Question

in each case given below find the value of k for which the given system of equation has infinitely many solutions 3x-(k-2) y=k and (k+1) x-18y=24​

Answer 1

Answer :

The value of k is either 8 or -9

Given :

The equations are :

• 3x - (k - 2)y = k
• (k + 1)x - 18y = 24

To Find :

• The value of k

Solution :

Let us consider from the given equations :

a₁= 3 , b₁ = -(k -2) and c₁ = k

a₂ = (k + 1) , b₂ = -18 and c₂ = 24

Since the equations has infinitely many solution so :

a₁/a₂ = b₁/b₂ = c₁/c₂

3/(k + 1) = -(k - 2)/(-18) = k/24

⇒ 3/(k + 1) = (k - 2)/18 = k/24

Now we have :

3/(k + 1) = k/24

⇒ k(k + 1) = 3×24

⇒ k² + k = 72

⇒ k² + k - 72 = 0

⇒ k² + 9k - 8k - 72 = 0

⇒ k(k + 9) -8(k + 9) = 0

⇒ (k - 8)(k + 9) = 0

Therefore ,

k - 8 = 0 and k + 9 = 0

→ k = 8 and → k = -9

Answer 2

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QUESTION -

in each case given below find the value of k for which the given system of equation has infinitely many solutions 3x-(k-2) y=k and (k+1) x-18y=24.

SOLUTION -

The given System of Equations is :

3x-(k-2) y=k and (k+1) x-18y=24

For the actual solution, see the above attachment...

ANSWER :

K = 8 or -9.