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:
[tex] \left\{ \begin{aligned}x - y &= 3 \\-2x + 4y &= -2\end{aligned}\right. [/tex]
If we multiply the first equation by 2, we get
[tex] 2x - 2y = 6 [/tex]
Now, if we replace the second equation with the sum of the second equation and [tex] 2x - 2y = 6 [/tex], we get
[tex] \left\{ \begin{aligned}x - y &= 3 \\(-2x+2x) + (4y - 2y) &= -2 + 6\end{aligned}\right. [/tex]
which simplifies to
[tex] \left\{ \begin{aligned}x - y &= 3 \\2y &= 4\end{aligned}\right. [/tex]
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
[tex] 4x - 4y = 12 [/tex]
If we replace the second equation with the sum of the second equation and [tex] 4x - 4y = 12 [/tex], we get
[tex] \left\{ \begin{aligned}x - y &= 3 \\(-2x+4x) + (4y-4y) &= -2 + 12\end{aligned}\right. [/tex]
which simplifies to
[tex] \left\{ \begin{aligned}x - y &= 3 \\2x &= 10\end{aligned}\right. [/tex]
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.
