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Given, the pair of linear equations are kx + 3y = k - 3 12x + ky = k We have to determine the value of k for which the pair of linear equations will have no solution. We know that, For a pair of linear equations in two variables be a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\), then the graph will be a pair of parallel lines and so the pair of equations will have no solution. Here, a₁ = k, b₁ = 3, c₁ = k - 3 a₂ = 12, b₂ = k, c₂ = k So, a₁/a₂ = k/12 b₁/b₂ = 3/k c₁/c₂ = (k - 3)/k For no solution, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\) So, k/12 = 3/k ≠ (k - 3)/k Case 1) k/12 = 3/k k(k) = 3(12) k2 = 36 k = ±6 Case 2) 3/k ≠ (k - 3)/k 3(k) ≠ k(k - 3) 3k ≠ k2 - 3k k2 - 3k - 3k ≠ 0 k2 - 6k ≠ 0 k(k - 6) ≠ 0 So, k = 6, 0 Therefore, for the value of k = -6, the pair of linear equations have no solution. ✦ Try This: For which value(s) of λ, do the pair of linear equations λx + y = 2λ/3 and x/2 + λy = 10 have no solution ☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 3 NCERT Exemplar Class 10 Maths Exercise 3.3 Problem 2 For which value(s) of k will the pair of equations kx + 3y = k - 3; 12x + ky = k have no solutionSummary: For the value of k = -6, the pair of linear equations kx + 3y = k - 3; 12x + ky = k has no solution ☛ Related Questions:
The given pair of linear equations is kx + 3y = k – 3 …(i) 12x + ky = k …(ii) On comparing the equations (i) and (ii) with ax + by = c = 0, We get, a1 = k, b1 = 3, c1 = -(k – 3) a2 = 12, b2 = k, c2 = – k Then, a1 /a2 = k/12 b1 /b2 = 3/k c1 /c2 = (k-3)/k For no solution of the pair of linear equations, a1/a2 = b1/b2≠ c1/c2 k/12 = 3/k ≠ (k-3)/k Taking first two parts, we get k/12 = 3/k k2 = 36 k = + 6 Taking last two parts, we get 3/k ≠ (k-3)/k 3k ≠ k(k – 3) k2 – 6k ≠ 0 so, k ≠ 0,6 Therefore, value of k for which the given pair of linear equations has no solution is k = – 6.
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For what values of k will the equations KX 3y K 3 0 and 12x Ky k 0 have infinitely solutions?For what values of k will the following pair of linear equations have infinitely many solutions? kx + 3y - (k – 3) = 0. 12x + ky - k = 0. The value of k which satisfies both the equations is 6.
What is the value of k for which the system of equations?Since the system of equations has a unique solution. Therefore, for all real values of k, (k≠10)given systems of equations have unique solutions. Or in other words, the value of k is all numbers except k = 10.
For what value of k the pair of linear equations KX Y 4 3 6 12 9 XY has an infinite number of solutions?Pair of Linear Equations in Two Variables. Find the value of k for which the pair of equations kx - 4y = 3; 6x - 12y = 9 has an infinite number of solution. Hence, the given system of equation has infinitely many solution if, k = 2.
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