Answer:
We can only be certain that <em>a</em> weighs 12.
There are infinitely many possiblities for <em>b</em> and <em>c</em>.
Step-by-step explanation:
We have the equation:
![a+b+c+a+c+b+a+c=12+a+a+b+b+c+c+c](https://tex.z-dn.net/?f=a%2Bb%2Bc%2Ba%2Bc%2Bb%2Ba%2Bc%3D12%2Ba%2Ba%2Bb%2Bb%2Bc%2Bc%2Bc)
Each variable indicates a weight.
We would like to determine the weights of each variable (if possible).
First, we can rearrange the equation to acquire:
![(a+a+a)+(b+b)+(c+c+c)=12+(a+a)+(b+b)+(c+c+c)](https://tex.z-dn.net/?f=%28a%2Ba%2Ba%29%2B%28b%2Bb%29%2B%28c%2Bc%2Bc%29%3D12%2B%28a%2Ba%29%2B%28b%2Bb%29%2B%28c%2Bc%2Bc%29)
We can combine like terms:
![3a+2b+3c=12+2a+2b+3c](https://tex.z-dn.net/?f=3a%2B2b%2B3c%3D12%2B2a%2B2b%2B3c)
Notice that both sides have 2<em>b</em> and 3<em>c</em>. Therefore, it is possible for us to cancel them since each nullify the other side. So, we will subtract 2<em>b</em> and 3<em>c</em> from both sides. This yields:
![3a=12+2a](https://tex.z-dn.net/?f=3a%3D12%2B2a)
Therefore, we can solve for <em>a</em>. Subtract 2<em>a</em> from both sides:
![a=12](https://tex.z-dn.net/?f=a%3D12)
Hence, the weight of <em>a</em> is 12.
Using the newly acquired information, we can go back to our simplified equation:
![3a+2b+3c=12+2a+2b+3c](https://tex.z-dn.net/?f=3a%2B2b%2B3c%3D12%2B2a%2B2b%2B3c)
Since <em>a</em> is 12:
![3(12)+2b+3c=12+2(12)+2b+3c](https://tex.z-dn.net/?f=3%2812%29%2B2b%2B3c%3D12%2B2%2812%29%2B2b%2B3c)
Evaluate:
![36+2b+3c=12+24+2b+3c](https://tex.z-dn.net/?f=36%2B2b%2B3c%3D12%2B24%2B2b%2B3c)
Simplify:
![36+2b+3c=36+2b+3c](https://tex.z-dn.net/?f=36%2B2b%2B3c%3D36%2B2b%2B3c)
We can subtract 36 from both sides:
![2b+3c=2b+3c](https://tex.z-dn.net/?f=2b%2B3c%3D2b%2B3c)
As you can see, this is a true statement.
Since this is a true statement, there are infinitely many possible values for <em>b</em> and <em>c</em>.
Therefore, the only weight we are <em>certain</em> of knowing is weight <em>a</em> weighing 12.