Question

Find the value of k for which the following equations have no solution. 3x + y = 1 and (2k – 1 )x + (k – 1)y = 2k + 1

Answer 1

Solution :

We have two equation :

• 3x + y = 1
• (2k - 1)x + (k - 1)y = 2k + 1

$\bigstar$The system of these equation is of the form :

$\bullet\:\sf{a_1x+b_1y+c_1=0}\\\bullet\sf{a_2x+b_2y+c_2=0}$

• 3x + y -1 = 0
• (2k-1)x + (K-1)y - 2k + 1 = 0

Using formula for no solution equation :

$\boxed{\bf{\frac{a_1}{a_2} =\frac{b_1}{b_2} \neq \frac{c_1}{c_2} }}}$

$\longrightarrow\sf{\dfrac{3}{(2k-1)} =\dfrac{1}{(k-1)}\neq\dfrac{-1}{-2k+1}}\\\\\\\longrightarrow\sf{\dfrac{3}{(2k-1)} =\dfrac{1}{(k-1)} }\\\\\\\longrightarrow\sf{3(k-1)=1(2k-1)}\\\\\longrightarrow\sf{3k-3=2k-1}\\\\\longrightarrow\sf{3k-2k=-1+3}\\\\\longrightarrow\bf{k=2}$

Thus;

The value of k will be 2 .