Question

PLZ HELP! WILL MARK THE BRAINIEST!!!!!!!

Answer 1

The answer is C. What I did what plug in numbers for M, N, P, R, S, and T. I got C doesn't work!

Answer 2

Answer: C.

$Mx+Ny=P\\(2M-R)x+(2N-S)y=P-2$

Step-by-step explanation:

Given: The system $Mx+Ny=P\\ Rx+Sy=T$ has the solution (1,3), where M,N,P,R,S and T are non zero real numbers.

A.

$Mx+Ny=P\\ 7Rx+7Sy=7T$

Divide 7 on both sides on the second equation , we will get

$Mx+Ny=P\\ Rx+Sy=T$

Thus, this system has solution (1,3)

B.

$(M+R)x+(N+S)y=P+T.......(1)\\\\ 7Rx+7Sy=7T................(2)$

Subtract equation (2) from equation (1), we get

$Mx+Ny=P\\ Rx+Sy=T$

Thus, this system has solution (1,3)

C.

$Mx+Ny=P...........(1)\\\\(2M-R)x+(2N-S)y=P-2T............(2)$

We can rewrite the equation (2) as

$2Mx-Rx+2Ny-Sy=P-2T............(3)$

Multiply 2 on both sides of equation (1), we get

$2Mx+2Ny=2P.........(4)$

Subtract equation (3) from (4), we get

$Rx+Sy=P+2T$

But $Rx+Sy=T$

Thus, this system does not have solution as (1,3).

D.

$\frac{M}{2}+\frac{N}{2}=\frac{P}{2}\\\\Rx+Sy=T$

Multiply 2 on both sides on the first equation , we will get

$Mx+Ny=P\\ Rx+Sy=T$

Thus, this system has solution (1,3)