Given the following equations:
With the following data:
Could you find the answer?
Note: the icons resemble the animals, you don't have to look at (for example) the colors how the icons are painted, or other details.
To show these cute little animals, I used the icons of the following website: Animals icons created by Freepik - Flaticon
Based on the chosen animals, it seems likely that two values play in: the number of legs and the number of wings of each animal. Using this hypothesis we can express each animal as a value using the placeholders $A$ and $B$ for the value of each leg and wing respectively:
\begin{array}{lll}\text{Elephant}=4A\qquad & \text{Dog}=4A\qquad & \text{Eagle} = 2A+2B \\ \text{Fly}=6A+2B & \text{Dragonfly} = 6A+4B & \text{Swallow} = 2A+2B \\ \text{Spider} = 8A & \text{Snake} = 0 \end{array} And then validating that each of the equations hold:
Horizontal equations (top-to-bottom): \begin{align}4A + 4A + (2A + 2B) &= 8A + (2A + 2B) &\checkmark \\(6A + 2B) + (6A + 4B) - (2A + 2B) &= 2(8A) &\Rightarrow B=\frac{3}{2}A \\8A + 0 - 4A &= 4A &\checkmark \\(2A+2B) + (3(4A)+4A) + \frac{1}{2}(6A+4B) &= 2(6A+2B) + (2A+2B) + 4A &\Rightarrow B=\frac{3}{2}A \\\end{align} Vertical equations (left-to-right): \begin{align}4A + (6A+2B) - 8A &= 2A+2B &\checkmark \\4A + (6A+4B) + 0 &= 3(4A) + 4A &\Rightarrow B=\frac{3}{2}A \\(2A+2B) + (2A+2B) - 4A &= \frac{1}{2}(6A+4B) &\Rightarrow B=\frac{3}{2}A \\(8A + (2A+2B)) + 2(8A) + 4A &= (6A+2B) + 2(6A+4B) &\Rightarrow B=\frac{3}{2}A\end{align} As we can see, our hypothesis not only holds for every provided equation - it also gives us a value for $B=\frac{3}{2}A$. Given the value of $\text{Elephant} = 4A=8$, we can therefore calculate that $$A = 2, \quad B = 3$$ Giving us a final answer for the pegasus of $$\text{Pegasus} = 4A + 2B = 4\cdot 2 + 2\cdot 3 = 14$$
