Question

Find those values of $m$ the system of equation
$2x+my-4=0$
$3x-7y-10=0$
for which (1) there is unique solution
and (2) there is no solution.

Answer 1

Given equations are
$2x+my-4=0$
$3x-7y-10=10$
Here $a_1=2,b_1=m,c_1=-4,a_2=3,b_2=-7,c_2=-10$
(i) When equations have a unique solution, then
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
$\therefore \frac{2}{3}\ne \frac{m}{-7}$
$\Rightarrow$ $m\ne \frac{-14}{3}$
(ii) When the equation have no solution, then
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\therefore \frac{2}{3}=\frac{m}{-7}\ne \frac{-4}{-10}$
$\Rightarrow$ $\frac{2}{3}=\frac{m}{-7}$ (from first two terms)
$\Rightarrow$ $m=\frac{-14}{3}$