Question

Find the value of k for which each of the following systems of equations have infinitely many solution:
x+(k+1)y=4(k+1)x+9y=5k+2

Answer 1

Answer:

i m not sure about that this is correct,

let assume x=0,y=1

1+(k+1)1=4(k+1)1+9*1=5k+2

2k=4k+9=7k

2k=11k=7k

2k=7k-11k

2k=4k

k=4k-2

k= 2k

k=2

Answer 2

For system of equation have infinite many solution , the value of k is 2

Step-by-step explanation:

Given as :

The linear equation are

x + ( k + 1 ) y - 4  = 0                                    .........1

( k + 1 ) x + 9 y - ( 5 k + 2 ) = 0                      ........2

According to question

The standard equation  $a_1 x + b_1y + c_1 = 0$

                                        $a_2 x + b_2y + c_2 = 0$

The system of equation have infinite many solution

The condition for infinite many solution

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

$\dfrac{1}{1+k}$ = $\dfrac{k+1}{9}$ = $\dfrac{-4}{-(5 k + 2)}$

by cross multiplication

9 × 1 = ( 1 + k) ( 1 + k)

Or, 1 + k² + 2 k = 9

Or,  k² + 2 k - 8 = 0

i.e k² + 4 k - 2 k - 8 = 0

Or, k ( k + 4 ) - 2 ( k + 4 ) = 0

∴   ( k + 4 ) ( k - 2 ) = 0

Or,  k + 4 = 0             k - 2 = 0

or,   k = - 4                   k = 2

So, The value of k = 2

Hence, for system of equation have infinite many solution , the value of k is 2       Answer