$\text{Visual approach}$

$\rule{325 pt}{0.1 pt}$

$\text{Algebraic Approach}$

$\text{Let one square be }\textcolor{#3D99F6}{\text{S}} \text{, one triangle be }\textcolor{#20A900}{\text{T}}\text{ and one circle be }\textcolor{#D61F06}{\text{C}}.$

$\begin{aligned} \begin{aligned} {\textcolor{#3D99F6}{\text{S}}}+{\textcolor{#20A900}{\text{T}}}&=18 \\ {\textcolor{#D61F06}{\text{C}}}+{\textcolor{#20A900}{\text{T}}}&=12 \\ {\textcolor{#3D99F6}{\text{S}}} +{\textcolor{#D61F06}{\text{C}}}&=16 \\ \implies 2{\textcolor{#3D99F6}{\text{S}}}+2{\textcolor{#D61F06}{\text{C}}}+2{\textcolor{#20A900}{\text{T}}}&=46 \\ \implies {\textcolor{#3D99F6}{\text{S}}}+{\textcolor{#D61F06}{\text{C}}}+{\textcolor{#20A900}{\text{T}}}&=23 \\ \end{aligned} \end{aligned}$

My approach was a little different. As long as you could find one of these values, you should be able to figure out whar s + c + t is, mainly because all three equations are more or less the sum of any of the other two, so you may as well reduce it to the sum of two values.

s + c + t is either equal to:

*c + 18

*s + 12

*t + 16

Now, if you want to look for s, you will get s=11.

(c + s) - (c + t) = 16 - 12

s - t = 4

(s-t) + (s + t) = 4 + 18

2s = 22

s = 11

So now you can find s + c + t as follows:

s + (c + t) = 11 + 12 = 23

Add all the top three scales, by adding LHS and RHS on each scales.

We get $2 (S + T + C) = 46$ , thus its half will be $\boxed{23}$

11+7+5=23 that is how I got the answer. After plugging in those numbers to the shapes.

Just cancel the 2 t's in two equations and therefore s > c and s - c = 6. With that in mind 11 + 5 = 16 for the next equation and there was a difference of 6 for the 11 and 5. So that's how I solved it.

Let's take the weight of the Square, Triangle and Circle as s, t and c respectively.

According to the question, we can make 3 formulae:

(1) $s+t=18$

(2) $c+t=12$

(3) $s+c=16$

Now we need to convert these formulae into equations in two variables for easier calculation. Starting with Eq. (1), we get

(4) $s=18-t$

Then if we substitute the value of $s$ in Eq.(3), we now get:

$18-t+c=16$

$-(t-c)=-2$

(5) $t-c=2$

Thus now, we have two main formulae to find out the answer. Now all we need to do is substitute the value of t in Eq.(2) or Eq.(5). I have done this in Eq.(5) for simplicity.

$t-c=2$

(6) $t=2+c$

Now with substituting the value of $t$ as $2+c$ in Eq.(2), we finally get:

$2+c+c=12$

$2c=10$

$c=5$

Now using the value of c as 5 in Eq. (6) and Eq.(3), we get:

$s=11$ AND $t=7$

To find the missing value, we just substitute the value of $s$ , $c$ and $t$ .

Therefore, we find the final answer as follows:

$s+c+t = 11+5+7 = 23$

I have also posted a question on Brilliant. Click here