No, we do not multiply the first equation by 2 and then add it to the second equation. We have to multiply the first equation by 4.
To illustrate this point, consider system A:
![\left\{ \begin{aligned}x - y &= 3 \\-2x + 4y &= -2\end{aligned}\right.](https://img.qammunity.org/2019/formulas/mathematics/high-school/n9co7wc3sm2jhnl9641r5p4nexzmeww7ba.png)
If we multiply the first equation by 2, we get
![2x - 2y = 6](https://img.qammunity.org/2019/formulas/mathematics/high-school/vno0fpfio22d8x7lmhsaxy5nool4x4qib9.png)
Now, if we replace the second equation with the sum of the second equation and
, we get
![\left\{ \begin{aligned}x - y &= 3 \\(-2x+2x) + (4y - 2y) &= -2 + 6\end{aligned}\right.](https://img.qammunity.org/2019/formulas/mathematics/high-school/xinb2i493jc8hrjm7kt82uev61qdu0n42s.png)
which simplifies to
![\left\{ \begin{aligned}x - y &= 3 \\2y &= 4\end{aligned}\right.](https://img.qammunity.org/2019/formulas/mathematics/high-school/ibwz2hmubdx13ccczvom75acmdf4fgq959.png)
This is not an equivalent system to System B. We can see that we ended up with a 2y = 4 equation.
In order to end up with a 2x = 10 second equation, we have to multiply the first equation of system A by 4 to get
![4x - 4y = 12](https://img.qammunity.org/2019/formulas/mathematics/high-school/u30fzfet8gtnbwm4urboh5w0zby8dl90un.png)
If we replace the second equation with the sum of the second equation and
, we get
![\left\{ \begin{aligned}x - y &= 3 \\(-2x+4x) + (4y-4y) &= -2 + 12\end{aligned}\right.](https://img.qammunity.org/2019/formulas/mathematics/high-school/kk62vrpom9o1cga9uxsu66dmcneq3w29sn.png)
which simplifies to
![\left\{ \begin{aligned}x - y &= 3 \\2x &= 10\end{aligned}\right.](https://img.qammunity.org/2019/formulas/mathematics/high-school/932jviamym0tx1ifc1vaig6d8yrktyp5hr.png)
Otherwise, you are correct. The solution to system B is the solution to system A. Adding an equation to another does not change the system.