Question

For what value of $a,$the system of linear equations; $ax+3y=a-3;12x+ay=a$ has no solution?

Answer 1

Step 1:If the pair of linear equations has no solutions

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Given equations,

$$\begin{gathered} ax+3y=a-3 \\ 12x+ay=a \end{gathered}$$

$$\begin{gathered} \text{here},a_1=a,b_1=3,c_1=-(a-3) \\ {a}_2=12,b_2=a,c_2=-a \\ \Rightarrow \frac{a}{12}=\frac{3}{a} \\ \Rightarrow a^2=36 \\ \Rightarrow a=\pm 6 \\ here\frac{c_1}{c_2}=\frac{-(a-3)}{-a}=\frac{a-3}{a} \end{gathered}$$

Step 3:When $a=6$

$$\begin{gathered} \frac{a_1}{a_2}=\frac{6}{12}=\frac{1}{2}; \\ \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2} \\ \frac{c_1}{c_2}=\frac{6-3}{6}=\frac{3}{6}=\frac{1}{2} \end{gathered}$$

Here,$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Step 4: When $a=-6$

$$\begin{gathered} \frac{a_1}{a_2}=\frac{-6}{12}=\frac{-1}{2}; \\ \frac{b_1}{b_2}=\frac{3}{-6}=\frac{-1}{2} \\ \frac{c_1}{c_2}=\frac{-6-3}{-3}=\frac{-9}{-6}=\frac{3}{2} \\ \text{here}\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2} \end{gathered}$$

Therefore the value of $a=-6$