Question

Which of the following systems of equations has no solution?
(A). $3x-y=2,9x-3y=6$
(B). $4x-7y+28=0,5x-7y+9=0$
(c). $3x-5y-11=0,6x-10y-7=0$
(D). None of these

Answer 1

Hint: Following system of equations are based on the Linear Equation in two variable, which can be represented as ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0~\text{and }\!\!~\!\!\text{ }{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0~$. As we know that Linear equation in two variable has 3 conditions that are
a) They are intersecting at point $=\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$
b) They are collinear $=\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$
c) They are parallel $=\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$

Complete step-by-step solution:
Since here the condition of intersecting means it had one solution, the condition of collinear means it had infinite solutions and parallel means these lines will not touch/intersect at any point so they will not have any solution. So we had to go with the third condition of the linear equation of two variables i.e. when they are parallel. In order to solve we just need to compare the value of ${{a}_{1}},{{b}_{1}},{{c}_{1}},{{a}_{2}},{{b}_{2}},{{c}_{2}}$ from the equation given and if the condition satisfied then the line will follow the condition and if not then they will not follow.
So Moving with our question, we had to go with all three options given
So, with option a; $3x-y=2,9x-3y=6$
Writing it in the form of ${{a}_{1}},~{{b}_{1}}\ldots .$ and so on
It will give $3x+\left( -y \right)-2=0,9x+\left( -3y \right)-6=0$
By comparing
Here ${{a}_{1}}=3,~{{b}_{1}}=-1,~{{c}_{1}}=-2~\text{and }\!\!~\!\!\text{ }{{a}_{2}}=9,\text{ }\!\!~\!\!\text{ }{{b}_{2}}=-3,{{c}_{2}}=-6$
Going with the condition i.e. $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$
So by putting value we will get
 $\begin{align}
  & \dfrac{3}{9}=\dfrac{-1}{-3} = \dfrac{-2}{-6} \\
 & \dfrac{1}{3}=\dfrac{1}{3} = \dfrac{1}{3} \\
\end{align}$
As these shows to be not valid, as it seems like i.e. $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$ and we want $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$. So, option a does not stand for no solution so the system of equations.
Going with option B. i.e. $4x-7y+28=0, 5x-7y+9=0$
Writing it in the form of ${{a}_{1}},~{{b}_{1}}\ldots .$ and so on
$4x+\left( -7y \right)+28=0,5x+\left( -7y \right)+9=0$
By comparing
Here ${{a}_{1}}=4,~{{b}_{1}}=-7,~{{c}_{1}}=28~\text{and }\!\!~\!\!\text{ }{{a}_{2}}=5,\text{ }\!\!~\!\!\text{ }{{b}_{2}}=-7,{{c}_{1}}=9$
Going with the condition i.e. $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$. So by putting value we will get
$\begin{align}
  & \dfrac{4}{5} \ne \dfrac{-7}{-7}\ne \dfrac{28}{9} \\
 & \dfrac{4}{5} \ne \dfrac{1}{1}\ne \dfrac{28}{9} \\
\end{align}$
But is also seems not to be like condition we want, rather it going like $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$. So, option B does not stand for no solution so the system of equations.
Going with the option C i.e. $3x-5y-11=0,6x-10y-7=0$.
Writing it in the form of ${{a}_{1}},~{{b}_{1}}\ldots .$ and so on
$3x+\left( -5y \right)+\left( -11 \right)=0,6x+\left( -10y \right)+\left( -7 \right)=0$.
By comparing
Here ${{a}_{1}}=3,~{{b}_{1}}=-5,~{{c}_{1}}=-11~\text{and }\!\!~\!\!\text{ }{{a}_{2}}=6,\text{ }\!\!~\!\!\text{ }{{b}_{2}}=-10,{{c}_{1}}=-7$
Going with the condition i.e. $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$. So by putting value we will get
$\begin{align}
  & \dfrac{3}{6}=\dfrac{-5}{-10}\ne \dfrac{-11}{-7} \\
 & \dfrac{1}{2}=\dfrac{1}{2}\ne \dfrac{11}{7} \\
\end{align}$
As it seems to follow the condition of $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$. So this system of equations will have no solution.
Hence the option C is correct.

Note: In order to deal with the type of question you need to know these three conditions very well. Moreover, while comparing the value of ${{a}_{1}},~{{b}_{1}}\ldots .$ and so on you need to go with the sign. As we had done with the negative value in the equation.