Flamingo Dance

From the question, suppose $x$ be the number of one-legged standers in the first photograph and $y$ the two-legged standers.

Then $x + 2y = 71$

And after the switch, we will obtain: $y + 2x = 58$ .

Adding both equations together, we will get:

$3x + 3y = 129$

$x+y = 43$

Therefore, even though we do not know which variable stands for which gender, we know that in total there are $\boxed{43}$ flamingos in the lake.

$M+2F=71$ ------->(1)

$2M+F=58$ ------>(2)

from (1), $M=71-2F$

then substitute in (2)

$2(71-2F)+F=58$

$F=28$

solve for $M$

$M=71-2F=71-2(28)=15$

$no.$ $of$ $flamingos=28+15=43$

Let $M=male$ and $F=female$

$M+2F=71$ $\implies$ $\color{#D61F06}\boxed{1}$

$2M+F=58$ $\implies$ $\color{#D61F06}\boxed{2}$

Solving for $M$ in $\color{#D61F06}\boxed{1}$ , we get

$M=71-2F$ $\color{#D61F06}\boxed{3}$

Substituting $\color{#D61F06}\boxed{3}$ in $\color{#D61F06}\boxed{2}$ , we get

$\boxed{F=28}$

Substituting $F=28$ to $\color{#D61F06}\boxed{3}$ , we get

$\boxed{M=15}$

$total~number~of~flamingos=28+15=$ $\color{plum}\large{\boxed{43}}$

Total number of legs 129. Divide by 3 equals 43 flamingos