Question

The number of values of k. for which the system or equations:
(k+1) x +8y = 4k
kx + (k+3) y = 3k-1
has no solution, is

• A. infinite
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

From the given system, we have
$\frac{k+1}{k}=\frac{8}{k+3}\ne \frac{4k}{3k-1}$ (\therefore\) system has no solution)
$\Rightarrow k^2+4k+3=8k$
$\Rightarrow k=1,3$
If k =1 then $\frac{8}{1+3}\ne \frac{4.1}{2}$ Which is false.
And if $k=3$
then $\frac{8}{6}\ne \frac{4.3}{9-1}$ which is true, $\therefore k=3$
Hence for only one value of k. System has no solution.