Answer 1

The given equations can be written in the matrix form as:

`[(3,-1),(4,1)][("x"),("y")]=[(1),(6)]`

By 4R₁ and 3R₂, we get,

`[(12,-4),(12,3)][("x"),("y")]=[(4),(18)]`

By R₂ – R₁, we get,

`[(12,-4),(0,7)][("x"),("y")]=[(4),(14)]`

∴ `[(12"x"-4"y"),(0+7"y")]=[(4),(14)]`

By equality of matrices,

12x − 4y = 4 ........(1)

7y = 14 .............(2)

From (2), y = 2

Substituting y = 2 in (1), we get,

12x − 8 = 4

∴ 12x = 12

∴ x = 1

Hence, x = 1, y = 2 is the required solution.