Question

the value of k for which the system of linear equations x+2y = k-11 an 2x+ky+10+k=0 has infinitely many solutions is
k=8
k= -4
k= 4
k= 10

Answer 1

Answer:

k=4

Step-by-step explanation:

see the picture......

Answer 2

Answer:

4

Step-by-step explanation:

Given :

x + 2 y = k - 11

Rewrite as

x + 2 y - k + 11 = 0 ......(i)

And

2 x + k y + 10 + k = 0 .....(ii)

We know formula for many solution

$\dfrac{a1}{a2} = \dfrac{b1}{b2} = \dfrac{c1}{c2}$

Comparing from (i) we get

a1 = 1

b1 = 2

c1 = - k + 11

Again from (ii) we get

a2 = 2

b2 = k

c2 = 10 + k

Now putting values in formula

$\dfrac{1}{2} = \dfrac{2}{k} = \dfrac{ - k + 11}{10 + k} \\ \\ \dfrac{1}{2} = \dfrac{2}{k} \\ \\ k = 4$

Thus we get answer 4 .