For what value of $\alpha$, the system of equations 
$\alpha x+3y=\alpha -3$
$12x+\alpha y=\alpha$
will have no solution?

Given: 

The given system of equations is:

$\alpha x+3y=\alpha -3$

$12x+\alpha y=\alpha$

To do: 

We have to find the value of $\alpha$ for which the given system of equations has no solution.

Solution:

The given system of equations is,

$\alpha x+3y-(\alpha -3)=0$

$12x+\alpha y-\alpha=0$

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 no solution 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,

$a_1=\alpha, b_1=3, c_1=-(\alpha-3)$ and $a_2=12, b_2=\alpha, c_2=-\alpha$

Therefore,

$\frac{\alpha}{12}=\frac{3}{\alpha}≠\frac{-(\alpha-3)}{-\alpha}$

$\frac{\alpha}{12}=\frac{3}{\alpha}≠\frac{\alpha-3}{\alpha}$

$\frac{\alpha}{12}=\frac{3}{\alpha}$ and $\frac{3}{\alpha}≠\frac{\alpha-3}{\alpha}$

$\alpha \times \alpha=12\times3$ and $3≠\alpha-3$

$(\alpha)^2=36$ and $\alpha≠3+3$

$\alpha=\sqrt{36}$ and $\alpha≠6$

$\alpha=6$ or $\alpha=-6$ and $\alpha≠6$

This implies,

$\alpha=-6$

The value of $\alpha$ for which the given system of equations has no solution is $-6$.  

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Updated on: 10-Oct-2022

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