Given: 3x + 5y = 12 – eq 1

5x + 3y = 4 – eq 2

Here,

a₁ = 3, b₁ = 5, c₁ = - 12

a₂ = 5, b₂ = 3, c₂ = - 4

= , =

Here, ≠

∴ given system of equations have unique solutions.

Now,

In – eq 1

3x = 12 – 5y

X =

Substitute x in – eq 2

we get,

5× + 3y = 4

= 4

60 – 25y + 9y = 12

60 – 16y = 12

16y = 60 – 12

16y = 48

y = = 3

∴ y = 3

Now, substitute y in – eq 1

We get,

3x + 5×3 = 12

3x + 15 = 12

3x = 12 – 15

3x = - 3

x = = - 1

∴ x = - 1

∴ x = - 1 and y = 3