For what value of k does the system of equation kx+2y-1=0 and 5x-3y+2=0 has no solution
Answers
Answered by
38
No solution means it is parallel lines
a1÷ a2= b1÷ b2
K/5 = 2/-3
K=5×2/-3
K=10/-3
Hence the equation will become
10x/-3 + 2y - 1=0
10x - 6y + 3 =0
10/5=-6/-3 is not equal to 3/2
Answered by
3
Given
- kx+2y-1=0
- 5x-3y+2=0
To find
- value of k
Solution
we are provided with two system of equations and are asked to fight the value of k so that the system will not have a solution. lt could be found by using the properties of parallel lines.
ie, a1/a2 = b1/b2
from given equations , a1 = k
a2 = 5, b1 = 2 & b2 = -3
using the above equation,
k/5 = 2/(-3)
or, k/5 = 2/-3
or, k = 5 × 2/-3
or, k = -10/3
Therefore, the value of k4 which the system will not have a solution would be -10/3
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