Question

$ax+2y=5$
$3x-6y=20$
In the system of equations above, $a$ is a constant. if the system has one solution, which of the following CANNOT be the value of $a$?

• A. $-1$
• B. $4$
• C. $1$
• D. $3$

Answer 1

The correct option is A $-1$

The given equations are :

$ax+2y+(-5)=0\dots \left(i\right)$

$3x+(-6)y+(-20)=0\dots \left(ii\right)$

As we can see they are of the form $ax+by+c=0$

Also the condition for some given system of equations to have solution is,

$\Rightarrow \frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

Here $a_1,a_2,b_1,b_2$ all are constants, and their values are respectively:

$a_1=a$

$a_2=3$

$b_1=2$

$b_2=-6$

So if the given system of equations has exactly one solution then,

$\Rightarrow \frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

$\Rightarrow \frac{a}{3}\ne \frac{2}{-6}$

$\Rightarrow a\ne -1$

Hence out of all the values given in the options $a$ can not be equal to $-1.$

Hence option $A$ is correct.