Question

For what value of c, the following system of linear equations has infinite number of solutions?

$cx+3y-(c-3)=0;12x+cy-c=0$

Answer 1

Step 1:If the pair of linear equations has infinite number of solutions

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

Given,

$$\begin{gathered} cx+3y-(c-3)=0.........\left(1\right) \\ 12x+cy-c=0.........\left(2\right) \end{gathered}$$

$$\begin{gathered} \text{here}{a}_1=c,b_1=3,c_1=-(c-3) \\ {a}_2=12,b_2=c,c_2=-c \end{gathered}$$

$$\begin{gathered} \frac{c}{12}=\frac{3}{c}=\frac{-(c-3)}{-c} \\ \text{Taking}\frac{c}{12}=\frac{3}{c} \\ \Rightarrow c^2=36 \\ \Rightarrow c=\pm 6 \end{gathered}$$

Step 2:When $c=6$

$$\begin{gathered} \frac{c}{12}=\frac{6}{12}=\frac{1}{2} \\ \frac{3}{c}=\frac{3}{6}=\frac{1}{2} \\ \frac{-(c-3)}{-c}=\frac{6-3}{6}=\frac{3}{6}=\frac{1}{2} \end{gathered}$$

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

Step 3: When $c=-6$

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

Therefore the value of $c=6$