Find the value of k so that the system of equations has no solution:
3x-y-5=0; 6x-2y-k=0
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Answers
Step-by-step explanation:
3x/6x = 1/2
-y/-2y = 1/2
-5/-k = ?
let put the value of k =10
-5/-10 = 1/2...
so the value of k will be 10....
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I am give you 3-4 answers
Choose by yourself :
1.
Answer:
The value of k = 10.
Step-by-step explanation:
Given two equations, 3x – y – 5 = 0 (i.e) 3x - y = 5 …. (1)
6x - 2y – k = 0 (i.e) 6x - 2y = k …. (2)
Divide equation (2) by 2, it becomes, 3x – y = … (3)
From equation (1) and (3) LHS are equal, hence consider RHS
= 5, k = 10 for the problem to have solution,
Condition for no solution be k ≠ 10.
2.
k can be any value because in no solution condition
a1/a2=b1/b2 is not equal to c1/c2
3.
To have no solution the two equations should have the form - a1/ a2 = b1/b2 not equals to c1/c2
So the value of k is any no. except 10 .
4.
Given Equation is 3x - y - 5 = 0 = > a1 = 3, b1 = -1, c1 = -5.
Given Equation is 6x - 2y - k = 0 = > a2 = 6, b2 = -2, c2 = -k.
Now,
Given that Equation has no solutions.
= > (a1/a2) = (b1/b2) ≠ (c1/c2)
= > (3/6) = (-1/-2) ≠ (-5/-k)
=. (1/2) = (1/2) ≠ (5/k)
= > (1/2) ≠ (5/k)
= > k ≠ 10.
Therefore, k can take all real values except 10
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