The correct option will be :
A. To get system B, the second equation in system A was replaced by the sum of that equation and the first equation multiplied by 5. The solution to system B will be the same as the solution to system A.
System A :
-x - 2y = 7 ..........(1)
5x - 6y = -3 ..........(2)
If we multiply the first equation by 5, then we will get :
5(- x - 2y) = 5(7)
⇒ - 5x - 10y = 35 ........... (3)
Now the Sum of equation (2) and equation (3) is:
![5x -6y = -3 \\ -5x -10y =35](https://tex.z-dn.net/?f=%205x%20-6y%20%3D%20-3%20%5C%5C%20-5x%20-10y%20%3D35%20)
⇒ Sum :
Now if we replace the second equation in system A with this
, then we will get the system B.
Solution of system B:
![-x -2y =7 \\ \\ -16y= 32](https://tex.z-dn.net/?f=%20-x%20-2y%20%3D7%20%5C%5C%20%5C%5C%20-16y%3D%2032%20%20)
First we will take the second equation as there is only one variable 'y'. So, we will solve that equation for 'y'
![-16y= 32\\\\ \frac{-16y}{-16} = \frac{32}{-16} \\ \\ y= -2](https://tex.z-dn.net/?f=%20-16y%3D%2032%5C%5C%5C%5C%20%5Cfrac%7B-16y%7D%7B-16%7D%20%3D%20%5Cfrac%7B32%7D%7B-16%7D%20%5C%5C%20%5C%5C%20y%3D%20-2%20)
Now for solving 'x', we will plug y= -2 into the first equation
![-x-2y= 7\\ \\ -x -2(-2) = 7\\ \\ -x +4 = 7 \\ \\ -x+4 -4 = 7-4 \\ \\ -x =3\\ \\ \frac{-x}{-1}= \frac{3}{-1}\\ \\ x= -3](https://tex.z-dn.net/?f=%20-x-2y%3D%207%5C%5C%20%5C%5C%20-x%20-2%28-2%29%20%3D%207%5C%5C%20%5C%5C%20-x%20%2B4%20%3D%207%20%5C%5C%20%5C%5C%20-x%2B4%20-4%20%3D%207-4%20%5C%5C%20%5C%5C%20-x%20%3D3%5C%5C%20%5C%5C%20%5Cfrac%7B-x%7D%7B-1%7D%3D%20%5Cfrac%7B3%7D%7B-1%7D%5C%5C%20%20%20%5C%5C%20x%3D%20-3%20)
So, the solution of system B is (-3, -2), that means the solution of both systems are same.