The system mx+ny=p has the solution (1,3), where rx+sy=t m, n,p, r,s and t are non zero real numbers. which of the following system

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The system of equations ( mx + ny = p ) has the solution ( (1, 3) ), where ( m, n, p, r, s, ) and ( t ) are non-zero real numbers. Which of the following systems would not have ( (1, 3) ) as a solution?

Vickie Rippin DVM · Accepted Answer

Given:The system Mx+Ny=P has a solution (1,3) where Rx+Sy=T; M, N, P, R,S and T are non-zero real numbers.Solve for M, N, R, P, S, T:M +3N = PR + 3S = T The given choices should simplify to the equations above. A) Mx +Ny = P     7Rx + 7Sy = 7T     7(Rx + Sy) = 7T     Rx + Sy = Tremarks: CORRECTB) (M+R)x + (N+S)y = P +  T    Rx + Sy = T     Mx + Rx + Ny + Sy = P + T     Mx + Ny + T = P + T     Mx + Ny = Premarks: CORRECTC) Mx + Ny = P    (2M - R)x + (2N - S)y = P - 2T    2Mx - Rx + 2Ny - Sy = P - 2T    2(Mx + Ny) - (Rx + Sy) = P - 2T     2P - (Rx + Sy) = P - 2Tremarks: INCORRECT

Anonymous · Answer

The answer is definitely C.  My friend took the test and got it correct.

Anonymous · Answer

C. Step-by-step explanation:Given: The system has the solution (1,3), where M,N,P,R,S and T are non zero real numbers. A. Divide 7 on both sides on the second equation , we will get Thus, this system has solution (1,3)B. Subtract equation (2) from equation (1), we getThus, this system has solution (1,3) C.We can rewrite the equation (2) asMultiply 2 on both sides of equation (1), we getSubtract equation (3) from (4), we getBut Thus, this system does not have solution as (1,3). D.Multiply 2 on both sides on the first equation , we will get Thus, this system has solution (1,3)

Anonymous · Answer

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!

Aron Pollich

The system mx+ny=p has the solution (1,3), where rx+sy=t m, n,p, r,s and t are non zero real numbers. which of the following system

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