Question

Find k for which the system 2x + 3y − 5 = 0 and 4x + ky − 10 = 0 has an infinite number of solutions.

Answer 1

The given system is
$$\begin{gathered} 2x+3y-5=0.....\left(i\right) \\ 4x+ky-10=0.....\left(ii\right) \end{gathered}$$
Here, $a_1=2,b_1=3,c_1=-5.a_2=4,b_2=k\mathrm{and}{c}_2=-10$.
For the system, to have an infinite number of solutions, we must have
$$\begin{gathered} \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ \Rightarrow \frac{2}{4}=\frac{3}{k}=\frac{-5}{-10} \\ \Rightarrow \frac{1}{2}=\frac{3}{k}=\frac{1}{2} \\ \Rightarrow k=6 \end{gathered}$$
Hence, k = 6.