Question

If the system of equations
3x + y = 1 and (2k-1)x+(k-1)y = 2k +1 is Inconsistent ,then find the value of k.

Answer 1

Answer:

For the given system to be inconsistent,  $k=2$ and $k\neq{-2}$

Step-by-step explanation:

Consider the general system of linear equation,

$a_1x+b_1y+c_1=0$ and

$a_2x+b_2y+c_2=0$

When the system of linear equations  represents a pair of Parallel Lines then there is no point of intersection and hence no values of x and y which satisfies both equation. Thus, system has no solution and such system of linear equation is  inconsistent pair of  linear equations

For the system to be inconsistent,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$  .............(A)

Now , For the given system of equation

$3x + y = 1$       ....(1) and,

$(2k-1)x+(k-1)y = 2k +1$      ........(2)

On comparing,

$a_1=3$, $b_1=1$, $c_1=-1$

$a_2=(2k-1)$, $b_2=(k-1)$, $c_2=-(2k+1)$

Putting above values in (A),

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

$\frac{3}{(2k-1)}=\frac{1}{(k-1)}\neq \frac{-1}{-(2k+1)}$

Considering, first two ratios,

$\frac{3}{(2k-1)}=\frac{1}{(k-1)}$

On solving , we get

${3}\times{(k-1)}={(2k-1)}$

$k=2$

Considering, last two ratios,

$\frac{1}{(k-1)}\neq \frac{-1}{-(2k+1)}$

${(k-1)}\neq {(2k+1)}$

$k\neq{-2}$

Thus, For the given system to be inconsistent,  $k=2$ and $k\neq{-2}$

Answer 2

The value of k = 2.

Step-by-step explanation:

$\frac{a_{1} }{a_{2} } = \frac{3}{2k - 1}$

$\frac{b_{1} }{b_{2} } = \frac{1}{k - 1}$

For Inconsistent lines,

$\frac{a_{1} }{a_{2} } = \frac{b_{1} }{b_{2} }$ ≠ $\frac{c_{1} }{c_{2} }$

⇒ 3/(2k-1) = 1/(k-1)

⇒ 3k - 3 = 2k - 1

⇒ 3k - 2k = -1 + 3

⇒ k = 2

Learn more: find the value of k

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