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