Question

If you are asked if a person has a certain number of cows and birds. They have $172$ eyes and $344$ legs. How many cows and birds does he have?

Answer 1

Hint: In this question it is given the total numbers of legs and eyes of cows and birds. A cow has four legs and two eyes. A bird has two legs and two eyes. So, we first assume total number of cows as $x$and total number of birds as $y$. Now make the equation using the information from the question. So, total numbers of legs are $4x+2y$ and total number of eyes are $2x+2y$.

Complete step-by-step answer:
Let us assume that total numbers of cows are $x$and total numbers of birds are $y$.
Now from question it is given that total numbers of eyes are $172$ and total numbers of legs are $344$
As we know that a cow has four legs and two eyes and a bird has two eyes and two legs. Hence, we can write equation for eyes as
$2x+2y=172------(1)$
And equation for legs as
$4x+2y=344-----(2)$
From equation (1) we can write
$2y=172-2x-----(3)$
Now substituting the value of $2y=172-2x$ in equation $(2)$ we can write
$ 4x+172-2x=344 \\
 \Rightarrow 2x=344-172 \\
\Rightarrow 2x=172 \\
 \Rightarrow x=\dfrac{172}{2} \\
 \Rightarrow x=86 \\ $
Hence, we get the number of cows as 86.
Now we can put the value of $x=86$ in equation $(3)$ so as to get the value of $y$
$ 2y=172-2(86) \\
\Rightarrow 2y=172-172 \\
 \Rightarrow y=0 \\ $
Hence, total numbers of birds are zero.


Note: The above question is based on a linear equation in two variables, so whenever we have to solve such equations, we have to check that its solution exists or not.
If $ax+by+c=0\text{ and }\alpha x + \beta y+ \lambda =0$ are two linear equation then
Case 1. It has unique solution if $\dfrac{a}{\alpha }\ne \dfrac{b}{\beta }$
Cae2. It has no solution if $\dfrac{a}{\alpha }=\dfrac{b}{\beta }$
Case 3. It has many solution if $\dfrac{a}{\alpha }=\dfrac{b}{\beta }=\dfrac{c}{\lambda }$