Math, asked by priyanshi8736, 1 year ago

find the value of k for which of the following system of linear equations have no solution 3x+y=1;(2k-1)x+(k-1)y=2k+1

Answers

Answered by Anonymous
12

GIVEN EQUATION:-

 \bf \: 3x + y - 1 = 0, \\

 \bf \: and \: (2k - 1)x + (k - 1)y - (2k + 1) = 0 \\

TO FIND OUT:-

 \textsf{Value of k=?} \\  \\

SOLUTION:-

 \bf \: The  \: given  \: equations \:are \: of \: the \: form

\boxed{ \boxed{ \bf \: a_1x+b_1y+c_1=0   \: \&   \: \: a_2x+b_2y+c_2=0} } \\

 \bf \: Where  \: a_1=3,b_1=1,c_1=-1 \\  \\

 \bf \: And \: a_2=(2k-1),b_2=(k-1),c_2=-(2k+1) \\  \\

In order that the given system has no solution, we must have

  \bf \:  \frac{a_1}{a_2} = \frac{b_1}{b_2}  \ne \frac{c_1}{c_2}  \\  \\

This happen when

 \bf \ \frac{3}{(2k - 1)}  =  \frac{1}{(k  - 1)} \ne \frac{1}{(2k + 1)}  \\  \\

 \bf When  \:  \frac{3}{(2k-1)}  =  \frac{1}{(k-1)}  \\

 \bf \: and \:  \frac{1}{(k - 1)}  \ne \frac{1}{(2k + 1)}  \\

 \bf \implies \:  \frac{3}{(2k - 1)}  =  \frac{1}{(k - 1)}  \implies3k - 3 = 2k - 1 \\  \\  \bf \implies \: k = 2

When k=2,then we have,

 \bf  \frac{1}{(k - 1)}   \ne \frac{1}{(2k + 1)} \:  \: \:  \:  \:  \:  --  \huge[ \small  \frac{1}{(2-1)} \ne \frac{1}{(4+1)} \huge]    \\  \\

 \bf \therefore \:  \frac{3}{(2k - 1)}  =  \frac{1}{(k - 1)}  \ne \frac{1}{(2k + 1)}  \\  \\

Hence,the given solution of equation has no solution when k=2

Similar questions