Question

For what value of k will the pair of equation have no solution?3x+y=1,(2k-1)x+(k-1)y=2k+1

Answer 1

HEY THERE!!!


Question:-

For what value of k will the pair of equation have no solution?

3x+y=1,(2k-1)x+(k-1)y=2k+1

Method Of Solution:-

For No Solution Must be!!

a1/a2 = b1/b2 ≠ c1/c2

Here, (a1 = 3 , a2 = 2k-1) , (b1 = 1 , b2 = (k-1) and (c1 = 1 , c2= 2k+1

Taken,


a1/a2 = b1/b2

Substitute the Given value in Equation!!


a1/a2 = b1/b2

=> 3/2k-1 = 1/k-1

=> 3(k-1) = 1(2k-1)

=> 3k-3 = 2k-1

=> 3k-2k = -1+3

=> k = 2

Hence, Value of k for this Equation = 2

Answer 2

$<b>$

HEY MATE HERE IS YOUR ANSWER

$\rule {168} {2}$

3x + y = 1,( 2k - 1 ) x + ( k - 1 ) y = 2 k + 1


$= > \frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}$


Here in question,

➡️ {a1 = 3, a2 = 2k - 1}

➡️ {b1 = 1, b2 = (k-1)}

➡️ {c1 = 1, c2 = 2k+1}

$= > took \\ \\ = > \frac{a1}{a2} = \frac{b1}{b2}$


Now, substitute values given in question

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


➡️ 3(k-1) = 1(2k-1)

➡️ 3k-3 = 2k-1

➡️ 3k - 2k= - 1 +3

➡️ $\huge {K = 2}$

$\rule {168} {2}$

Hope it will help you

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Ansh as ans81