Find $ c $ if the system of equations $ c x+3 y+3-c=0, 12 x+cy-c=0 $ has infinitely many solutions?

Given: 

The given system of equations is:

\( c x+3 y+3-c=0, 12 x+cy-c=0 \)

To do: 

We have to find the value of $c$ for which the given system of equations has infinitely many solutions.

Solution:

The given system of equations can be written as:

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

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

The condition for which the above system of equations has infinitely many solutions is

$\frac{a_{1}}{a_{2}} \ =\frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}} \ $

Comparing the given system of equations with the standard form of equations, we have,

$a_1=c, b_1=3, c_1=3-c$ and $a_2=12, b_2=c, c_2=-c$

Therefore,

$\frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c}$

$\frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c}$

$\frac{3}{c}=\frac{3-c}{-c}$

$3=-(3-c)$

$3=-3+c$

$c=3+3$

$c=6$

The value of $c$ for which the given system of equations has infinitely many solutions is $6$. 

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Updated on: 10-Oct-2022

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