Question

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$$\begin{gathered} 5x+2y=2k, \\ 2(k+1)x+ky=(3k+4). \end{gathered}$$

Answer 1

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

Now, we have the following three cases:
Case I:
$\frac{5}{2\left(k+1\right)}=\frac{2}{k}$
$\Rightarrow 2\times 2\left(k+1\right)=5k\Rightarrow 4\left(k+1\right)=5k$
$\Rightarrow 4k+4=5k\Rightarrow k=4$

Case II:
$\frac{2}{k}=\frac{2k}{\left(3k+4\right)}$
$\Rightarrow 2k^2=2\times \left(3k+4\right)$
$\Rightarrow 2k^2=6k+8\Rightarrow 2k^2-6k-8=0$
$\Rightarrow 2\left(k^2-3k-4\right)=0$
$\Rightarrow k^2-4k+k-4=0$
$\Rightarrow k(k-4)+1(k-4)=0$
$\Rightarrow \left(k+1\right)\left(k-4\right)=0$
$\Rightarrow \left(k+1\right)=0\mathrm{or}\left(k-4\right)=0$
$\Rightarrow k=-1\mathrm{or}k=4$

Case III:
$\frac{5}{2\left(k+1\right)}=\frac{2k}{3k+4}$
$\Rightarrow 15k+20=4k^2+4k$
$\Rightarrow 4k^2-11k-20=0$
$\Rightarrow 4k^2-16k+5k-20=0$
$\Rightarrow 4k\left(k-4\right)+5\left(k-4\right)=0$
$\Rightarrow \left(k-4\right)\left(4k+5\right)=0$
$\Rightarrow k=4\mathrm{or}\mathrm{k}=\frac{-5}{4}$

Hence, the given system of equations has an infinite number of solutions when k is equal to 4.