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larisa86 [58]
3 years ago
6

Need help what are the classifications of each system

Mathematics
1 answer:
ratelena [41]3 years ago
5 0
You can graph them on Desmos, then use this attachment to find out the classification.

In the first system, they're the same line and have infinitely many solutions, therefore the answer is consistent, dependent.

In the second system, they're parallel lines with no solution, therefore the answer is inconsistent.

In the third system, they intersect and they have one solution, therefore the answer is consistent, independent.

In the fourth system, they're the same line and have infinitely many solutions, therefore the answer is consistent, dependent.

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Step-by-step explanation:

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Answer:

<u>Options 1 and 3</u>

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.

<u>System A)</u>  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)

<u>System B)</u> 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)

<u>System C)</u> (T-P)x + (U-Q)y = V-R  ⇒(5)

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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)

<u>System D)</u> (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) <u>OR </u>(T-P) and (Q-U)

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

<u>System E)</u> (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:

<u>Options 1 and 3</u>

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