Question

find k ,if given linear equation has an infinite solutions 2x + (k-2y)= k, 6 x + (2k - 1)y = (2k + 5)

Answer 1

Answer:

k = 5

Step-by-step explanation:

Given a pair of linear equations such that,

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

6x + (2k-1)y = (2k+5) .......(2)

Also, given that,

They have Infinite number of solutions.

To find the value of k.

We know that,

For infinite number kf solutions,

• a1/a2 = b1/b2 = c1/c2

Substituting the values, we get,

=> 2/6 = (k-2)/(2k-1) = k/(2k+5)

=> 1/3 = (k-2)/(2k-1) = k/(2k+5)

Therefore, we will get,

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

=> 2k - 1 = 3k - 6

=> 3k - 2k = 6-1

=> k = 5

And

=> 2k+5 = 3k

=> 3k-2k = 5

=> k = 5

Hence, required value of k = 5.

Answer 2

${ \huge{ \bold{ \underline{ \underline{ \pink{Question:-}}}}}}$

Find k , if given linear equation has an infinite solutions 2x + (k-2y) = k, 6 x + (2k - 1)y = (2k + 5) ..

______________________

${ \huge{ \bold{ \underline{ \underline{ \orange{Answer:-}}}}}}$

✎Given :

$\implies{2x+(k-2y) = k .... (i)}$

$\implies{6x+(2k-1) y=2k+5}$

✎To Find :

• Value of k ..

✎Now Comparing :

$\implies{\dfrac{{a}_{1}}{{a}_{2}}=\dfrac{{b}_{1}}{{b}_{2}}=\dfrac{{c}_{1}}{{c}_{2}}}$

✎By Putting Values :

$\implies{\cancel\dfrac{2}{6}=\dfrac{k-2y}{2k-1}=\dfrac{k}{2k+5}}$

$\implies{\dfrac{1}{3}=\dfrac{k-2}{2k-1}=\dfrac{k}{2k+5}}$

✎By this we will get :

$\implies{3(k-2) =1(2k-1)}$

$\implies{3k-6=2k-1}$

$\implies{3k-2k=-1+6}$

$\implies{1k=5}$

$\implies{{ \large{ \bold{ \bold{ \bold{ \purple{k=5}}}}}}}$

✎Now,

$\implies{3k=2k+5}$

$\implies{3k-2k=5}$

$\implies{1k=5}$

$\implies{{ \large{ \bold{ \bold{ \bold{ \blue{k=5}}}}}}}$

✎So, the value of k is 5 ..