The total number of lions and peacocks in a certain zoo is 50 . The total number of their legs is 140 . Then find the number of lions and peacocks in the zoo.

Let the number of lions in the zoo be ' $x$ ' and the number of peacocks be ' $y$ '.
According to the first condition,
the total number of lions and peacocks is 50 .
$$\therefore x+y=50\text{...(i)}$$
Lion has 4 legs and Peacock has 2 legs.
According to the second condition,
the total number of their legs is 140 .
$$\therefore 4x+2y=140$$
Dividing both sides by 2 ,
$$2x+y=70\dots \text{(ii)}$$
Subtracting equation (i) from (ii),
$$\begin{array}{cc}2x+y& =70\\ x+y& =50\\ --& -\\ x=20& \end{array}$$
Substituting $x=20$ in equation (i),
$$\begin{array}{cc}& x+y=50\\ & \therefore 20+y=50\\ & \therefore y=50-20=30\end{array}$$
$\therefore$ The number of lions and peacocks in the zoo are 20 and 30 respectively.