Question

if the system of equations 4x+y=3and (2k-1)x+(k-1)y =2k+1 is inconsistent, then find the value of k

Answer 1

here is your answer by Sujeet,

Given that equation,

4x+y=3
(2k-1)+(k-1)y=2k+1
For no solution we must be have,

A1/A2=b1/b2≠c1/c2

4/(2k-1)=1/(k-1)
4k-4=2k-1
4k-2k=-1+4
2k=3
k=3/2...


that's all

Answer 2

The value of k is 3/2.

Step-by-step explanation:

Given that,

4x + y = 3

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

A.T.Q.

The Equation is inconsistent, then

$a_{1}$/$a_{2}$ = $b_{1}$/$b_{2}$ ≠ $c_{1}$/$c_{2}$

so, as per the equation,

$a_{1}$ = 4, $b_{1}$ = 1, $c_{1}$ = -3

$a_{2}$ = 2k-1, $b_{2}$ = (k-1), $c_{2}$ = -(2k+1)

so,

a1/a2 = b1/b2

4/(2k-1) = 1/(k-1)

4k-4 = 2k-1

4k-2k = -1 + 4

2k = 3

k = 3/2

Thus, k = 3/2

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