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

Given: 

The given system of equations is:

\( 6 x+3 y=c-3 \)
\( 12 x+c y=c \)

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:

$6x+3y-(c-3)=0$
$12x+cy-c=0$

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,

$a_1=6, b_1=3, c_1=-(c-3)$ and $a_2=12, b_2=c, c_2=-c$

Therefore,

$\frac{6}{12}=\frac{3}{c}=\frac{-(c-3)}{-c}$

$\frac{1}{2}=\frac{3}{c}=\frac{c-3}{c}$

$\frac{1}{2}=\frac{3}{c}$ and $\frac{3}{c}=\frac{c-3}{c}$

$c=3\times2$ and $3\times c=c \times(c-3)$

$c=6$ and $3=c-3$

$c=6$ and $c=3+3$

$c=6$

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

Hence, for $c=6$ the given system of equations has infinitely many solutions.

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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