Question

For which values of k does the pair of equations given below has no solution?

2x + 3y = 5

4x – (k – 1) y = 3.​

Answer 1

Solution:

Consider 2 linear equations of the form:

• a₁x + b₁y + c₁ = 0
• a₂x + b₂y + c₂ = 0

For a pair of linear equations to have no solution,

$\bigstar \: \: \boxed{ \sf{ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} }}$

Compare the above condition with the linear equations.

$\dashrightarrow \: \: \sf{ \dfrac{2}{4} = \dfrac{3}{ - (k - 1)} }$

$\implies \sf{ \dfrac{2}{4} = \dfrac{3}{1 - k} }$

On cross multiplying,

→ 2 (1 - k) = 3 (4)

⇒ 2 - 2k = 12

⇒ -2k = 12 - 2

⇒ -2k = 10

⇒ k = 10 ÷ -2

⇒ k = -10 ÷ 2

⇒ k = -5

∴ The value of k = -5