - 3 = 13 and 16 - 9 = 7, neither or which are divisible by 2.
3y = 15 is no good because you only have 1 left for 2x and 2z which is not enough, and it is not divisible by 2 anyway.
your options are therefore 3y = 6 or 3y = 11
if 3y = 6, then you get 2x + 6 + 2z = 16
subtract 6 from both sides of the equation to get:
2x + 2z = 10
your possible values are:
2x = 2, 2z = 8
2x = 4, 2z = 6
2x = 6, 2z = 4
2x = 8, 2z = 2
these are all possibilities, but there is not one unique solution.
if 3y = 12, then you get 2x + 12 + 2z = 16
subtract 12 from both sides of the equation to get:
2x + 2z = 4
2x can be equal to 2
2z can be equal to 2
this, of course, assumes that none of 2x or 3y or 2z can be equal to 0.
you get a unique solution when x = 1 and y = 4 and z = 1
you then get:
2x + 3y + 2z = 2 + 12 + 2 = 16.
i would go with that.
x = 1
y = 4
z = 1
2x + 3y + 2z = 2 + 12 + 2 = 16
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The equation obviously has an infinite number of solutions; choose any values you want for two of the variables and determine the corresponding value of the third.
So for the problem to make sense, we need to assume we are looking for non-negative integer solutions, or perhaps for positive integer solutions. Allowing a value of 0 for any of the variables produces a large number of additional solutions, so we will assume we are looking for solutions in positive integers.
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2x, 2z, and 16 are all even; that means 3y must be even; and that means y most be even. Since the sum is 16, that means the only possible values for y are 2 and 4.
If y=2, then
In positive integers, that gives us 4 solutions:
(x,y,z)=(1,2,4)
(x,y,z)=(2,2,3)
(x,y,z)=(3,2,2)
(x,y,z)=(4,2,1)
If y=4, then
In positive integer, that gives us only one solution:
(x,y,z)=(1,4,1)
So we have found a total of five solutions to the equation in positive integers.
Since there are no instructions in the statement of the problem that let us choose one of the five solutions, all five of them should be considered acceptable.
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