Question

Find the value of K for which the system of equations are 3x+y=1 and (2K-1)x+(K-1)y=2(K+1) has no solution

Answer 1

$\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}$
$\frac{3}{(2k - 1)} = \frac{1}{(k - 1)} ≠ \frac{1}{(2k - 1)}$
L.H.S
$\frac{3}{(2k - 1)} = \frac{1}{(k - 1)}$
$3k - 3 = 2k - 1$
$3k - 2k = 3 - 1$
$k = 2$
R.H.S
$\frac{1}{(k - 1)} ≠ \frac{1}{(2k - 1)}$
$2k + 1≠k - 1$
$2k - k≠ - 1 - 1$
$k≠ - 2$
Hence, k =2

Answer 2

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