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
