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Options 1 and 3

Step-by-step explanation:

We should know that, the system of linear equations can be treated as matrices, i.e: we can modify or make any operations provide that we must apply the same operation for all terms of each equation.

Given:  the solution for the following system is (2,9)

Px + Qy = R  ⇒(1)

Tx + Uy = V  ⇒(2)

We will check which system of equation has the same solution.

System A)  Px + Qy = R

                  (P+T)x + (Q+U)y = R+V  ⇒(3)

So, By summing (1) and (2) we will get the equation (3)

So, system A has the same solution (2,9)

System B) Px + Qy = R

                 (P+2T)x + (Q+2U)y = R-2V  ⇒(4)

By multiplying equation (2) by 2 and add with equation (1), we will get:

 (P+2T)x + (Q+2U)y = R+2V

Which is not the same as equation (4)

So, system B has not the same solution (2,9)

System C) (T-P)x + (U-Q)y = V-R  ⇒(5)

                  Tx + Uy = V  

By multiplying equation (1) by -1 and add with equation (2), we will get the equation (5)

So, system C has the same solution of (2,9)

System D) (T-P)x + (Q+U)y = V-R  ⇒(6)

                  Tx + Uy = V  

We cannot get equation (6) by the same operation over equation (1)

Note the coefficient of x and y⇒ (T-P) and (Q+U)

They must be (T+P) and (Q+U) OR (T-P) and (Q-U)

So, system D has not the same solution of (2,9)

System E) (5T-P)x + (5U-Q)y = V-5R ⇒ (6)

                  Tx + Uy = V  

By subtract equation (1) from 5 times equation (2), we will get:

(5T-P)x + (5U-Q)y = 5V-R

Which is not the same as equation (6)

So, system E has not the same solution (2,9)

As a conclusion, the systems which have the same solution are:

Options 1 and 3

1. credits

2. dual - aspect

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