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
Liliyana25:
R u sure it is (k-3)x+3y=K.....or 0..
Answers
Answered by
3
(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.
====================
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.
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.
====================
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.
click on thanks button above ;;select best answer
Similar questions