Question

2. Find the value of k for which the following pair of equations has no solution:
x+ 2y =3 ;(k-1)x+ (k+1)y = (k+2)​

Answer 1

Given system of equations is,

x+2y=3..........[1]

and,(k+1)x+(k+1)y=k+2

Every set of equations would be of the form,

a1x+b1y=c1

and,a2x+b2y=c2

On comparison we obtain,

a1=1,a2=k+1,b1=2,b2=k+1,c1=3 and c2=k+2

We need to find out the value of k for which the given system of equations has no solution,

a1/a2=b1/b2≠c1/c2

→1/k-1=2/k+1≠3/k+2

→1/k-1=2/k+1 or 1/k-1≠3/k+2

→2k-2=k+1 or 3k-3≠k+2

→k=3 or k≠5/2

For k=3,the given system of equations have no solutions

Answer 2

Answer:

Value of k = 3

Step-by-step explanation:

Given Problem:

Find the value of k for which the following pair of equations has no solution:x+ 2y =3 ;(k-1)x+ (k+1)y = (k+2)

​Solution:

To Find:

Value of k

---------------------------

Method:

$For \ x \ + \ 2y\ = 3 \ or\ x\ + \ 2y\ - 3\ = 0$

$\implies\ a_1 = 1, b_1 = 2 , c_1 = -3$

$\implies For \ no \ solution,\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

$\implies \dfrac{1}{k- 1} = \dfrac{2}{k+1} \neq \dfrac{3}{k+2}$

$\implies\ And,\ \dfrac{1}{k-1} = \dfrac{2}{k+1}$

$\implies\ k+1 = 2k - 2$

$\implies\ 3 = k$

$\implies\ Hence, \ Value \ of \ k \ =3$