# Determine the value of k for which the following systemic linear equations has infinite number of solutions (k-3)x+3y=k, kx+ky=12

1
by anudeepd2
R u sure it is (k-3)x+3y=K.....or 0..

2015-10-17T17:46:39+05:30

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(k - 3)  x  + 3 y = k     --- (1)
k  x  + k  y = 12        --- (2)

(1) * k  - (2) * 3  =>
k(k - 3) x  - 3 k x = k² - 36
x =  (k² - 36) / (k² - 6 k)  =  (k - 6) (k + 6) / [ k (k - 6) ]

x = (k + 6) / k        if  k ≠ 6
There is only one solution.

If k = 6, then:  the equations are :
3 x + 3 y = 6
6 x + 6 y = 12
Then there are infinite solutions.

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The determinant of the following matrix of the coefficients of x and y:

|  k-3      3    |
|  k        k    |

is equal to:    k² - 3 k - 3 k     =  k²  - 6 k
Determinant = 0  for   k = 0  and  k = 6.

check  if the equations given have  infinite solutions for these two values of k.

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