Question

x+(k+1)y=5and (k+1)x+9y=(8k-1).find k

Answer 1

Hey!!

I've attached pics Plz refer them

k = 2 or k = -4

Hope this helps

Answer 2

Although a part of your question is missing, you may be referring to this full question:

Find the value of k for which the following system of linear equations has infinite solutions:

x+(k+1)y=5

(k+1)x+9y=(8k-1)

Given: Two equations x+(k+1)y=5 and (k+1)x+9y=(8k-1) have infinite solutions.

To find: Value of k

Solution: For two linear equations ax+by+c=0 and px+qy+r=0, the necessary condition for infinite solutions is:

$\frac{a}{p} = \frac{b}{q} = \frac{c}{r}$

Here, equation (i) is x+(k+1)y-5=0. Therefore, a = 1, b= (k+1) and c= -5.

Equation (ii) is (k+1)x+9y-(8k-1)=0. Therefore, p= (k+1), q= 9 and c= -(8k-1)

Now, putting the values:

$\frac{1}{k + 1} = \frac{k + 1}{9} = \frac{ - 5}{ - (8k - 1)}$

Taking the first two equations:

$\frac{1}{k + 1} = \frac{k + 1}{9}$

$= > {(k + 1)}^{2} = 9$

$= > k + 1 = \sqrt{9}$

$= > k + 1 = 3 \: or \: k + 1 = - 3$

$= > k = 2 \: or \: k = - 4$

Solving the first and last equations:

$\frac{1}{k + 1} = \frac{ - 5}{ - (8k - 1)}$

$= > - (8k - 1) = - 5(k + 1)$

$= > 8k - 1 = 5k + 5$

=> 8k-5k = 5+1

=> 3k = 6

=> k = 6/3

=> k = 2

The value of k should satisfy all the equations to be the correct value. Since k= -4 is not a solution to these 2 equations, the final value of k =2.

Therefore, for the equations to have infinite solutions, the value of k is 2.