Question

Question 2
For which value (s) of k will the pair of equations
kx + 3y = 5 – 3,
12x + 5y = k
has no solution?

Answer 1

Given pair of linear equations is
kx + 3y = k – 3
and 12x + ky = k
on comparing with ax + by + c = 0 we get
$a_1=k,b_1=3and{c}_1=1(k–3)a_2=12,b_2=kand{c}_2=-k$
For no solution of the pair of linear equations
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}\frac{k}{12}=\frac{3}{k}\ne \frac{-(k-3)}{-k}$
Tasking first two parts we get
$\Rightarrow \frac{k}{12}=\frac{3}{k}\Rightarrow k^2=36\Rightarrow k=\pm 6$
Taking last two parts, we get
$\frac{3}{k}\ne \frac{k-3}{k}\Rightarrow 3k\ne k(k-3)\Rightarrow 3k-k(k-3)\ne 0\Rightarrow k(3-k+3)\ne 0\Rightarrow k\ne 0andk\ne 6$
Hence, required value of k for which the given pari of linear equations has no solutions is – 6