Question

3. For what value of k, 2x + 3y = 4 and (k + 2)x + 6y = 3k + 2 will have infinitely many solutions.

Answer 1

Answer:

For infinite solution a1/a2=b1/b2=c1/c2

We have a1= 2x,a2=2/K+2,b1=3,b2=6 and c1=4,c2=3K+2

2/K+2=3/6=4/3k+2

2/k+2=4/3k+2

6k+4=4k+8

2k=4

So,k =2

Answer 2

Given :-

The pair of linear equations are:-

• 2x + 3y = 4
• (k + 2)x + 6y = 3k + 2

To Find :-

• Value of k.

Solution :-

Condition for infinitely many solutions is,

⇒ a₁/a₂ = b₁/b₂ = c₁/c₂

Where,

• (a₁, a₂) = (2, k + 2)
• (b₁, b₂) = (3, 6)
• (c₁, c₂) = (4, 3k + 2)

Putting these values, we get,

⇒ 2/k + 2 = 3/6 = 4/3k + 2

Taking the first two terms, we get,

⇒ 2/k + 2 = 3/6

⇒ 2/k + 2 = 1/2

⇒ k + 2 = 4

⇒ k = 4 - 2

⇒ k = 2

Hence,

• Value of k = 2.