Question

(i) Prove that there is a value of c(≠ 0) for which the system
6x + 3y = c − 3
12x + cy = c has infinitely many solutions. Find this value.

(ii) Find c if the system of equations cx + 3y + 3 – c = 0, 12x + cy – c = 0 has infinitely many solutions?

Answer 1

(i)

To find: To determine for what value of c the system of equation has infinitely many solution

We know that the system of equations

For infinitely many solution

Here

Consider the following

Now consider the following for c

But it is given that c ≠ 0. Hence c = 6

Hence for the system of equation have infinitely many solutions.

(ii) If the system of equations $a_{1}x+b_{1}y+c_1=0\mathrm{and}{a}_{2}x+b_{2}y+c_2=0$ has infinitely many solutions, then $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$.

Since, the system of equations cx + 3y + 3 – c = 0, 12x + cy – c = 0 has infinitely many solutions
Therefore,
$$\begin{gathered} \frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c} \\ \Rightarrow \frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c} \\ \Rightarrow \frac{c}{12}=\frac{3}{c} \\ \Rightarrow c\times c=3\times 12 \\ \Rightarrow c^2=36 \\ \Rightarrow c=\pm 6 \\ \mathrm{For}c=6, \\ \frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c} \\ \Rightarrow \frac{6}{12}=\frac{3}{6}=\frac{3-6}{-6}=\frac{1}{2} \\ \mathrm{Hence},\mathrm{it}\mathrm{holds}\mathrm{for}c=6. \\ \mathrm{For}c=-6, \\ \frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c} \\ \Rightarrow \frac{-6}{12}=\frac{3}{-6}=\frac{3-\left(-6\right)}{6}\ne \frac{-1}{2} \\ \mathrm{Hence},\mathrm{it}\mathrm{does}\mathrm{not}\mathrm{holds}\mathrm{for}c=-6. \end{gathered}$$

Hence, the value of c is 6.