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For which value(s) of k will the pair of equations kx + 3y = k - 3; 12x + ky = k have no solution

Solution:

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 solution

Summary:

For the value of k = -6, the pair of linear equations kx + 3y = k - 3; 12x + ky = k has no solution

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