Question

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

Answer 1

Hint: In the given question, we are required to find the system of equations in two variables that possess no solution. We must know the condition for the system of linear equations in two variables to have no solution at all to solve the problem. We first convert the given two equations into a standard form of linear equation in two variables and then fulfill the requirements.

Complete step by step answer:
There are three types of systems of linear equations in two variables. One is the system of linear equations in two variables where there is no solution of the equations. Both the equations represent parallel lines in such a case. Second is the system of linear equations in two variables where only one unique solution of the equations is present.

Such equations represent two intersecting straight lines on the Cartesian plane. Third is the system of linear equations in two variables where there are infinitely many solutions of the equations. Such equations represent the same line on the Cartesian plane.We will check all the options one by one for the equations that possess no solution.

Analyzing option (A):
We have, $3x - y = 2$ and $9x - 3y = 6$
So, for the equations to have no solution, the equations must represent parallel lines on the Cartesian plane. Now, we know that the general form of a linear equation in two variables is $ax + by + c = 0$. The equations given to us when represented in standard form become: $3x - y - 2 = 0$ and $9x - 3y - 6 = 0$.
So, if a system has no solution, then,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$

Here, for the equation $3x - y - 2 = 0 - - - - \left( 1 \right)$ , we have,
${a_1} = 3$, ${b_1} = - 1$ and ${c_1} = - 2$.
Also, for the equation $9x - 3y - 6 = 0 - - - - \left( 2 \right)$, we have,
${a_2} = 9$, ${b_2} = - 3$ and ${c_2} = - 6$.
Now, calculating the ratios of coefficients and constants of both the equations, we get,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{3}{9} = \dfrac{1}{3}$, $\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{ - 1}}{{ - 3}} = \dfrac{1}{3}$ and $\dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{ - 2}}{{ - 6}} = \dfrac{1}{3}$
So, $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.
So, the given system possesses infinite solutions. So, option (A) is not the correct answer.

Analyzing option (B):
We have, $4x - 7y + 28 = 0$ and $5y - 7x + 9 = 0$
Now, we know that the general form of a linear equation in two variables is $ax + by + c = 0$.
The equations given to us are already represented in standard form. So, if a system has no solution, then,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$
Here, for the equation $4x - 7y + 28 = 0 - - - - \left( 1 \right)$ , we have,
${a_1} = 4$, ${b_1} = - 7$ and ${c_1} = 28$.
Also, for the equation $ - 7x + 5y + 9 = 0 - - - - \left( 2 \right)$, we have,
${a_2} = - 7$, ${b_2} = 5$ and ${c_2} = 9$.
Now, calculating the ratios of coefficients and constants of both the equations, we get,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{4}{{ - 7}}$, $\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{ - 7}}{5}$ and $\dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{28}}{9}$
So, $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$.
So, the given system possesses one unique solution. So, option (B) is not the correct answer.

Analyzing option (C):
We have, $3x - 5y - 11 = 0$ and $6x - 10y - 7 = 0$
The equations given to us are already represented in standard form. So, if a system has no solution, then, $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$
Here, for the equation $3x - 5y - 11 = 0 - - - - \left( 1 \right)$ , we have,
${a_1} = 3$, ${b_1} = - 5$ and ${c_1} = - 11$.
Also, for the equation $6x - 10y - 7 = 0 - - - - \left( 2 \right)$, we have,
${a_2} = 6$, ${b_2} = - 10$ and ${c_2} = - 7$.
Now, calculating the ratios of coefficients and constants of both the equations, we get,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{3}{6} = \dfrac{1}{2}$, $\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{ - 5}}{{ - 10}} = \dfrac{1}{2}$ and $\dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{ - 11}}{{ - 7}} = \dfrac{{11}}{7}$
So, $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$.
So, the given system possesses no solution.

Hence, option C is the correct answer.

Note: We must have a grip over the topic of linear equations in two variables in order to solve such a question within a limited time frame. We must ensure accuracy in arithmetic and calculations to be sure of our answer. One should know about the transposition rule so as to shift terms from one side of the equation to another and solve the equation. General form of a linear equation in two variables must be remembered.