Question

Find the value of C which has no solution :Cx+3y=3,12x+Cy=6

Answer 1

Answer:

According to the condition for no solution

a1/a2 =b1/b2≠ c1/c2

c/12 =3/c

c2 =36

c =+6 or  -6    Eq 1

(3/c)≠1/2

c≠6           Eq.2

Combining 1 and 2 equations c =-6

Step-by-step explanation:

Answer 2

Answer:

The given system of equations has no solution when C is equal to -6.

Step-by-step explanation:

A system of equations, also understood as a group of concurrent or an equation system exists as a finite set of equations for which we selected the common answers.

The given system of equations:

$Ck x+3 y=3$

$C x+3 y-3=0 \ldots \text { (i) }$

$&12 x+C y=6$

$&12 x+C y-6=0 \ldots \text { (ii) }$

These equations are of the following form:

$a_{1} x+b_{1} y+c_{1}=0, a_{2} x+b_{2} y+c_{2}=0$

where, $a_{1}=k, b_{1}=3, c_{1}=-3$ and $a_{2}=12, b_{2}=k, c_{2}=-6$

So that the presented system has no solution, we must hold:

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

$&\text { i.e., } \frac{C}{12}=\frac{3}{C} \neq \frac{-3}{-6}$

$&\frac{C}{12}=\frac{3}{C} \text$

and  $\frac{3}{C} \neq \frac{1}{2}$

$&\Rightarrow \mathrm{C}^{2}=36 \text { and } \mathrm{C} \neq 6$

$&\Rightarrow \mathrm{C}=\pm 6 \text { and } \mathrm{C} \neq 6$

Hence, the given system of equations has no solution when C is equal to -6.