Question

Find the value of $a$, the pair of linear equations $ax+3y=a-3,12x+ay=a$ has no solution?

Answer 1

$ax+3y-(a-3)=0$

$12x+ay-a=0$

System of equations of form

$p_{1}x+q_{1}y+r_1=0$

$p_{2}x+q_{2}y+r_2=0$

$\frac{p_1}{p_2}=\frac{q_1}{q_2}\ne \frac{r_1}{r_2}$

$\frac{a}{12}=\frac{3}{a}\ne \frac{-(a-3)}{-a}$

$\frac{3}{a}\ne \frac{-(a-3)}{-a};\frac{a}{12}=\frac{3}{a}.$

$3\ne a-3;a^2=36$

$3+3\ne a;a=\sqrt{36}$

$a\ne 6;a=\pm 6$

$[\because a\ne 6]$

$\therefore$ Value of a = 6.