NIELIT Scientist B 2020 November: 27

Question

 

What is the value of $k$ for which the following system of equations has no solution:

$\text{2x-8y=3 and kx+4y=10}$

• A. $-2$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Suppose we have system of linear equations $,a_{1}x + b_{1}y + c_{1} = 0,$ and $a_{2}x + b_{2}y + c_{2} = 0.$

• If the system has unique solution $:\dfrac{a_{1}}{a_{2}} \neq \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$
• If the system has no solution $:\dfrac{a_{1}}{a_{2}}  =  \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}$
• If the system has infinite many solutions $:\dfrac{a_{1}}{a_{2}}  =  \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$

Now, given that the system of linear equations $,2x – 8y = 3 \quad \rightarrow(1),$ and $kx + 4y = 10 \quad \rightarrow(2)$ 

If the system of the equation has no solution, then $\dfrac{2}{k} = \dfrac{-8}{4} \implies k  = -1.$

So, the correct answer is $(C).$