Question 1184906
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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.<br>
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.<br>
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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.<br>
If y=2, then<br>
{{{2x+3(2)+2z=16}}}
{{{2x+2z=10}}}
{{{x+z=5}}}<br>
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)<br>
If y=4, then<br>
{{{2x+3(4)+2z=16}}}
{{{2x+2z=4}}}
{{{x+z=2}}}<br>
In positive integer, that gives us only one solution:
(x,y,z)=(1,4,1)<br>
So we have found a total of five solutions to the equation in positive integers.<br>
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.<br>