Question

Q4. Find the value of 'p' for which the pair of equations :
2x + py = 1 and 3x + 5y = 7 will have NO solution?​

Answer 1

Hyy Dude

Let ,

Let ,the given system of linear equations are of the form

Let ,the given system of linear equations are of the form2x+ky=1     a₁x +b₁y+c₁ = 0                  

Let ,the given system of linear equations are of the form2x+ky=1     a₁x +b₁y+c₁ = 0                   3x-5y=7     a₂x +b₂y+c₂ = 0

Let ,the given system of linear equations are of the form2x+ky=1     a₁x +b₁y+c₁ = 0                   3x-5y=7     a₂x +b₂y+c₂ = 0the condition for unique solution is a₁/a₂≠  b₁b₂

Let ,the given system of linear equations are of the form2x+ky=1     a₁x +b₁y+c₁ = 0                   3x-5y=7     a₂x +b₂y+c₂ = 0the condition for unique solution is a₁/a₂≠  b₁b₂here ,

Let ,the given system of linear equations are of the form2x+ky=1     a₁x +b₁y+c₁ = 0                   3x-5y=7     a₂x +b₂y+c₂ = 0the condition for unique solution is a₁/a₂≠  b₁b₂here ,a₁ = 2, a₂ = 3

Let ,the given system of linear equations are of the form2x+ky=1     a₁x +b₁y+c₁ = 0                   3x-5y=7     a₂x +b₂y+c₂ = 0the condition for unique solution is a₁/a₂≠  b₁b₂here ,a₁ = 2, a₂ = 3b₁ = k ,b₂ = -5

Let ,the given system of linear equations are of the form2x+ky=1     a₁x +b₁y+c₁ = 0                   3x-5y=7     a₂x +b₂y+c₂ = 0the condition for unique solution is a₁/a₂≠  b₁b₂here ,a₁ = 2, a₂ = 3b₁ = k ,b₂ = -5⇒ 2/3 ≠k/ -5

Let ,the given system of linear equations are of the form2x+ky=1     a₁x +b₁y+c₁ = 0                   3x-5y=7     a₂x +b₂y+c₂ = 0the condition for unique solution is a₁/a₂≠  b₁b₂here ,a₁ = 2, a₂ = 3b₁ = k ,b₂ = -5⇒ 2/3 ≠k/ -5⇒-10 ≠ 3k

Let ,the given system of linear equations are of the form2x+ky=1     a₁x +b₁y+c₁ = 0                   3x-5y=7     a₂x +b₂y+c₂ = 0the condition for unique solution is a₁/a₂≠  b₁b₂here ,a₁ = 2, a₂ = 3b₁ = k ,b₂ = -5⇒ 2/3 ≠k/ -5⇒-10 ≠ 3k⇒k ≠ -10/3

Let ,the given system of linear equations are of the form2x+ky=1     a₁x +b₁y+c₁ = 0                   3x-5y=7     a₂x +b₂y+c₂ = 0the condition for unique solution is a₁/a₂≠  b₁b₂here ,a₁ = 2, a₂ = 3b₁ = k ,b₂ = -5⇒ 2/3 ≠k/ -5⇒-10 ≠ 3k⇒k ≠ -10/3therefore , the value of k is any real number other then -10/3

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Answer 2

Answer:

yes it has no solution.....