Total price of $6$ shirts and $7$ trousers is $\text{Rs. }4,130$ while that of $4$ shirts and $9$ trousers is $\text{Rs. }4,270$. Find out the price of $3$ shirts.

Let the cost of $1$ shirt be represented by variable $S$ and $1$ trouser by $T$.
From the data given, we form two simultaneous linear equations:

1. $6S+7T=4130$
2. $4S+9T=4270$

To solve by elimination, multiply equation (1) by $9$ and equation (2) by $7$:

$$54S+63T=37170\text{— (3)}$$

$$28S+63T=29890\text{— (4)}$$

Subtract equation (4) from equation (3):

$$(54S-28S)=37170-29890$$

$$26S=7280$$

$$S=\frac{7280}{26}=280$$

The single item price of $1$ shirt is $\text{Rs. }280$. Therefore, calculate the value for $3$ shirts:

$$\text{Price of 3 shirts}=3\times 280=\text{Rs. }840$$