Question

Find the values of $k$ for which the system
$2x+ky=1$
$3x-5y=7$
will have (i) a unique solution, and (ii) no solution. Is there a value of $k$ for which the system has infinitely many solutions?

Answer 1

The straight lines :

$2x+ky=13x-5y=7$

Comparing $2x+ky=1$

with $a_{1}x+b_{1}y+c_1=0$

$a_1=2,b_1=k,c_1=-1$

Comparing $3x-5y=7$

with $a_{2}x+b_{2}y+c_2=0$

$a_2=3,b_2=-5,c_2=-7$

$\therefore$ For a unique solution ,

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}\Rightarrow \frac{2}{3}\ne \frac{k}{-5}\therefore k\ne \frac{-10}{3}$

For infinite solutions ,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\Rightarrow \frac{2}{3}=\frac{k}{-5}=\frac{1}{7}thereisnosuchvalueofkwhichwillsatisfytheequation$

For no solutions :

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}\Rightarrow \frac{2}{3}=\frac{k}{-5}\therefore k=-\frac{10}{3}$