7,041 views
22 votes
22 votes

Consider the following set of equations

  • $x+2y=5$
  • $4x+8y=12$
  • $3x+6y+3z=15$

This set

  1. has unique solution
  2. has no solution
  3. has finite number of solutions
  4. has infinite number of solutions

3 Answers

Best answer
41 votes
41 votes

There are no solutions.

If we multiply $1^{\text{st}}$ equation by $4,$ we get

  • $4x + 8y = 20$

But $2^{\text{nd}}$ equation says

  • $4x + 8y = 12$

Clearly, there can not be any pair of $(x,y),$ which satisfies both equations.

Correct Answer: B.

edited by
14 votes
14 votes

First transformed the matrix into echelon form and rest can seen from the pic.

0 votes
0 votes

(B) Thus, The R1 is Dependent Upon R2 And as per calculation |A| = 0 So, It is linearly dependent then it is Having No solution

Answer:

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