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I have a system of two equations with three unknowns.

$$x+y+z=7$$ $$x+2y+3z=10$$

On solving, I got the following values. $$ y = a$$ $$x = (11-a)/2$$ $$z = (3-a)/2$$

How would I go about finding the number of solutions possible for this system? Also, is there a finite set of solutions?

Note: I made the original set of equations to solve another problem in which one has to find the number of seven digit numbers with sum of digits equal to 10 and made by using the digits 1, 2, 3 only.

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  • $\begingroup$ Integral, rational, real or complex solutions ? $\endgroup$ – Dietrich Burde Mar 28 '16 at 17:37
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Assuming you were asked to solve in real numbers, there are infinitely many solutions, one for every value of $a$. Often in such situations, "how many" also means, the general solution depends on "how many" parameters. In your case the answer is one.

However, in your "original" problem you were looking for solutions where $x, y, z$ are all non-negative integers. See for which values of $a$ (also an integer) that happens. Since $y=a$ you don't have to try too many values. However, you will have an additional combinatorial problem (numbers depend on the order of digits, not just on the count of $1$'s, $2$'s and $3$'s).

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