The sum of $9$ shapes is $30$. There are $6$ circles and $3$ squares. What are the values of the shapes?

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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

2

There are 2 best solutions below

1
On

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 $2$.

3
On

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.