Question

Find the value of k if the system of equations have no solution. 3x + y = 1; (2k – 1)x + (k – 1)y = 2k + 1

Answer 1

Answer:

k=2,k is not Equal to -3

Step-by-step explanation:

3/2k-1=1/(k-1) is not Equal to 1/2k+1

3k-3=2k-1

solving this,k=2

2k+1 is not Equal to k-1

which is not Equal to-2

Answer 2

AnsWer :

• k = 2.
• k ≠ - 2.

SolutioN :

Condition : No Solution.

$\tt \dagger \: \: \: \: \: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

Where as,

• a1 = 3.
• a2 = 2k - 1.
• b1 = 1.
• b2 = k - 1
• c1 = 1.
• c2 = 2k + 1.

Now,

☛ Condition 1st :

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

$\tt : \implies \dfrac{3}{2k - 1} = \dfrac{1}{k - 1}$

$\tt : \implies 3(k - 1) = 2k - 1.$

$\tt : \implies 3k - 3= 2k - 1.$

$\tt : \implies k= 3 - 1$

$\tt : \implies k= 2.$

☛ Condition 2nd :

$\tt : \implies \dfrac{1}{k - 1} \neq \dfrac{1}{2k + 1}$

$\tt : \implies k - 1\neq 2k + 1$

$\tt : \implies 0\neq k + 1 + 1$

$\tt : \implies k\neq - 2.$

✡ Therefore, the value of k is 2 and not equal to - 2.

$\rule{200}3$

MorE InformatioN :

★ Many Solution :

$\tt \dagger \: \: \: \: \: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

★ Unique Solution :

$\tt \dagger \: \: \: \: \: \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$

★ No Solution :

$\tt \dagger \: \: \: \: \: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$