Question

4) The value of K for which the system : 4x + 2y = 3,
(k-1) x-6y = 9 has no unique solution is :
a) -13 b)9 c) -11 d) 13​

Answer 1

-11

Step-by-step explanation:

4x+2y=3

(k-1)x-6y=9

given the equation has no solution then,

a1/a2=b1/b2is not equal to c1/c2,

=4/k-1=2/-6

=-12=k-1

k=-11

Answer 2

The value of k is c)-11

Step-by-step explanation:

Given:

equations are 4x+2y = 3 & (k-1)x-6y =9

The equation has no unique solution

To find:

value of k

Solution:

The equations 4x + 2y = 3 & (k-1) x-6y = 9 can be written as

4x + 2y -3 =0 & (k-1) x-6y-9 = 0

We compare the equations to the forms  $a_{1} x +b_{1}y+c_{1}$ & $a_{2} x +b_{2}y+c_{2}$

we get,  $a_{1}$ = 4, $b_{1}$ = 2, $c_{1}$ = -3

$a_{2}$= (k-1), $b_{2}$= -6, $c_{2}$ = -9

As given, the equations have no unique solution

∴ $\frac{a_{1} }{a_{2} } = \frac{b_{1} }{b_{2} } \neq \frac{c_{1} }{c_{2} }$

∴ $\frac{4}{(k-1)} = \frac{2}{-6}\neq \frac{-3}{-9}$

∴ $\frac{4}{(k-1)} = \frac{2}{-6}$

∴ $-24 =$ $2(k-1)$

∴ $-24 = 2k -2$

∴ $-24+2 = 2k$

∴ $-22 = 2k$

∴ $k = \frac{-22}{2}$

∴ $k = -11$

∴ k = -11

Thus, the value of k for which the system has no unique solution is c)-11

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