Answers

2016-04-22T19:58:33+05:30
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



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2016-04-22T19:59:24+05:30
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}

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