Question

2. For what values of k will the following pair of linear equations have infinitely many solutions?


kx + 3y – (k – 3) = 0 and 12x + ky – k = 0

Class - 10

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Answer 1

$\textbf{Answer}$

Given linear equations are,
kx + 3y - (k-3) = 0
12x + ky - k = 0

General form of linear equation is,
$\textbf{ax + by + c = 0}$
$\textbf{Ax + Bx + C = 0}$

The pair of general linear equations will have infinitely many solutions if and only if,
$\textbf{a/A = b/B = c/C}$

So by comparing these general linear eqautions with the given linear equation we get,

a = k
A = 12
b = 3
B = k
c = - (k-3)
C = - k

Since a/A = b/B = c/C

=> k / 12 = 3 / k = - (k-3) / - k
=> k / 12 = 3 / k ----------------(1)
&
=> 3 / k = - (k-3) / - k ---------(2)

$\textbf{Solving equation (1)}$
k / 12 = 3 / k
=> k^2 = 12×3
=> k^2 = 36
=> k = √36
=> k = +6 and k = -6

$\textbf{Let solve equation (2) now,}$
3 / k = - (k-3) / - k
=> 3 / k = (k-3) / k
=> k ( k-3) = 3k
=> k^2 - 3k = 3k
=> k^2 - 6k = 0
=> k(k-6) = 0
=> k = 6 and k = 0

$\textbf{Since k = 6 satisfies both equation (1) and (2)}$
=> k = 6

So given pair of linear equations will have infinitely many solutions if the value of k is 6

$\textbf{Hope It Helps}$
$\textbf{Thanks}$