Question

Find c if the system of equations $cx+3y+\left( 3-c \right)=0$ and $12x+cy-c=0$ has infinitely many solutions.

Answer 1

This is a simple question of system of equations. It is a property based question. We know that if a system of two linear equations, have different type of solutions. If the two linear equations are, ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$, then it will have unique solutions when, $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$, it will have infinitely many solutions when, $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$ and it will have no solutions when, $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$. We will solve this question keeping this property in mind.

Complete step-by-step solution:
In this question, we have been given the two equations as, $cx+3y+\left( 3-c \right)=0$ and $12x+cy-c=0$. Now, if we compare them with the general form of a system of two linear equations, that is, ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$, then we have the values as follows,
$\begin{aligned} & {{a}_{1}}=c,{{b}_{1}}=3,{{c}_{1}}=\left( 3-c \right) \\\ & {{a}_{2}}=12,{{b}_{2}}=c,{{c}_{2}}=\left( -c \right) \\\ \end{aligned}$
Now, we know that for infinitely many solutions, it must satisfy the condition, $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$. So, we will substitute the values and check. So, we have,
$\begin{aligned} & \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}} \\\ & \Rightarrow \dfrac{c}{12}=\dfrac{3}{c}=\dfrac{\left( 3-c \right)}{\left( -c \right)} \\\ \end{aligned}$
Let us take only the first and second term. So, we get,
$\begin{aligned} & \dfrac{c}{12}=\dfrac{3}{c} \\\ & \Rightarrow {{c}^{2}}=12\times 3 \\\ & \Rightarrow {{c}^{2}}=36 \\\ & \Rightarrow c=\pm 6\ldots \ldots \ldots \left( i \right) \\\ \end{aligned}$
Now let us take the second and third term, so we get,
$\begin{aligned} & \dfrac{3}{c}=\dfrac{\left( 3-c \right)}{\left( -c \right)} \\\ & \Rightarrow -3=3-c \\\ & \Rightarrow c=6\ldots \ldots \ldots \left( ii \right) \\\ \end{aligned}$
So, if we look at the values obtained in equations (i) and (ii), then c = - 6 will be rejected. Therefore, we will get the value of c as 6.

Note: We must note that the system of linear equations which has infinitely many solutions represent two parallel lines that lie on each other. These lines are also called as coincident lines. If the linear system has no solutions, it represents two parallel lines which are not coincident.