Question

For what value of k will the following pair of linear equations have infinitely many solutions?
$$\begin{gathered} 2x-3y=7 \\ (k+1)x+(1-2k)y=(5x-4) \end{gathered}$$

Answer 1

The given equations are:
2x - 3y - 7 = 0
(k + 1)x + (1 - 2k)y - (- 5k + 4) = 0
The given equations are of the form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, where
a₁ = 2, b₁ = - 3, c₁ = - 7 and a₂ = (k + 1), b₂ = (1 - 2k) and c₂ = (4 - 5k)
The given pair of equations can have infinitely many solutions if
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
Or
$\frac{2}{k+1}=\frac{-3}{\left(1-2k\right)}=\frac{-7}{\left(4-5k\right)}$
$\Rightarrow \frac{2}{k+1}=\frac{3}{\left(2k-1\right)}\mathrm{and}\frac{3}{\left(2\mathrm{k}-1\right)}=\frac{7}{\left(5\mathrm{k}-4\right)}$
⇒ 4k - 2 = 3k + 3 and 15k - 12 = 14k - 7
⇒ k = 5 and k = 5
Hence, k = 5.