Question

find the value of k , for which the system of equations has infinity many solutions: 2x-3y=7; (k+2)x-(2k+1)y-3(2k-1)

Answer 1

here is your answer by Sujeet,

we must for infinite many solution,
a1/a2=b1/b2=c1/c2
2/k+2=7/-(2k+1)
-4k+2=7k+14
-4k-7k=14-2
-11k=12
k=-12/11

that's all Sujeet...

Answer 2

Answer:

The value of k is 2

Step-by-step explanation:

Given system of linear equation:

2x -3y = 7

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

As system of equations has infinity many solutions if

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

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

Cross multiply

$-(2k+1) \times 2= -3 \times (k+2)$

$-4k-4= -3k-6$

simplify the above

$-4k+3k= 4-6$

$-1k= -2$

$k= 2$

Hence, the value of k is 2