Answer:
1 and 3
Step-by-step explanation:
Given the system of two equations
[tex]\left\{\begin{array}{l}Px+Qy=R\\ \\Tx+Uy=V\end{array}\right.[/tex]
When you add, subtract, multiply the equations of the system, you get equivalent system of equations (with the same solution).
1. Add these two equations to get the system
[tex]\left\{\begin{array}{l}Px+Qy=R\\ \\Px+Tx+QY+Uy=R+V\end{array}\right.\Rightarrow \left\{\begin{array}{l}Px+Qy=R\\ \\(P+T)x+(Q+U)y=R+V\end{array}\right.[/tex]
Hence, option 1 is correct.
2. Multiply the second equation by 2 and add two equations:
[tex]\left\{\begin{array}{l}Px+Qy=R\\ \\2Tx+2Uy=2V\end{array}\right.\Rightarrow \left\{\begin{array}{l}Px+Qy=R\\ \\(P+2T)x+(Q+2U)y=R+2V\end{array}\right.[/tex]
Since [tex]R+2V\neq R-2V,[/tex] this option is false
3. Subtract two equations:
[tex]\left\{\begin{array}{l}Tx-Px+Uy-Qy=V-R\\ \\Tx+Uy=V\end{array}\right.\Rightarrow \left\{\begin{array}{l}(T-P)x+(U-Q)y=V-R\\ \\Tx+Uy=V\end{array}\right.[/tex]
Option 3 is correct.
4. Since [tex]U-Q\neq Q+U[/tex], option 4 is false
5. Multiply the second equation by 5 and subtract the first equation:
[tex]\left\{\begin{array}{l}Px+Qy=R\\ \\5Tx+5Uy=5V\end{array}\right.\Rightarrow \left\{\begin{array}{l}(5T-P)x+(5U-Q)y=5V-R\\ \\Tx+Uy=V\end{array}\right.[/tex]
Since [tex]V-5R\neq 5V-R[/tex], this option is false.
