You can use the constant which is multiplied with the coefficient of x in the first equation, which ends up making it negative and of equal magnitude to the coefficient of x variable in the second equation.
The first equation should be multiplied by the constant 1/12.
Given
The System of the equation is;
[tex]\rm 3x-6y=-8\\\\ \dfrac{-1}{4}x+4y=6[/tex]
How to choose what quantity to multiply the first equation?
This method is actually called the method of elimination to solve a system of linear equations.
We make one specific variable's coefficients of equal magnitude so that we can subtract or add the equations and eliminate that variable to make it easy to get the value of the other variable which will then help in getting the value of the first variable (if working in dual variable system).
Then,
[tex]\rm \dfrac{-a_2}{a_1} = \dfrac{-\dfrac{-1}{4}}{3}\\\\\dfrac{-a_2}{a_1} = \dfrac{1}{4}\times \dfrac{1}{3}\\\\\dfrac{-a_2}{a_1} = \dfrac{1}{12}[/tex]
The system of the equation we've got is;
[tex]\rm \dfrac{1}{4}x-\dfrac{1}{2}y=-\dfrac{2}{3}\\\\ \dfrac{-1}{4}x+4y=6[/tex]
Thus,
The first equation should be multiplied by the constant 1/12.
Learn more about the system of linear equations here:
brainly.com/question/13722693
