Question

1. Find the value of k for which the system of equations kx +3y = k-3, 12x + ky =k has a unique solution.
2. Find the values of k for which the pair of linear equations (3k+1)x+3 y -2=0 and ( k2 +1)x +(k -2)y = 5 has no solution.
3. Find the value of k if the following system of equations has infinitely many solutions:
1)( k-1)x - y = 5, ( k+1) x +(1-k)y = 3k +1
2) 2x + ( k-2) y = k, 6 x +(2k-1)y = 2k +5
4. If 2x + 3y = 5, create an equation which is (i) intersecting (ii) parallel to the given line.( Put your thinking cap to answer this question)

Plzzz answer these questions if you know step by step . I will mark u brainliest .

Answer 1

Given :  System of Equation  kx +3y = k-3, 12x + ky =k

To find :  Value of k for unique & no solution

Step-by-step explanation:

kx +3y = k-3,

12x + ky = k

has a unique solution.

=> k/12  ≠ 3/k  ≠ (k - 3)/k

=> k  ≠ ± 6

k² ≠ 12k - 36

=> (k - 6)² = 0

=> k ≠ 6

3 ≠ k - 3

=> k ≠ 6

All values of k  other than ± 6 has unique solution

(3k+1)x+3 y -2=0 and

( k² +1)x +(k -2)y = 5 has no solution.

=> (3k + 1)/(k² + 1)  =  3/(k - 2)  ≠ 2/5

=> 3k² -5k - 2 = 3k² + 3

=> k = - 1

3/(k - 2)   ≠ 2/5

3/(-3) ≠ 2/5

=> k = -1   no solution

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