Question

for what value of k will the system of linear equations 2x+ky=1 ; 3x-5y=7 has a unique solution..

Answer 1

let ,
the given system of linear equations are of the form
2x+ky=1     a₁x +b₁y+c₁ = 0                  
3x-5y=7     a₂x +b₂y+c₂ = 0
the condition for unique solution is a₁/a₂≠  b₁b₂
here ,
a₁ = 2, a₂ = 3
b₁ = k ,b₂ = -5
⇒ 2/3 ≠k/ -5
⇒-10 ≠ 3k
⇒k ≠ -10/3
therefore , the value of k is any real number other then -10/3

Answer 2

The equations are:

2x + ky = 1
3x - 5y = 7

For the system of equations to have a unique solution,

a₁/a₂ ≠ b₁/b₂
∴2/3 ≠ k/-5
∴2/3 ≠ -k/5
∴2/3 * 5 ≠ -k
∴10/3 ≠ - k
∴k ≠ -10/3

Thus, k can be any real value except -10/3.

Thus, k ∈ R - {-10/3}