Option A
The solution is [tex](0, \frac{-1}{4})[/tex]
Solution:
Given system of equations are:
3x + 6y = 1 ------ eqn 1
x - 4y = 1 ------ eqn 2
We have to find solution to system of equations
We can use substitution method
From eqn 2,
x = 1 + 4y -------- eqn 3
Substitute eqn 3 in eqn 1
3(1 + 4y) + 6y = 1
3 + 12y + 6y = 1
18y = 1 - 3
18y = -2
Divide both sides by 18
[tex]y = \frac{-2}{18}\\\\y = \frac{-1}{4}[/tex]
Substitute the above value of y in eqn 3
[tex]x = 1 + 4 \times \frac{-1}{4}\\\\x = 1-1\\\\x = 0[/tex]
Thus solution is [tex](0, \frac{-1}{4})[/tex]
