Question

Find the value of k for which the given system of equations has infinitely many equations: kx + 3y = k - 3 , 12 x+ ky = k

Answer 1

Given equations :-

• kx + 3y = k - 3
• 12x + ky = k

For a pair of equations to have infinitely many solutions, a1/a2 = b1/b2 = c1/c2

Now, comparing :-

k/12 = 3/k = (k - 3)/k

k/12 = 3/k

k(k) = 3(12)

k^2 = 36

k = 6

•°• For k = 6, the given equations have infinitely many solutions.

Answer 2

$\rule{200}3$

$\huge\tt{GIVEN:}$

• System of equations has infinitely many equations: kx + 3y = k - 3 , 12 x+ ky = k

$\rule{200}3$

$\huge\tt{TO~FIND:}$

• The value of k

$\rule{200}3$

$\huge\tt{SOLUTION:}$

As we know , for a infinitely many solutions,

a1/a2 = b1/b2 / c1/c2

Now, if we compare them,

⇝k/12 = 3/k = (k - 3)/k

⇝ k/12 = 3/k

⇝k × k = 3 × 12

⇝k² = 36

⇝k = √36

⇝k = 6

So, k = 6 have infinitely many solutions...

$\rule{200}3$