Answer: C.
[tex]Mx+Ny=P\\(2M-R)x+(2N-S)y=P-2[/tex]
Step-by-step explanation:
Given: The system [tex]Mx+Ny=P\\ Rx+Sy=T[/tex] has the solution (1,3), where M,N,P,R,S and T are non zero real numbers.
A.
[tex]Mx+Ny=P\\ 7Rx+7Sy=7T[/tex]
Divide 7 on both sides on the second equation , we will get
[tex]Mx+Ny=P\\ Rx+Sy=T[/tex]
Thus, this system has solution (1,3)
B.
[tex](M+R)x+(N+S)y=P+T.......(1)\\\\ 7Rx+7Sy=7T................(2)[/tex]
Subtract equation (2) from equation (1), we get
[tex]Mx+Ny=P\\ Rx+Sy=T[/tex]
Thus, this system has solution (1,3)
C.
[tex]Mx+Ny=P...........(1)\\\\(2M-R)x+(2N-S)y=P-2T............(2)[/tex]
We can rewrite the equation (2) as
[tex]2Mx-Rx+2Ny-Sy=P-2T............(3)[/tex]
Multiply 2 on both sides of equation (1), we get
[tex]2Mx+2Ny=2P.........(4)[/tex]
Subtract equation (3) from (4), we get
[tex]Rx+Sy=P+2T[/tex]
But [tex]Rx+Sy=T[/tex]
Thus, this system does not have solution as (1,3).
D.
[tex]\frac{M}{2}+\frac{N}{2}=\frac{P}{2}\\\\Rx+Sy=T[/tex]
Multiply 2 on both sides on the first equation , we will get
[tex]Mx+Ny=P\\ Rx+Sy=T[/tex]
Thus, this system has solution (1,3)
