Question

1. Find the value(s) of k for which the following simultaneous linear equations have a
unique solution : (k – 1) x – 3y = 4, 3x – (4k + 1) y = 5.
1

Answer 1

Solution :

We have equation :

$\bullet\:\sf{(k-1)x-3y=4}\\\bullet\sf{3x-(4k+1)y=5}$

Let the system of equation are;

$\mapsto\sf{a_1x+b_1y+c_1=0}\\\mapsto\sf{a_2x+b_2y+c2=0}$

We know that formula of the unique solution :

$\boxed{\bf{\frac{a_1}{a_2} \neq \frac{b_1}{b_2} }}}}$

$\longrightarrow\sf{(k-1)x-3y-4=0...............(1)}\\\\\longrightarrow\sf{3x-(4k+1)y-5=0.............(2)}$

So;

$\longrightarrow\sf{\dfrac{(k-1)}{3} \neq \dfrac{-3}{-(4k+1)} }\\\\\\\longrightarrow\sf{\dfrac{(k-1)}{3} \neq \dfrac{3}{(4k+1)}}\\\\\\\longrightarrow\sf{(4k+1)(k-1)\neq 9}\\\\\longrightarrow\sf{4k^{2} -4k+k-1\neq 9}\\\\\longrightarrow\sf{4k^{2}-3k\neq10}\\\\\longrightarrow\sf{4k^{2}-3k-10\neq0}\\\\\longrightarrow\sf{4k^{2} -8k+5k-10\neq0}\\\\\longrightarrow\sf{4k(k-2)+5(k-2)\neq0}\\\\\longrightarrow\sf{(k-2)(4k+5)\neq0}\\\\\longrightarrow\sf{k-2\neq0\:\:Or\:\:4k+5\neq0}$

$\longrightarrow\sf{k\neq2\:\:\:Or\:\:\:4k\neq-5}\\\\\longrightarrow\bf{k\neq2\:\:\:Or\:\:\:k\neq-5/4}$