We have been given the first equation as [tex]2x+4y=0[/tex]
Now, we find the second equation from the given table. From the table, we have two points [tex](-1,8),(3,-4)[/tex]
The slope of the line is given by
[tex]m=\frac{y_2-y_1}{x_2-x_1}\\ \\ m=\frac{-4-8}{3+1}\\ \\ m=\frac{-12}{4}\\ \\ m=-3[/tex]
Thus, the equation of the line is given by
[tex]y-y_1=m(x-x_1)\\ y-8=-3(x+1)\\ y-8=-3x-3\\ y=-3x+5[/tex]
Therefore, we have the system of equations
[tex]2x+4y=0...............(1)\\ y=-3x+5................(2)[/tex]
Substitute the value of y in equation 1, we get
[tex]2x+4(-3x+5)=0\\ 2x-12x+20=0\\ -10x=-20\\ \\ x=\frac{-20}{-10} \\ \\ x=2[/tex]
The x value is given by 2
