Determine the value of k so that the following linear equations have no solution: (3k + 1) x + 3y − 2 = 0 (k^2 + 1) x + (k − 2) y − 5 = 0

xcopyv4n

xcopyv4n

Answered question

2023-02-28

Determine the value of k so that the following linear equations have no solution: ( 3 k + 1 ) x + 3 y 2 = 0 ( k 2 + 1 ) x + ( k 2 ) y 5 = 0

Answer & Explanation

Nhluvukoj6m

Nhluvukoj6m

Beginner2023-03-01Added 6 answers

The given equational system is
(3k + 1) x + 3y − 2 = 0
( k 2 + 1 ) x + ( k 2 ) y 5 = 0
Here, a1 = 3k + 1, b1 = 3, c1 = −2
a2 = k2 + 1, b2 = k − 2, c2 = −5
There is no answer to the given system of equations.
a1a2=b1b2c1c23k+1k2+1=3k-2-2-53k+1k2+1=3k-2and3k-225
Now,
3k+1k2+1=3k-23k+1k-2=3k2+13k2-5k-2=3k2+3-5k=5
k=-1
When k = −1,
3k-2=3-1-2=3-3=-1
Thus, for k = −1, 3k-225
Therefore, the given system of equations has no solution when k = −1.

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