Question

For what value of k does the pair of equations $5x+2y=2k\mathrm{and}2(\mathrm{k}+1)\mathrm{x}+\mathrm{ky}=(3\mathrm{k}+4)$ have an infinite number of solutions?

(a) k = 5
(b) k = 4
(c) $k=\frac{2}{3}$
(d) $k=\frac{-2}{3}$

Answer 1

The correct option is (b).

The given system of equations can be written as follows:
5x + 2y − 2k = 0 and 2(k + 1)x + ky − (3k + 4) = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 5, b₁ = 2, c₁ = −2k and a₂ = 2(k + 1), b₂ = k and c₂ = −(3k + 4)
∴ $\frac{a_1}{a_2}=\frac{5}{2\left(k+1\right)},\frac{b_1}{b_2}=\frac{2}{k}\mathrm{and}\frac{c_1}{c_2}=\frac{-2k}{-\left(3k+4\right)}=\frac{2k}{\left(3k+4\right)}$
For an infinite number of solutions, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
∴ $\frac{5}{2\left(k+1\right)}=\frac{2}{k}=\frac{2k}{\left(3k+4\right)}$
⇒ $\frac{5}{2(k+1)}=\frac{2}{k}\mathrm{and}\frac{2k}{(3k+4)}=\frac{2}{k}$
⇒ 5k = 4(k + 1) ...(i)
And, 2(3k + 4) = 2k² ...(ii)
From (i) we have:
5k = 4k + 4 ⇒ 5k − 4k = 4 ⇒ k = 4
From (ii) we have:
6k + 8 = 2k² ⇒ k² − 3k − 4 = 0 ⇒ (k − 4)( k + 1) = 0 ⇒ k = 4, −1
Hence, from (i) and (ii), we get:
k = 4