The sum of $9$ shapes is $30$. There are $6$ circles and $3$ squares. What are the values of the shapes? Hello mathematicians!
Today I was caught off guard with a 4th grade question and I think I solved it. Now after taking a look at this and thinking, I was really shocked at how stupid I was, is or still am. Since I am in a college mathematics, doing functions and stuff but this had me the most stressed out as I could ever be. Because it was my little cousin's homework! I mean 14th grader can't even do 4th grades mathematics? Am I stupid?
Question: 

The sum of $9$ shapes is $30$. There are $6$ circles and $3$ squares. What are the values of the shapes? (nothing else is given)

$6$ Circles + $3$ Squares = $30$ 
My calculation:
Now I thought, oh hey it's a easy question and I did this.
$$6a + 3b = 30$$
$$6a = 30-3b$$
$$a = 5-0.5b$$
$$6 \times (5-0.5b) + 3b = 30$$
However, I get zero trying to solve for $a$ and $b$. 
So my theory is that there is no right answer, you can simply say the value for each circle is $3$, then $3 \times 6 = 18$ and $30 - 18 = 12$, so the value for each square is $4$. And you can find it with any numbers.
So is my theory right? or am I stupid because this is a very laughable equation for a college student? But for me, well I think I am stupid.
Appreciate your time in reading and solving my question.
-Thanks
 A: You get $2a + b = 10$, so you should get the solution set of $(a,b)$ to be $S = \{(1,8),(2,6),(3,4),(4,2)\}$, assuming $a$ and $b$ are positive integers. 
A: Your solution doesn't work because you isolated $a$ in the equation then substituted it back into the same equation.  The result is a tautology (not zero, that wouldn't satisfy $6a+3b=30$.)
There isn't a second equation given to do an algebraic solution.  So forget you know algebra; just start guessing and checking.  A multiple of $6$ plus a multiple of $3$ equals $30$.  I stumbled on $a=4$ and $b=2$ after a bit.  Since $6$ is a multiple of $3$, you can find more solutions by decreasing $a$ by $1$ and increasing $b$ by $2$.
A: Firstly, we can clearly divide everything by 3 to get the numbers nicer. With obviously named variables, we then wish to find solutions to $2c + s = 10$. There is a whole line of solutions (indeed, what we just wrote down the equation $s = -2c + 10$ of the line with slope $-2$ and $s$-intercept $10$). Since we want natural-number solutions, we can now just brute force it: we must have $c \leq 5$ and $s \leq 10$ (otherwise $2c + s > 10$), and we know that $s$ must be even (else $2c + s$ is odd). Further, we know that there's at most one $s$ for each $c$ (since our solutions form a line), so we have only six things to check: we just try each value of $c$ and see if there's a suitable $s$: 
If $c = 0$, then our equation is $s = 10$, which has a very obvious solution.
If $c = 1$, then our equation is $2 + s = 10$, with solution $s = 8$.
If $c = 2$, then our equation is $4 + s = 10$, with solution $s = 6$.
If $c = 3$, then our equation is $6 + s = 10$, with solution $s = 4$.
If $c = 4$, then our equation is $8 + s = 10$, with solution $s = 2$.
If $c = 5$, then our equation is $10 + s + 10$, with solution $s = 0$.  
Thus, the full set of solutions is $(c,s) \in \{(0,10),(1,8),(2,6),(3,4),(4,2),(5,0)\}$.
