Question

find the value of k if the system of linear equations (3k+4)x + 2y = -(k+3) ; (2k-5)x + 4y = 7-k has no solution

Answer 1

we have to find the value of k for which system of linear equations ; (3k + 4)x + 2y = -(k + 3) and (2k - 5)x + 4y = (7 - k) has no solution.

solution : we know, the system of two equations

a₁x + b₁y = c₁ and a₂x + b₂y = c₂ has no solution only if a₁/a₂ = b₁/b₂ ≠ c₁/c₂

applying above condition,

(3k + 4)/(2k - 5) = 2/4 ≠ -(k + 3)/(7 - k)

now (3k + 4)/(2k - 5) = 2/4 = 1/2

⇒2(3k + 4) = (2k - 5)

⇒6k + 8 = 2k - 5

⇒4k = -13

⇒k = -13/4

again, 2/4 = 1/2 ≠ -(k + 3)/(7 - k)

for k = -13/4,

-(-13/4 + 3)/(7 + 13/4) = -(-1/4)/(41/4) = 1/41 ≠ 1/2

therefore the value of k = -13/4 for which system of equations has no solution.