Question

PLEASE HELP , TRYING TO BE ON HONOR ROLL !

Answer 1

Option A

The solution is $(0, \frac{-1}{4})$

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

$y = \frac{-2}{18}\\\\y = \frac{-1}{4}$

Substitute the above value of y in eqn 3

$x = 1 + 4 \times \frac{-1}{4}\\\\x = 1-1\\\\x = 0$

Thus solution is $(0, \frac{-1}{4})$