Question

The value of k for which the system of equations kx + (k + 4)y – 7 = 0 and (k + 4)x + 9ky – 10 = 0 has no solution, can be

Answer 1

Answer:

k=-1 or -2

Step-by-step explanation:

as the question states that there are no real solutions

the condition for inconsistence roots is

A1/A1= B1/B2  $\neq$ C1/C1

according to the question

k/k+4=k+4/9k

by cross multiplication

8$k^{2}$-8k-16

now solve this quadratic equation to get the ans

Answer 2

The answer is k = -1 or 2

Given:  kx + (k + 4)y – 7 = 0 _(1)

           (k + 4)x + 9ky – 10 = 0_(1)  has no solutions

To find: the value of k

Solution: Compare given equations with

a₁x+b₁y -c₁ = 0 and a₂x+b₂y -c₂= 0

When we compare kx + (k + 4)y – 7 = 0 with a₁x+b₁y -c₁ = 0

⇒ a₁ = k, b₁ = k+4 and c₁ = 7  

When we compare (k + 4)x + 9ky – 10 = 0 with a₂x+b₂y -c₂ = 0

⇒ a₂ = k+4, b₂ = 9k and c₂ = 10

If given system of equations has no solutions then the following condition will be followed by the equations

⇒  $\frac{a_{1} }{a_{2} } = \frac{b_{1} }{b_{2} } \neq \frac{c_{1} }{c_{2} }$    

⇒ From above values   $\frac{k }{k+4 } = \frac{k+4 }{9k} \neq \frac{7 }{10 }$

From  _(2)   $\frac{k }{k+4 } = \frac{k+4 }{9k}$

⇒ 9k(k) =(k+4)(k+4)  

⇒ 9k² = (k+4)²

⇒ 9k² = k² + 16 +2(k)(4)  

⇒ 9k² - k² - 16 - 8k = 0

⇒ 8k²- 8k - 16 = 0

⇒ k²-  k - 2 = 0       [write as product of factors ]

⇒ k²+ k - 2k - 2 = 0

⇒ k( k + 1 ) - 2(k + 1) = 0

⇒  (k + 1) (k - 2) = 0

⇒  (k +1)  = 0                               ⇒  (k-2) = 0

     k = - 1                                           k =  2

⇒ k values are  -1 and 2

#SPJ2